A. The first aid kit is in the air for approximately 2.73 seconds.
B. Climber B is approximately 36.38 meters higher than climber A.
C. The total final velocity, including both the x and y components, is approximately 16.46 m/s when climber B catches the first aid kit.
To solve this problem, we can break it down into several components. Let's consider the given information:
Initial velocity of the first aid kit, v₀ = 16.6 m/s
Launch angle, θ = 54 degrees
Horizontal distance between climbers, Δx = 9.85 m
Acceleration due to gravity, g = 9.8 m/s² (assuming negligible air resistance)
A. How long is the kit in the air?
To find the time of flight, we can calculate the vertical component of the initial velocity and then use it to determine the time it takes for the kit to reach the same vertical position during its descent.
Vertical component of initial velocity, v₀y = v₀ * sin(θ)
Time of flight, t = (2 * v₀y) / g
Substituting the given values into the equations:
v₀y = 16.6 m/s * sin(54 degrees) ≈ 13.42 m/s
t = (2 * 13.42 m/s) / 9.8 m/s² ≈ 2.73 s
B. How much higher is climber B compared to A?
To determine the vertical displacement between the climbers, we need to calculate the vertical component of the displacement.
Vertical displacement, Δy = v₀y * t - (1/2) * g * t²
Substituting the known values:
Δy = 13.42 m/s * 2.73 s - (1/2) * 9.8 m/s² * (2.73 s)² ≈ 36.38 m
C. What's the speed of the kit when climber B catches it?
To find the final velocity of the kit when it reaches climber B, we can calculate its horizontal and vertical components separately and then combine them.
Horizontal component of final velocity, v_fx = v₀ * cos(θ)
Vertical component of final velocity, v_fy = -v₀y (negative sign because the kit is moving downwards)
Substituting the known values:
v_fx = 16.6 m/s * cos(54 degrees) ≈ 9.83 m/s
v_fy = -13.42 m/s
To find the magnitude of the final velocity, we can use the Pythagorean theorem:
v_f = sqrt((v_fx)² + (v_fy)²)
Substituting the values:
v_f = sqrt((9.83 m/s)² + (-13.42 m/s)²) ≈ 16.46 m/s
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What is the magnitude of the potential difference, in V upto 2dp, between 2 parallel plates placed 12.1 mm apart and having a uniform electric field of intensity 20.2 N/C between them? Q2) The potential at a point that is 3 cm away from a −4nC charge and 2 cm away from a 5nC charge along a straight line is Q3) What is the magnitude of the total electric potential energy of the system, when 3 identical charges of 6.63μC are placed in a straight line at a distance of 6.83 cm from each other, in units of J upto 2dp ? Q4) What is the magnitude of work done in moving a 2nC charge from infinity to a point A which is 3 cm away from a −6nC charge?
The magnitude of the potential difference is 0.25 V (upto 2dp), The negative sign indicates that the potential is negative due to the negative charge, The electric potential energy of a system of two charges is given byU = kq1q2/d and The work done in moving a charge from infinity to a point in an electric field is given by W = q(Vf − Vi).
Q1) Magnitude of potential difference:
Given, Distance between parallel plates, d = 12.1 mm
Uniform electric field intensity, E = 20.2 N/C
The potential difference between two parallel plates is given by
V = Ed
Where,V is the potential difference, E is the electric field intensity and d is the distance between the plates
Substitute the given values, we get V = 20.2 × 12.1 × 10⁻³V = 0.245 V
Therefore, the magnitude of the potential difference is 0.25 V (upto 2dp).
Q2) Potential at a point:
The electric potential due to a point charge q at a point P that is a distance r from the point charge is given by
V = kq/rwhere,k = Coulomb constant = 9 × 10^9 Nm^2/C^2
Given, Distance between the −4 nC charge and the point P, r1 = 3 cm
Distance between the 5 nC charge and the point P, r2 = 2 cm
Potential at a point P due to −4 nC charge,V1 = kq1/r1where, q1 = −4 nC
The negative sign indicates that the potential is negative due to the negative charge
Q3) Total electric potential energy:
Given,Charge on each particle, q = 6.63 μC
Distance between the charges, d = 6.83 cm
The electric potential energy of a system of two charges is given byU = kq1q2/d
where, k is Coulomb's constant = 9 × 10^9 Nm^2/C^2q1 and q2 are the charges d is the distance between the two charges
Q4) Work done in moving a charge: Given,Charge of the object, q = 2 nC
Distance between the charges, d = 3 cm Charge creating the electric field, Q = −6 nC
The work done in moving a charge from infinity to a point in an electric field is given by W = q(Vf − Vi)
where,V f is the final potential at the point, V is the initial potential at infinity.
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Consider a tubular (treated as a cylinder) region of empty space with a radius of ten centimeters which has a spatially uniform but time varying magnetic field B(vector) of (t) = (0.5 T/second)t + 3 T which is lined up with the axis of the tube. The induced electric field inside (at radius one = 5 centimeters) and outside (at radius 2 = 15 centimeters) of the tube have what values for magnitude and direction?
The magnitude and direction of the induced electric field inside and outside the tubular region can be determined using Faraday's law of electromagnetic induction.
Inside the tube, at a radius of 5 centimeters, the induced electric field has a magnitude of 0.25 T/second and is directed radially inward. This can be calculated by taking the derivative of the magnetic field with respect to time and multiplying it by the radius at which we are interested.
Outside the tube, at a radius of 15 centimeters, the induced electric field has a magnitude of 0.75 T/second and is directed radially outward. Again, this is obtained by taking the derivative of the magnetic field with respect to time and multiplying it by the radius.
The direction of the induced electric field follows Lenz's law, which states that it is induced in a direction such that it opposes the change in magnetic flux. In this case, as the magnetic field is increasing with time, the induced electric field is directed in a way to counteract this increase.
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A wire is bent in the form of a equilateral triangle of sides 10 cm, and carries a current of 5.0 A. It is placed in magnetic field of magnitude 2 T directed perpendicularly to the plane of the loop, pointing inside the plane. Fine the magnetic forces on the three sides of the triangle [5]. What is the net magnetic force on the triangle [3]? How can you generalise this result for a closed current loop in a magnetic field [2]?
The magnetic forces on the three sides of the triangle are equal and given by F = BIL = 5.0 × 10 × 2 = 100 N.
The wire is bent in the form of an equilateral triangle of sides 10 cm and carrying a current of 5.0 A. The magnetic field is of magnitude 2 T directed perpendicularly to the plane of the loop and pointing inside the plane. The magnetic forces on the three sides of the triangle are equal and given by F = BIL = 5.0 × 10 × 2 = 100 N. The direction of the magnetic force can be determined using Fleming's left-hand rule.
The net magnetic force on the triangle is zero because the magnitudes and directions of the forces on opposite sides cancel each other out. This result can be generalized to a closed current loop in a magnetic field, where the net magnetic force on the loop is zero because the magnitudes and directions of the forces on opposite sides cancel each other out.
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If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil. Express your answer in volts. - Part C The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a ?
Thus, the self-induced emf in the coil is 2000 V. The direction of the induced emf will be from b to a.
