The magnitude of the magnetic field is 0.00171 T.
The magnitude of the magnetic field when an airplane flies north in a magnetic field pointing 30 degrees down from north can be calculated using the formula;
emf = Bℓv Sinθ
Where;
emf = 0.15VB = ?ℓ = 30mv = 175m/sSinθ = sin 30º = 1/2
Substitute the given values into the formula;0.15V = B × 30m × 175m/s × (1/2)
Rearrange the formula to isolate B;B = 0.15V / (30m × 175m/s × (1/2)) = 0.00171 T
Therefore, the magnitude of the magnetic field is 0.00171 T.
Another way to approach the question is using Faraday’s law of electromagnetic induction which states that the emf generated in a coil is proportional to the rate of change of magnetic flux through the coil. In this case, the magnetic field is perpendicular to the velocity vector of the airplane, therefore the magnetic flux is given by;
Φ = B × A
Where;Φ = BA = wingspan = 30m
Substituting into the formula;
emf = dΦ / dtBut dA/dt = 0 because the wingspan is constant
Therefore;
emf = B dA / dt = BA (d/dt) sinθ = BA cosθ (dθ/dt)The induced emf is 0.15V, the wingspan is 30m and the velocity of the plane is 175m/s.
Because the plane is flying north, the angle of the magnetic field with the horizontal is 30º and it is pointing down from the north pole, therefore the angle between the magnetic field and the velocity vector of the plane is 60º. Therefore;
emf = BA cos60º (dθ/dt)0.15V = (30m × B × (1/2)) (175m/s / 30m)B = 0.00171 TTherefore, the magnitude of the magnetic field is 0.00171 T.
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The work done by an ideal monatomic gas undergoing one complete Carnot cycle is 9936 J. The Carnot engine operates between TH=2990 K and TL=1140 K. The high temperature expansion happens between V1=301 L and V2=899 L. How many mol of gas participate in this process?
The number of moles of gas participating in the Carnot cycle can be determined by calculating the net work done and using the ideal gas law. In this case, the number of moles is approximately 11.94.
The net work done by an ideal gas in a Carnot cycle can be calculated using the formula:
[tex]W_{net} = nRT_{H}ln(V2/V1)[/tex]
where [tex]W_{net[/tex] is the net work done, n is the number of moles of gas, R is the gas constant, [tex]T_H[/tex] is the high temperature in Kelvin, V1 is the initial volume, and V2 is the final volume.
In this case, the net work done is given as 9936 J. The gas constant R is a constant value. The high temperature [tex]T_H[/tex] is given as 2990 K, and the initial and final volumes V1 and V2 are given as 301 L and 899 L, respectively.
By rearranging the formula, we can solve for the number of moles:
[tex]n = W_{net} / RT_{H}ln(V2/V1)[/tex]
Substituting the given values, we can calculate the number of moles of gas participating in the process, which is approximately 11.94 moles.
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A soccer ball is kicked from the ground with an initial velocity of 17.9 m/s, at an angle of 20.5 degrees with respect to the horizontal. Once the ball reaches its maximum height, what is the velocity in the x direction at that moment? Ignore air resistance
A soccer ball is kicked from the ground with an initial velocity of 17.9 m/s, at an angle of 20.5 degrees with respect to the horizontal. Once the ball reaches its maximum height, its velocity is 16.8 m/s.
When a projectile reaches its maximum height, its vertical velocity component becomes zero. Therefore, at the maximum height, the velocity in the x-direction (horizontal velocity) remains constant.
Given:
Initial velocity ([tex]V_o[/tex]) = 17.9 m/s
Launch angle (θ) = 20.5 degrees
To find the velocity in the x-direction at the maximum height, we need to determine the horizontal component of the initial velocity ([tex]v_x[/tex]).