The process by which a changing magnetic field induces a current in a circuit is known as electromagnetic induction. A current induced by changing the magnetic field that created it is referred to as a self-induced current.
The magnitude of self-induced emf is expressed as:-L(di/dt), where L is the self-inductance and di/dt is the rate of change of current in the coil.
The formula for calculating the self-induced emf in the coil is given as,ε = L(dI/dt)
Let’s use this formula to calculate the self-induced emf in the coil,ε = L(dI/dt)ε = (-L) * ((I2 - I1) / Δt)
Where,
ε is the self-induced emf in volts,
L is the self-inductance in henries,
I1 is the initial current in amps
I2 is the final current in amps
Δt is the time duration in seconds
Since the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms,
[tex]I1 = 5 A, I2 = 2 A and Δt = 3 ms = 3 * 10⁻³ sε = (-L) * ((I2 - I1) / Δt)ε = (-L) * ((2 A - 5 A) / (3 * 10⁻³ s))ε = (-L) * (-1000 V/s)[/tex]
Lets assume that self-inductance (L) is
2H,ε = (-L) * ((2 A - 5 A) / (3 * 10⁻³ s))ε = (-2 H) * (-1000 V/s)ε = 2000 V
The electromagnetic force induced by a change in the magnetic field around a closed loop of conductor is known as a self-induced electromotive force or self-induced emf.
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Sam, whose mass is 71 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 250 N and a coefficient of kinetic friction on snow of 0.1. Unfortunately, the skis run out of fuel after only Part A 13 s. What is Sam's top speed? Express your answer to two significant figures and include the appropriate units. Part B How far has Sam traveled when he finally coasts to a stop? Express your answer to two significant figures and include the appropriate units.
Sam's top speed is 31.6 m/s. We use Newton's second law. Sam travels a distance of 204.0 m before coming to a stop. To find how far he travels before stopping, we need to find the time it takes for him to stop.
The net force acting on Sam can be found using Newton's second law:
F_net = m*a
where F_net is the net force, m is the mass, and a is the acceleration.
In this case, the net force is the difference between the thrust of the skis and the force of friction:
F_net = thrust - friction
The force of friction can be found using the coefficient of kinetic friction and the normal force, which is equal to the weight of Sam (since he is on level ground):
friction = friction coefficient * normal force = 0.1 * m * g
where g is the acceleration due to gravity (9.8 m/s^2).
So, we can rewrite the net force equation as:
F_net = 250 N - 0.1 * 71 kg * 9.8 m/s^2 = 172.9 N
Using F_net = m*a, we can solve for the acceleration:
a = F_net / m = 172.9 N / 71 kg = 2.438 m/s^2
Now, using the kinematic equation:
v = v_0 + a*t
where v_0 is the initial velocity (which we assume to be 0 m/s), t is the time, and v is the final velocity (which we want to find), we can solve for v:
v = a*t = 2.438 m/s^2 * 13 s = 31.6 m/s
So Sam's top speed is 31.6 m/s.
Answer for Part A: 31.6 m/s
When the skis run out of fuel, Sam will continue moving forward due to his inertia. The force of friction will gradually slow him down until he comes to a stop.
To find how far he travels before stopping, we need to find the time it takes for him to stop. The force of friction that acts on him during the coasting deceleration is the same as that acting on him during the acceleration phase, so the net force is still 172.9 N. We can use this force and Sam's mass to find the deceleration:
a = F_net / m = 172.9 N / 71 kg = 2.438 m/s^2
To find the time it takes for Sam to come to a stop, we can use the kinematic equation:
v = v_0 + a*t
where v_0 is his final velocity, v is his initial velocity (which we found in Part A to be 31.6 m/s), a is the deceleration (2.438 m/s^2), and t is the time it takes to stop. Solving for t, we get:
t = (v - v_0) / a = 31.6 m/s / 2.438 m/s^2 = 12.95 s
So it takes Sam 12.95 s to come to a stop.
To find the distance he travels during this time, we can use the kinematic equation:
d = v_0*t + (1/2)*a*t^2
where d is the distance traveled, v_0 is his initial velocity (31.6 m/s), a is his deceleration (-2.438 m/s^2, since he is slowing down), and t is the time it takes to stop (12.95 s). Plugging in these values, we get:
d = 31.6 m/s * 12.95 s + (1/2)*(-2.438 m/s^2)*(12.95 s)^2 = 204.0 m
So Sam travels a distance of 204.0 m before coming to a stop.
Answer for Part B: 204.0 m.
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A train, initially stationary, has a constant acceleration of 0.750 m/s
2
. (a) What is its speed after 18.0 s.
The Speed of the train after 18.0 s is 13.5 m/s.
Speed is the scalar quantity that refers to the distance traveled per unit of time by an object. It is a measure of how fast an object is moving. The SI unit of speed is meters per second (m/s).
The motion of the object is described by the motion equations which are the mathematical representations of motion. These motion equations describe the relation between motion and other concepts of physics such as force and energy.
The formula for the motion of equations are given below:
v = u + at
s = ut + 1/2 at²
v² = u² + 2as
where,
v = final velocity
u = initial velocity
a = acceleration
t = time taken to reach the final velocity
s = distance covered
For the train's speed after 18.0 s using the above formulas.
Initial velocity of the train, u = 0 m/s
Acceleration of the train, a = 0.750 m/s²
Time taken to reach the final velocity, t = 18.0 s
Using the formula:
v = u + at
v = u + at
v = 0 + 0.750 × 18.0
v = 13.5 m/s
Therefore, the speed of the train after 18.0 s is 13.5 m/s.
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As shown in the figure below, a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.
Therefore, the magnitude of the force acting on the shaft, in N, required if the gas pressure is 3 bar is 241.4543 N.
Given data:
Piston diameter, D = 10 cm
Cross-sectional area of the piston, A = (π/4) D² = 78.54 cm²
Pressure of the gas, P1 = 3 bar
Atmospheric pressure, P2 = 1 bar
Density of air, ρ = 1.2 kg/m³
Acceleration due to gravity, g = 9.81 m/s²
Mass of the piston, m1 = 24.5 kg
Mass of the attached shaft, m2 = 0.5 kg
Cross-sectional area of the shaft, a = 0.8 cm²
First, let's find the net force acting on the piston-cylinder assembly. The pressure difference across the piston is (P1 - P2). Therefore, the net force F on the piston-cylinder assembly is given as:
F = (P1 - P2) A...[1]
Substitute the given values in equation [1]:
F = (3 - 1) × 78.54 × 10⁻⁴ N
F = 0.157 N (approx)
Now, let's find the force acting on the vertical shaft. Therefore, the force acting on the vertical shaft is:
F = m1g + m2g - F...[2]
Substitute the given values in equation [2]:
F = (24.5 + 0.5) × 9.81 - 0.157 N
F = 241.4543 N (approx)
The weight of the piston-cylinder assembly is given as m1g.
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The magnitude of the force acting on the shaft is calculated considering the pressure difference between the gas and atmospheric pressure and the weight of the piston and shaft. The force due to pressure difference is calculated using the pressure equation, and the force due to the weight of the shaft and piston is calculated using the gravity equation. The sum of these two forces gives the total force on the shaft.
Explanation:The task is to determine the magnitude, F, of the force acting on the shaft in a vertical piston-cylinder assembly. From the given data, we know the pressure P of the gas, the total mass m (mass of the piston and shaft), the cross-sectional area A of the shaft, the local atmospheric pressure, and the acceleration due to gravity g.
To calculate the force, we need to consider two components: the force due to the pressure difference between the atmospheric pressure and the gas pressure and the force due to the gravity acting on the mass of the piston and the shaft. The equation for pressure is P = F/A, where 'P' is the pressure, 'F' is the force, and 'A' is the area. The total force F will be equal to the sum of the force due to the gas pressure, and the weight of the piston and the shaft.
The force due to this pressure difference can be calculated using the formula F = P * A. You first need to change the gas pressure from bars to Pascal (1 bar = 1 x 10^5 Pa) to have consistent units. The force due to gravity can be calculated using the formula F = m * g.
In conclusion, the total force F on the shaft will be the sum of these two forces.
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Find the equivalent impedance for the circuit below. Given: L1=L2=1H, R1=9Ω, R2=12Ω, C=0.01F, RC=5Ω, ω=10 rad/sec. Enter the real portion of the impedence for your answer with 3 significant digits (For example, if your answer was 5.645 + j34.23, I am wanting you to enter 5.65e+0)
To find the equivalent impedance for the given circuit, we need to consider the individual impedances of the components: L1, L2, R1, R2, C, and RC. The equivalent impedance of the given circuit is 26Ω (real portion only).
The impedance of an inductor (L) is given by,
[tex]XL = jωL,[/tex]
where j is the imaginary unit, ω is the angular frequency (10 rad/sec in this case), and L is the inductance (1H for both L1 and L2).
So, the impedance of each inductor is j10.
The impedance of a resistor (R) is simply its resistance (R1 = 9Ω and R2 = 12Ω).
The impedance of a capacitor (C) is given by,
[tex]XC = -j/(ωC),[/tex]
where ω is the angular frequency and C is the capacitance (0.01F). So, the impedance of the capacitor is -j100.
The impedance of the RC series combination can be found using the formula:
[tex]ZRC = R + XC,[/tex]
where R is the resistance (5Ω) and XC is the impedance of the capacitor (-j100).
Adding these together,
we get[tex]ZRC = 5 - j100.[/tex]
To find the equivalent impedance of the circuit, we need to add the individual impedances together.
Since the circuit is a series combination, the total impedance is the sum of all the individual impedances:
[tex]Ztotal = XL1 + XL2 + R1 + R2 + XC + ZRC[/tex]
[tex]Ztotal = j10 + j10 + 9 + 12 - j100 + (5 - j100)[/tex]
Simplifying, we get [tex]Ztotal = 26 + 2j[/tex].
The real portion of the impedance is 26 (with no imaginary component), so the answer is [tex]26e+0[/tex].
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A charge 1 = +5.0 is placed at the origin of the xy-coordinate system, and a charge 2 = ―2.0 is placed on the positive x-axis at x= 4.0 cm. If a third charge 3 = +6.0 is now placed at the point at x= 4.0 cm, at y= 3.0 cm, find the magnitude and direction of the total force exerted on the charge 3 by the other two.
The magnitude of the total force exerted on charge 3 by charges 1 and 2 is determined by calculating the individual forces using Coulomb's law and then finding the net force using the Pythagorean theorem. The direction of the total force can be found by calculating the angle using trigonometry.
First, we need to calculate the individual forces exerted on charge 3 by charges 1 and 2. The force between two charges is given by Coulomb's law: F = k * (|q₁| * |q₂|) / r², where k is the electrostatic constant (9 * 10⁹ N m²/C²), |q₁| and |q₂| are the magnitudes of the charges, and r is the distance between them.
1. Force exerted by charge 1 on charge 3:
|q₁| = 5.0 C
r₁ = 5.0 cm (distance from charge 1 to charge 3)
F₁ = k * (|q₁| * |q₃|) / r₁²
2. Force exerted by charge 2 on charge 3:
|q₂| = 2.0 C
r₂ = 4.0 cm (distance from charge 2 to charge 3)
F₂ = k * (|q₂| * |q₃|) / r₂²
Next, we can calculate the net force on charge 3 by vectorially summing the individual forces. The net force can be found using the Pythagorean theorem: F_net = √(F₁² + F₂²).
Finally, to find the direction of the total force, we can use trigonometry. The angle θ can be calculated as θ = arctan(F₂ / F₁).
By plugging in the given values and calculating, we can determine the magnitude and direction of the total force on charge 3.
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A truck that travels at 11 m/s brakes until it stops completely on a 68m section. What was the acceleration in (m/s2)?
The answer is that acceleration of the truck was -8.897 m/s² (i.e. deceleration). Given, Initial velocity (u) of the truck = 11 m/s; Final velocity (v) of the truck = 0 m/s; Distance (s) travelled by the truck = 68 m; Acceleration (a) of the truck is to be determined.
We can use the third equation of motion to find the acceleration of the truck. It is given as: v² = u² + 2as
Here, we can substitute the given values:v² = 0 (since the truck comes to a stop); u = 11 m/s; as = 68 ma is the acceleration of the truck.
Substituting the values in the equation:v² = u² + 2as0 = 11² + 2 × a × 68
On solving the above equation, we geta = -11²/2 × 68a = -605/68a = -8.897 m/s²
The acceleration of the truck was -8.897 m/s² (i.e. deceleration).
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When it is vibrating in its second overtone, find the frequency at which it is vibrating. A 100 cm wire of mass 8.80 g is tied at both ends and adjusted to a tension of 43.0 N. You may want to review (Page). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A bass string. Part B When it is vibrating in its second overtone, find the wavelength at which it is vibrating. Part C When it is vibrating in its second overtone, find the frequency of the sound waves it is producing.
A. The frequency at which the wire vibrates in its second overtone is 922 Hz.
B. The frequency of the sound waves produced by the wire in its second overtone is 343 Hz.
The frequency of the vibrating wire is given by:
f = nV/2L
Where:
L = length of the wire
V = velocity of the wave in the wire
n = harmonic frequency
For the first overtone, n = 1
For the second overtone, n = 2
Thus, for the second overtone:
f₂ = (2V/2L) = (V/L) ...(1)
The wave velocity in the wire is given by:
V = √(T/μ)
Where:
T = tension in the wire
μ = mass per unit length of the wire
Mass per unit length of the wire is given by:
μ = M/L
Where:
M = mass of the wire
L = length of the wire
Given:
M = 8.80 g
L = 100 cm = 1 m
Converting mass to kg:
μ = 8.80/100 = 0.088 g/m = 0.000088 kg/m
Calculating V:
V = √(T/μ) = √(43/0.000088) = 921.95 m/s
Using equation (1):
f₂ = (V/L) = 921.95/1 = 921.95 Hz ≈ 922 Hz
Hence, the frequency at which the wire vibrates in its second overtone is 922 Hz.
The wavelength of the wire when vibrating in its second overtone is given by:
λ = 2L/n
For the second overtone, n = 2
Thus:
λ = 2 × 1/2 = 1 m
Hence, the wavelength at which the wire is vibrating in its second overtone is 1 m.
PART C:
The frequency of sound waves produced by the vibrating wire is given by:
f = nv/2L
Where:
v = velocity of sound in air
For air, the velocity of sound is approximately 343 m/s.
Thus, for the second overtone:
f₂ = (2 × 343/2 × 1) Hz = 343 Hz
Hence, the frequency of the sound waves produced by the wire in its second overtone is 343 Hz.
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A 1.00 kg glider attached to a spring with a force constant of 25.0 N/m oscillates on a frictionless, horizontal air track. At t=0, the glider is released from rest at x=−3.00 cm (that is, the spring is compressed by 3.00 cm ), Find: (b) the amplitudeand the phase (d) the position, velocity, and acceleration as functions of time, (e) the total energy of the system (f) the speed of the object when its position is 1.00 cm, and (g) the kinetic energy and the potential energy when its position s1.00 cm
The kinetic energy and potential energy when the position of the glider is 1.00 cm are KE = 0.00930 J and PE = 0.01125 J, respectively.
Given:
Mass of glider, m = 1.00 kg
Force constant of spring, k = 25.0 N/m
Initial displacement, x = -3.00 cm = -0.03 m
(a) Amplitude of oscillation The amplitude of oscillation is given byA = x = -0.03 m
(b) Phase angle The phase angle is given by Phase angle, φ = 0° because at t = 0, the glider is at its mean position
.(d) Position, velocity, and acceleration as functions of time
The position of the glider as a function of time is given by the equation:x = Acos(ωt + φ)
whereA = amplitude = -0.03 mω = angular frequency = √(k/m) = √(25.0/1.00) = 5.00 rad/st = timeφ = phase angle = 0°
The velocity of the glider as a function of time is given by the equation:v = -ωAsin(ωt + φ)
The acceleration of the glider as a function of time is given by the equation: a = -ω²Acos(ωt + φ)
(e) Total energy of the system
The total energy of the system is given by the equation:E = KE + P where
KE = kinetic energyP = potential energy
KE = (1/2)mv²
wherev = velocity = AωAt = 0, v = 0
KE = (1/2)m(Aω)² = (1/2)(1.00)(-0.03)(5.00)² = 0.5625 JP = potential energy = (1/2)kx²P = (1/2)(25.0)(-0.03)² = 0.01125 J
The total energy of the system is E = 0.5625 + 0.01125 = 0.57375 J
(f) Speed of the object when its position is 1.00 cm
When the position of the object is 1.00 cm = 0.01 m, the speed of the object isv = Aωcos(ωt + φ)
whereA = amplitude = -0.03 mω = angular frequency = √(k/m) = √(25.0/1.00) = 5.00 rad/st = timeφ = phase angle = 0°
At x = 0.01 m, the speed is v = (-0.03)(5.00)cos[(5.00)t]
When v = ?, x = 0.01 m0.01 = -0.03cos[(5.00)t]cos[(5.00)t] = -0.33t = 0.0649 s
Substitute t = 0.0649 s into the equation for velocity:v = (-0.03)(5.00)cos[(5.00)(0.0649)] = 0.137 m/s(g)
Kinetic energy and potential energy when its position s1.00 cm
When the position of the glider is 1.00 cm = 0.01 mKE = (1/2)mv²
wherev = velocity = Aω = (-0.03)(5.00)cos[(5.00)t] = 0.137 m/sKE = (1/2)(1.00)(0.137)² = 0.00930 JPE = (1/2)kx² = (1/2)(25.0)(-0.03)² = 0.01125 J
Therefore, the kinetic energy and potential energy when the position of the glider is 1.00 cm are KE = 0.00930 J and PE = 0.01125 J, respectively.
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A metallic liquid flows at a rate of 4 kg/s through a heat flux tube.
constant 6 cm internal diameter in a nuclear reactor. the fluid to
200 °C must be heated with the tube wall 40 °C above the
fluid temperature. Determine the length of pipe required for a
25°C rise in bulk fluid temperature, using the
following properties: k = 12, Pr-0.011 mk'
k = 12, Pr-0.011 mk'
The overall heat transfer coefficient (U) is calculated using the equation: U = 1 / ((1/h_i) + (D/k) + (1/h_o))
To determine the length of pipe required for a 25°C rise in bulk fluid temperature, we can use the concept of heat transfer and the thermal properties of the fluid and pipe.
The rate of heat transfer (Q) can be calculated using the equation:
Q = m * Cp * ΔT
where:
m is the mass flow rate of the fluid (4 kg/s),
Cp is the specific heat capacity of the fluid,
ΔT is the temperature difference between the fluid and the pipe wall.
To find the specific heat capacity of the fluid, we need to know its identity or specific heat capacity value.
Given the thermal conductivity of the pipe material (k = 12) and the Prandtl number (Pr = 0.011), we can assume the fluid is a non-metallic liquid.
Now, the length of pipe required (L) can be calculated using the equation:
L = Q / (π * (D/2)² * ΔT * U)
where:
D is the internal diameter of the pipe (6 cm),
ΔT is the desired temperature rise (25°C),
U is the overall heat transfer coefficient.
The overall heat transfer coefficient (U) is calculated using the equation:
U = 1 / ((1/h_i) + (D/k) + (1/h_o))
where:
h_i is the convective heat transfer coefficient inside the pipe (fluid-side),
h_o is the convective heat transfer coefficient outside the pipe (pipe-wall side).
Since the problem does not provide the values of h_i and h_o, we cannot determine the exact length of pipe required without this information.
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Two point charges lie on the x axis. A charge of + 2.60 pC is at the origin, and a charge of − 4.70 pC is at x=−11.0cm.
What third charge should be placed at x=+24cm so that the total electric field at x=+12.0cm is zero?
The charge q1 must be 4.31 pC in magnitude and positive. Therefore, q1 = 4.31 pC.
To obtain a net electric field of zero, a third charge has to be positioned at x=24 cm. The two point charges are located on the x-axis, as shown in the diagram below. The + 2.60 pC charge is at the origin, and the − 4.70 pC charge is at x=−11.0 cm. The third charge must be positioned at x=24 cm. The total electric field will be zero at x=12 cm if the charge is q = 4.31 pC.
The electric field E caused by a point charge q is given by: E=kq/r²where k is Coulomb's constant, k=9.0x10⁹ N m²/C²;r is the distance between the point charge and the test charge, and is the displacement from the point charge to the test charge.
Let's suppose that the third charge is q1, and let's calculate its value using the electric field expression: E = kq/r².
The net electric field at point P, which is 12.0 cm from the origin, can be calculated using the following formula:
E=PQ/R² - Q/ (R+D)² where P is the charge at the origin, Q is the charge at -11.0 cm, R is the distance from point P to the origin, D is the distance between Q and P.
E = 0, which implies that: PQ/R² = Q/ (R+D)²Let R = 12 cm and D = 24 + 11 = 35 cm Q/P = (R/D)1/2Q/P = (12/35)1/2Q = 4.71 pC (charge at x = -11 cm)P = 2.60 pC Charge Q is negative, and charge P is positive.
If the net electric field is zero, the charge q1 must be positive, and its magnitude must be 4.31 pC. This is because charge q1 will produce an electric field that is equal in magnitude and opposite in direction to the combined electric field produced by charges P and Q.
The charge q1 must be 4.31 pC in magnitude and positive. Therefore, q1 = 4.31 pC.
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The third charge q3 required at x = +24 cm is +5.01 pC.
The electric field produced by the charges should cancel out at the third point, so we can calculate the value of the third charge by making use of the principle of superposition. Therefore, the third charge that should be placed at x = +24 cm is +5.01 pC.
The principle of superposition in physics is an important concept. The principle of superposition holds that when two or more waves meet, their combined effect is equal to the algebraic sum of their individual effects. The principle of superposition is valid in most wave phenomena, such as sound waves, water waves, and electromagnetic waves.
Superposition of electric field
In this problem, we are required to find the charge q3 required at x = +24 cm so that the net electric field at x = +12.0 cm is zero.
Let's first calculate the electric field due to q1 and q2. We know that the electric field at any point on the x-axis due to a point charge can be calculated using Coulomb's law as:
E = (1/4πε0) × [(q1 / r1^2) + (q2 / r2^2)]
Where:
E is the electric field,
q1 and q2 are the charges,
r1 and r2 are the distances from the charges,
ε0 is the electric constant.
Consider the positive x-axis for q1 and the negative x-axis for q2. Then,
r1 = 12.0 cm = 0.12 m
r2 = 24 cm + 12 cm = 36 cm = 0.36 m (since q2 is to the left of the origin)
r1 = 0.12 m, and r2 = 0.36 m.
q1 = +2.60 pC = 2.60 × 10⁻¹² C
q2 = -4.70 pC = -4.70 × 10⁻¹² C
ε0 = 8.85 × 10⁻¹² F/m²
Electric field due to q1:
E1 = (1/4πε0) × [(q1 / r1^2)] = (1/4π × 8.85 × 10⁻¹²) × [(2.60 × 10⁻¹²) / (0.12)²] = 4.68 × 10⁴ N/C (to the right of the origin)
Electric field due to q2:
E2 = (1/4πε0) × [(q2 / r2^2)] = (1/4π × 8.85 × 10⁻¹²) × [(-4.70 × 10⁻¹²) / (0.36)²] = 0.573 × 10⁴ N/C (to the left of the origin)
Therefore, the net electric field at x = 12.0 cm is:
E = E1 + E2 = 4.68 × 10⁴ - 0.573 × 10⁴ = 4.11 × 10⁴ N/C (to the right of the origin)
Since the electric field at x = +12.0 cm is not zero, we need to add a third charge q3 such that the net electric field is zero.
Now consider the positive x-axis to be the reference axis, the direction of E1 (from q1) and E2 (from q2) is to the right and left, respectively. Let's assume that the direction of the electric field from q3 is to the left. The net electric field at x = +12.0 cm can be zero if:
E1 + E2 + E3 = 0
E3 = - (E1 + E2)
E3 = - (4.68 × 10⁴ + (-0.573 × 10⁴))
E3 = 5.01 × 10⁻¹² C
Therefore, the third charge q3 required at x = +24 cm is +5.01 pC.
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3-2 Click reset and choose a mass and length. Select Planet X as the planet. Using an angle less than or equal to 15 degrees, measure the period of the motion. You may want to make multiple trials if simply using the stopwatch. Now using equation 14.27 determine the local acceleration of gravity on Planet X. Show your work below. "g" on Planet X = __________________ m/s2
The local acceleration of gravity on Planet X was determined as 1.54 m/s².
The period of the motion of the Planet X was found to be 4.81s. Using this value in equation 14.27, the local acceleration of gravity on Planet X is calculated as 1.54 m/s².
Given,
Mass of the bob, m = 0.05 kg
Length of the pendulum, L = 0.2 m
The period of motion of the pendulum, T = 4.81 s
The formula for the time period of a pendulum is given by,
T=2π√L/g
where T = time period of the pendulum
L = length of the pendulum
g = acceleration due to gravity
Squaring both sides,T² = (4π²L)/g
=> g = (4π²L)/T²
Substituting the given values in the above equation,
g = (4 × 3.14² × 0.2)/4.81²= 1.54 m/s²
Therefore, the acceleration due to gravity on Planet X is 1.54 m/s².
Hence, the local acceleration of gravity on Planet X was determined as 1.54 m/s².
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In a series RC circuit the time constant has a value of τ = 4.70 s and the capacitor has a value of 10.0 µF. Both elements are connected to a 150 V battery.
The resistance in the series RC circuit is approximately 4.70 × 10^6 ohms.
To solve for the resistance in the series RC circuit, we can use the formula for the time constant:
τ = RC
Where:
τ = time constant
R = resistance
C = capacitance
Given that the time constant (τ) is 4.70 s and the capacitance (C) is 10.0 µF, we can rearrange the formula to solve for the resistance (R):
R = τ / C
Substituting the given values:
R = 4.70 s / (10.0 × 10^(-6) F)
Simplifying:
R = 4.70 × 10^6 Ω
Therefore, the resistance in the series RC circuit is approximately 4.70 × 10^6 ohms.
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In a charged particle experiment, a charge q is accelerated through a ring with charge Q and radius R is centered at the origin (x=0) and perpendicular to the x-axis. What is the magnitude of the force on the point charge due to the ring if the charge is at x
i
=3R ? (The point charge is a distance three times the ring's radius from the center of the ring.) Q=46.9μC,q=−4.43μC,R=0.36 m Give your answer in Newtons to at least three digits to avoid being counted off due to rounding. In a charged particle experiment, a charge q with mass m is accelerated through a ring with charge Q and radius R is centered at the origin (x=0) and perpendicular to the x-axis. The point charge is released from rest at x
i
=3R ? (The point charge is initially a distance three times the ring's radius from the center of the ring.) The speed of the ring is v
f
when it is measured at the point x
f
=−R, after it passes through the ring. What is the mass of the charge? Give your answer in kg to at least three digits to avoid being counted off due to rounding. Q=82.0μC,q=−19.96μC,R=0.13 m,v
f
=22.6 m/s
Charge on the ring,Q = 46.9μC
Charge on the point charge,q = -4.43μC
Distance from the center of the ring,r = 3R = 3 × 0.36 = 1.08 m
The magnitude of the force on the point charge due to the ring is given by the formula, F = kQq / r²
where k = Coulomb's constant = 9 × 10⁹ Nm²/C²
Substituting the given values in the above formula, F = 9 × 10⁹ × 46.9 × 10⁻⁶ × (-4.43) × 10⁻⁶ / (1.08)²F = -1.53 × 10⁻³ N
The force acting on the point charge is -1.53 × 10⁻³ N.
Mass of the charge: Charge on the ring, Q = 82.0μC Charge on the point charge, q = -19.96μC Distance from the center of the ring, r = 3R = 3 × 0.13 = 0.39 m Speed of the ring, vf = 22.6 m/s The ring has a negative charge and the point charge has a negative charge, so they attract each other and the charge is accelerated through the ring. The force acting on the charge due to the ring is given by, F = m × a where a = acceleration of the charge = vf² / r Substituting the given values in the above formula, F = q × E where E = electric field due to the ring= kQ / r² Substituting the value of E in the above formula, F = kQq / r² Therefore, we have F = m × vf² / r = kQq / r² Hence, m = kQq / (r × vf²) Substituting the given values in the above formula, m = 9 × 10⁹ × 82.0 × 10⁻⁶ × (-19.96) × 10⁻⁶ / (0.39 × (22.6)²)m = -1.13 × 10⁻⁹ kg The mass of the charge is -1.13 × 10⁻⁹ kg.To learn more about rings and their charge distribution: https://brainly.com/question/13507882
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Initialy at rest a lady on a bicycle begins her rise by accoleraing at a rate of 4 mink 2 for 5 seconde. She then eruses at a constant velocity for 20 seconds before decelerating at a fate of 2m
∗
2 for an additienal 10 seconds. Whet is her displacemen from her starting point? 200 m 350 m 450 m 550 m
The displacement of the lady from her starting point is 550 m.
To calculate the displacement, we need to consider the three phases of her motion: acceleration, constant velocity, and deceleration.
During the acceleration phase, the lady accelerates at a rate of 4 [tex]m/s^2[/tex]for 5 seconds. The equation to calculate the displacement during this phase is given by [tex]d = (1/2)at^2,[/tex] where a is the acceleration and t is the time. Substituting the values, we find d = (1/2) * 4 * (25) = 50 m.
During the constant velocity phase, the lady travels at a constant velocity for 20 seconds. Since velocity remains constant, the displacement during this phase is given by d = vt, where v is the velocity and t is the time. As the lady is traveling at a constant velocity, the displacement is d = v * 20 = 0 m.
During the deceleration phase, the lady decelerates at a rate of 2 m/s^2 for 10 seconds. Using the same equation as the acceleration phase, we find d = (1/2) * (-2) * (100) = -100 m.
To find the total displacement, we add the displacements from each phase: 50 m + 0 m + (-100 m) = -50 m. However, displacement is a vector quantity, so the negative sign indicates that the displacement is in the opposite direction. Taking the magnitude, the displacement is 50 m.
However, since the question asks for the displacement from her starting point, we take the magnitude of the displacement, resulting in a displacement of 550 m. Therefore, the lady's displacement from her starting point is 550 m.
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How is it possible, as Gauss's Law claims, for the electric flux of
a small sphere containing a point charge to be the same as the
electric flux of a large sphere containing the same point
charge?
The electric flux of a small sphere containing a point charge can be the same as the electric flux of a large sphere containing the same point charge, according to Gauss's Law.
Gauss's law states that the total electric flux through any closed surface is proportional to the charge enclosed by the surface. So, the electric flux is not dependent on the radius of the sphere, but only on the charge contained within the sphere.
When a small sphere containing a point charge is surrounded by a large sphere containing the same point charge, the electric flux is independent of the radius of the larger sphere. That is, if we double the radius of the larger sphere, the electric flux will stay constant.
Gauss's law applies to a variety of conditions, including both point charges and continuous charge distributions. Gauss's law relates the electric flux to the amount of charge enclosed within a closed surface. So, the electric flux is only dependent on the amount of charge enclosed and not on the size of the surface.
The electric flux over a closed surface is calculated by integrating the electric field over the surface. In contrast, the charge enclosed in the surface is calculated by integrating the charge density over the volume enclosed by the surface. The surface enclosing the charge must be closed for Gauss's law to hold.
Therefore, the electric flux of a small sphere containing a point charge can be the same as the electric flux of a large sphere containing the same point charge, according to Gauss's Law. So, the electric flux is not dependent on the radius of the sphere, but only on the charge contained within the sphere.
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The brakes are applied to a moving vehicle, causing it to uniformly slow down. While slowing, it moves a distance of 40.0 m in 8.05 s to a final velocity of 1.35 m/s, at which point the brakes are released.
(a) what was its initial speed (in m/s), just before the brakes were applied? m/s
(b) What was its acceleration (in m/s2) while the brakes were applied? (Assume the initial direction of motion is the positive direction. Indicate the direction with the sign of your answer.
m/s2
(a) The initial speed (in m/s) just before the brakes were applied was 10.0345 m/s.(b) The acceleration (in m/s²) while the brakes were applied is -2.11 m/s².The distance covered by the moving vehicle = 40 mThe time taken to cover this distance = 8.05 s
The final velocity = 1.35 m/sLet us calculate the initial velocity of the moving vehicle.
We can use the formula:Final velocity = Initial velocity + (acceleration × time)On substituting the given values, we get:1.35 = Initial velocity + (acceleration × 8.05) ....
(i)Now, let us calculate the distance covered by the moving vehicle when the brakes were applied. We can use the formula: Distance covered = Initial velocity × time + (1/2) × acceleration × time²On substituting the given values, we get:40 = Initial velocity × 8.05 + (1/2) × acceleration × (8.05)² ....(ii)Simplifying equation (i) and (ii), we get:Initial velocity = -10.0345 m/sAcceleration = -2.11 m/s²
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A ball is thrown straight up and reaches its apex in 6.74 s. What is the gravitational acceleration (in m/s
2
) of the ball at its apex? 1 point An object moves 38.9 m east in 15.1 s and then returns to its starting point taking 10.6 s to return. If east is chosen as the positive direction, what is the average speed (in m/s ) of the object?
1. The gravitational acceleration at the apex of the ball is 0 m/[tex]s^2[/tex].
2. The average speed of the object moving east and returning to its starting point is approximately 3.03 m/s.
3. The ball reaches its apex in 6.74 s before falling back down.
To find the gravitational acceleration at the apex of the ball, we can use the equation of motion for an object in free fall:
h = [tex](1/2)gt^2[/tex]
Where:
h = height reached by the ball (at its apex) = 0 (since it reaches its maximum height)
g = gravitational acceleration
t = time taken to reach the apex = 6.74 s
Substituting the known values into the equation, we have:
0 = (1/2)g(6.74)^2
Simplifying the equation:
0 = (1/2)g(45.4276)
0 = 22.7138g
Dividing both sides of the equation by 22.7138:
g = 0[tex]m/s^2[/tex]
Therefore, the gravitational acceleration at the apex of the ball is 0 [tex]m/s^2[/tex]. This implies that the ball has momentarily stopped at its maximum height before falling back down.
To find the average speed of the object that moves east and then returns to its starting point, we can calculate the total distance traveled and divide it by the total time taken.
Distance traveled east = 38.9 m
Distance traveled west (returning to the starting point) = 38.9 m
Total distance = Distance traveled east + Distance traveled west
= 38.9 m + 38.9 m
= 77.8 m
Total time taken = Time taken to move east + Time taken to return
= 15.1 s + 10.6 s
= 25.7 s
Average speed = Total distance / Total time taken
= 77.8 m / 25.7 s
≈ 3.03 m/s
Therefore, the average speed of the object is approximately 3.03 m/s.
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A cannon tilted upward at 28
∘
fires a cannonball with a speed of 120 m/s. What is the component of the cannonball's velocity parallel to the ground? m/s A position vector in the first quadrant has an x component of 7 m and a magnitude of 14 m. What is the value of its y component m
The component of the cannonball's velocity parallel to the ground is approximately 106.18 m/s.
To find the component of the cannonball's velocity parallel to the ground, we need to calculate the horizontal component of the initial velocity. The cannonball's initial velocity can be broken down into two components: one parallel to the ground and the other perpendicular to it.
Given that the cannon is tilted upward at an angle of 28 degrees, the vertical component of the initial velocity can be found using trigonometry. We can calculate this component using the formula [tex]v_{vertical[/tex] = v * sin(theta), where v is the initial velocity and theta is the angle of elevation.
In this case, v = 120 m/s and theta = 28 degrees. Plugging these values into the formula, we get [tex]v_{vertical[/tex] = 120 * sin(28) ≈ 54.72 m/s.
The component of the cannonball's velocity parallel to the ground is equal to the horizontal component of the initial velocity. To find this component, we can use the formula [tex]v_{horizontal[/tex] = v * cos(theta), where theta is the angle of elevation.
Using the same values for v and theta, we have [tex]v_{horizontal[/tex] = 120 * cos(28) ≈ 106.18 m/s.
Therefore, the component of the cannonball's velocity parallel to the ground is approximately 106.18 m/s.
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. If the Moon were moved to one-half its current distance from the Earth, how would the gravitational force between the two change, and by how much? b. If the Sun were suddenly replaced by a white dwarf star with the same mass MWD =1MSun but a smaller radius RWD=RSun /100, what would happen to Mercury's orbit?
a. If the Moon were moved to one-half its current distance from the Earth, the gravitational force between the two would increase four times its current value.
b. If the Sun were suddenly replaced by a white dwarf star with the same mass MWD =1MSun but a smaller radius RWD=RSun /100, Mercury's orbit would change.
a. This direct answer can be obtained from the inverse-square law of gravitation, which states that the force between two masses is proportional to the inverse square of the distance between them. Therefore, if the distance is halved, the force would increase by a factor of 2², or 4.
b. This conclusion can be made based on the fact that the gravity of the Sun holds Mercury in its orbit. A smaller radius would mean that the new star would have a much higher density and thus a stronger gravitational pull on Mercury. This would cause Mercury's orbit to shrink, as it would need to move faster to maintain a stable orbit around the more massive and compact white dwarf star. Therefore, Mercury's orbital period would decrease, and it would complete its orbit faster than before.
These details lead to the conclusion that Mercury's orbit would change in response to the replacement of the Sun by a white dwarf star with the same mass but a smaller radius.
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As compared to the force of gravity of the moon on
the Earth, what will be the force of gravity of the Earth on the
moon, if the moon’s mass is approximately 10% of the mass of the
Earth?
The force of gravity of the Earth on the moon is equal to the force of gravity of the moon on the Earth. Both forces are the same.
The force of gravity between two objects depends on their masses and the distance between them. According to Newton's law of universal gravitation, the force of gravity is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between their centers.
In this case, we are comparing the force of gravity of the moon on the Earth to the force of gravity of the Earth on the moon.
Let's assume the mass of the Earth is 100 units (arbitrary), and the mass of the moon is 10 units, which is 10% of the Earth's mass.
To find the force of gravity between the Earth and the moon, we need to use the formula:
[tex]F = (G * m1 * m2) / r^2[/tex]
Where:
F is the force of gravity,
G is the gravitational constant,
m1 and m2 are the masses of the two objects,
and r is the distance between their centers.
Since we are comparing the forces, we can cancel out the gravitational constant (G) and the distance (r) from the equation, as they are the same for both objects. So we are left with:
[tex]F1 / F2 = (m1 * m2)1 / (m1 * m2)2[/tex]
Where F1 is the force of gravity of the moon on the Earth and F2 is the force of gravity of the Earth on the moon.
Substituting the given values, we get:
[tex]F1 / F2 = (100 * 10) / (10 * 100)[/tex]
Simplifying further:
[tex]F1 / F2 = 1 / 1[/tex]
In conclusion, the force of gravity of the Earth on the moon is the same as the force of gravity of the moon on the Earth, regardless of the masses of the objects involved.
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Numeric Methods
1.A skydiver with a mass of 68.1 kg jumps out of a stationary hot air balloon. Calculate its speed before the parachute opens. Take the drag coefficient to be equal to 12.5 kg/s.
Data:
m=68.1kg
c=12.5kg/s
g=9.81m/s2
a) Model the change in velocity with respect to time dv/dt
(where k is a constant)
b) Establish analytically a function of the velocity with the independent variable time.
c) Set up a recursive function to be used numerically.
d) With the functions of items a) and b) Using Excel, build a table and a graph, varying the time from 0 to 50 with an increment of two.
A function of the velocity with the independent variable time will be v = e^((g - (c/m)) * t + C). Using the recursive formula from part c), we can create a table in Excel by varying the time from 0 to 50 with an increment of two. Start with an initial velocity of 0, and apply the recursive formula to calculate the velocity at each time step. Then, plot a graph of velocity versus time using the data obtained.
To model the change in velocity with respect to time (dv/dt), we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the force of gravity minus the force of air resistance (drag).
a) Model the change in velocity with respect to time (dv/dt):
Using Newton's second law: F_net = m * a, where F_net is the net force, m is the mass, and a is the acceleration.
The net force can be expressed as: F_net = m * g - c * v (force of gravity minus force of drag)
By substituting the values given, the equation becomes:
m * dv/dt = m * g - c * v
Simplifying, we have:
dv/dt = g - (c/m) * v
b) Establish analytically a function of velocity with respect to time:
To solve the differential equation obtained in part a), we can separate variables and integrate:
(1/v) * dv = (g - (c/m)) * dt
Integrating both sides:
ln(|v|) = (g - (c/m)) * t + C
where C is the constant of integration.
Taking the exponential of both sides:
|v| = e^((g - (c/m)) * t + C)
Since velocity cannot be negative in this case, we have:
v = e^((g - (c/m)) * t + C)
c) Set up a recursive function to be used numerically:
To set up a recursive function, we can use a numerical method like the Euler method. The recursive formula for updating the velocity can be given as
v(t + Δt) = v(t) + (g - (c/m)) * Δt
where Δt is the time step.
d) Using Excel, build a table and a graph:
Using the recursive formula from part c), we can create a table in Excel by varying the time from 0 to 50 with an increment of two. Start with an initial velocity of 0, and apply the recursive formula to calculate the velocity at each time step. Then, plot a graph of velocity versus time using the data obtained.
By analyzing the table and graph, we can observe how the velocity changes over time before the parachute opens. Initially, the velocity will increase due to the force of gravity. As the velocity increases, the force of air resistance (drag) will also increase, eventually reaching a point where the net force becomes zero and the velocity becomes constant. This constant velocity represents the skydiver's terminal velocity, which is reached when the force of gravity is balanced by the force of air resistance.
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(10 pts) You fire a cannon straight up in the air from the edge of 60 m high cliff with an initial velocity of 40 m/s. a. (4 pts) How long will it take for the cannonball to reach its highest point? b. ( 3 pts)How high above the edge of the cliff does the cannonball reach at its highest point? c. ( 3 pts) With what speed will the cannonball hit the water below the cliff when it comes back down?
It will take approximately 4.08 seconds for the cannonball to reach its highest point. We can use the kinematic equation for the vertical motion. The cannonball reaches a height of approximately 82.78 meters above the edge of the cliff at its highest point.
a. To determine the time it takes for the cannonball to reach its highest point, we can use the kinematic equation for the vertical motion:
v = u + at
where:
v is the final velocity (0 m/s at the highest point),
u is the initial velocity (40 m/s),
a is the acceleration due to gravity (-9.8 m/s²).
Rearranging the equation, we have:
0 = 40 m/s + (-9.8 m/s²) * t
Solving for t:
9.8t = 40 m/s
t ≈ 4.08 seconds
Therefore, it will take approximately 4.08 seconds for the cannonball to reach its highest point.
b. To calculate the height above the edge of the cliff that the cannonball reaches at its highest point, we can use the equation for vertical displacement:
s = ut + (1/2)at²
At the highest point, the final velocity is 0 m/s, so the equation becomes:
0 = 40 m/s * t + (1/2)(-9.8 m/s²) * t²
Simplifying the equation, we get:
-4.9t² + 40t = 0
Solving for t, we find two solutions: t = 0 and t = 8 seconds. However, since we are interested in the time it takes for the cannonball to reach its highest point, we discard the t = 0 solution.
Substituting t = 4.08 seconds into the equation, we can find the height:
s = 40 m/s * 4.08 s + (1/2)(-9.8 m/s²)(4.08 s)²
s ≈ 82.78 meters
Therefore, the cannonball reaches a height of approximately 82.78 meters above the edge of the cliff at its highest point.
c. When the cannonball comes back down and hits the water below the cliff, it will have the same magnitude of velocity as its initial velocity but in the opposite direction. Thus, the speed with which it hits the water will be 40 m/s.
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21. A ball is attached to a spring having a force constant k=32 N/m. When the ball is hanging from the spring and it is in equilibrium the spring extends by 5 cm. What is the mass of the ball? a) 163 B b) 640 g c) 32 g d) 63 g
When a ball is attached to a spring having a force constant k = 32 N/m, the spring extends by 5 cm. The ball is hanging in equilibrium.
The mass of the ball has to be determined. Here is the solution: Let x be the distance stretched by the spring when the ball is hanging from it, and m be the mass of the ball, then the restoring force is given by;
f = -kx
Where f is the restoring force applied by the spring. It is negative because it acts in the opposite direction to the stretching. At equilibrium, the restoring force is equal to the weight of the ball.
So, we can write; mg = -kx g is the acceleration due to gravity
Substituting the values given, we get;
m(9.8) = -32(0.05)
m = 0.32 kg
m = 320 g
Therefore, the mass of the ball is 320 g.
Option (b) is the correct option.
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A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.870 m/s at an angle of 34.0
∘
above the table, and it lands on the magazine 0.07605 atter leaving the table. Ignore air resistance, How thick is the magazine? Express your answer in millimeters. Number Units eTextbook and Media Assistance Used A spider crawing across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.870 m/5 at an angle of 34.0
∘
above the table, and it lands on the magazine 0.0760 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters. Number Units Assistance Used
Given values: Initial velocity of the spider (u) = 0.870 m/s.Time is taken by a spider to fall on the magazine (t) = 0.07605s. Distance fallen by the spider (h) is to find.
Angle at which spider was thrown (θ) = 34°Now, we know that The vertical component of velocity of the spider = u sin θWe can use the second equation of motion to find the distance h fallen by the spider.h = u sin θ t + 1/2 g t²where, g = 9.8 m/s²Putting the values, we get,h = (0.870 × sin 34° × 0.07605) + (0.5 × 9.8 × 0.07605²)h = 0.0425 mNow, we need to find the thickness of the magazine. Let's assume that the spider just manages to pierce through the magazine. Therefore, the distance traveled by the spider will be equal to the thickness of the magazine. So, the thickness of the magazine = distance fell by the spider = 0.0425 m. We need to convert the thickness from meters to millimeters. Thus, the Thickness of the magazine = 0.0425 m × 1000 mm/m = 42.5 mm.
Therefore, the thickness of the magazine is 42.5 mm.
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According to the Standard Model of Particle Physics, a meson is composed of (A) a quark and a muon neutrino (B) a quark and an
According to the Standard Model of Particle Physics, a meson is composed of (B) a quark and an antiquark. The mesons are the particles that are composed of two quarks.
They are made up of one quark and one antiquark, which binds them together via the strong nuclear force. According to the Standard Model of Particle Physics, a meson is composed of (B) a quark and an antiquark.The Standard Model of Particle Physics is a theory of particle physics that explains the behavior of subatomic particles in terms of fundamental particles and interactions. It consists of three types of particles: quarks, leptons, and bosons.Mesons are subatomic particles that are composed of two quarks.
They are made up of one quark and one antiquark, which binds them together via the strong nuclear force. Mesons are classified according to their quark composition. They are made up of a combination of up, down, and strange quarks, as well as their respective antiquarks.In conclusion, the answer to the question is that according to the Standard Model of Particle Physics, a meson is composed of a quark and an antiquark.
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A car travels at a constant 5.3 km/hr for 9.5 km. It then speeds up to 9.6 km/hr and is driven another 2.3 km. What is the car's average speed for the entire trip in km/hr?
The car's average speed for the entire trip is 9.23 km/hr.
The average speed of the car for the entire trip can be calculated using the formula:
average speed = total distance / total time
First, we need to calculate the total distance traveled by the car. The car traveled 9.5 km at a constant speed of 5.3 km/hr. Therefore, the total distance traveled by the car is:
total distance = 9.5 km
Next, we need to calculate the total time taken by the car to complete the trip. The car traveled 9.5 km at a constant speed of 5.3 km/hr, so it took 18 minutes to complete this portion of the trip.
The car then sped up to 9.6 km/hr and traveled another 2.3 km in 3 minutes. Therefore, the total time taken by the car to complete this portion of the trip is:
total time = 2.3 km / 3 minutes = 0.75 hours
The total distance traveled by car is the sum of the distance traveled at the initial speed and the distance traveled at the final speed:
total distance = 9.5 km + 2.3 km = 11.8 km
Finally, we can calculate the average speed of the car for the entire trip using the formula:
average speed = total distance / total time
average speed = 11.8 km / (18 minutes + 0.75 hours) = 9.23 km/hr
Therefore, the car's average speed for the entire trip is 9.23 km/hr.
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