[tex]V_x[/tex]= [tex]V_o[/tex]* cos(θ)
Substituting the values:
[tex]V_x[/tex]= 17.9 m/s * cos(20.5°)
[tex]V_x[/tex]= [tex]V_o[/tex]* cos(θ)
Given:
[tex]V_o[/tex]= 17.9 m/s
θ = 20.5°
Using the formula, we can substitute the values and calculate [tex]V_x[/tex]:
[tex]V_x[/tex]= 17.9 m/s * cos(20.5°)
Taking the cosine of 20.5°, we get:
cos(20.5°) ≈ 0.9383
Substituting this value back into the equation:
[tex]V_x[/tex]= 17.9 m/s * 0.9383
Calculating this expression gives us the velocity in the x-direction at the maximum height:
[tex]V_x[/tex]≈ 16.8 m/s
Therefore, the velocity in the x-direction at the moment the soccer ball reaches its maximum height is approximately 16.8 m/s.
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An electric dipole is formed from ±1.00nC charges spaced 2.00 mm apart. The dipole is at the origin, oriented along the x-axis.
What is the electric field strength at the points (x,y)=(7.00cm, 0cm)?
The electric field strength at the point (7.00 cm, 0 cm) due to the electric dipole is approximately 8.99 × 10⁹ N/C directed along the positive x-axis.
To calculate the electric field strength at the point (x, y) = (7.00 cm, 0 cm) due to an electric dipole, we can use the formula for the electric field of a dipole:
E = (1 / (4πε₀)) * ((2p) / r³) * cosθ
Where:
- E is the electric field strength.
- ε₀ is the permittivity of free space (ε₀ ≈ 8.854 × 10⁻¹² C²/Nm²).
- p is the magnitude of the dipole moment (p = q * d).
- r is the distance from the dipole to the point.
- θ is the angle between the dipole axis and the line connecting the dipole to the point.
In this case, the dipole is oriented along the x-axis, so θ = 0°. The distance from the dipole to the point is:
r = √((x - x₀)² + (y - y₀)²)
Substituting the values:
x₀ = 0 cm (x-coordinate of the dipole)
y₀ = 0 cm (y-coordinate of the dipole)
x = 7.00 cm
y = 0 cm
r = √((7.00 cm - 0 cm)² + (0 cm - 0 cm)²)
= √(49.00 cm²)
= 7.00 cm
The magnitude of the dipole moment is given by:
p = q * d
Where:
- q is the magnitude of each charge in the dipole (q = 1.00 nC = 1.00 × 10⁻⁹ C).
- d is the separation between the charges in the dipole (d = 2.00 mm = 2.00 × 10⁻³ m).
p = (1.00 × 10⁻⁹ C) * (2.00 × 10⁻³ m)
= 2.00 × 10⁻¹² Cm
Now we can calculate the electric field strength:
E = (1 / (4πε₀)) * ((2p) / r³) * cosθ
θ = 0°
ε₀ ≈ 8.854 × 10⁻¹² C²/Nm²
E = (1 / (4π(8.854 × 10⁻¹² C²/Nm²))) * ((2 * 2.00 × 10⁻¹² Cm) / (7.00 cm)³) * cos(0°)
≈ (1 / (4π(8.854 × 10⁻¹² C²/Nm²))) * (4.08 × 10⁻¹² Cm / (7.00 cm)³)
Evaluating the expression:
E ≈ 8.99 × 10⁹ N/C
Therefore, the electric field strength at the point (7.00 cm, 0 cm) is approximately 8.99 × 10⁹ N/C directed along the positive x-axis.
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A proton is moving due south, in an area where the Earth's magnetic field is directed straight down toward the ground. Calculate the magnetic force on the proton if the field has strength 6μT and the proton initially moves at speed 550 m/s. F=5.28∗10
−22
N East Q2: At a certain location, the horizontal component of the earth's magnetic field is 2.5×10
−5
T, due north. An electron moves parallel to the ground with just the right speed for the magnetic force on it to balance its weight. Assuming the electron is perpendicular to B, calculate the velocity of the electron. Mass of electron =9.1
∗
10
∧
−31 kg v=2.2295∗10
−6
s
m
West
1. The magnetic for a proton is: 5.28 × 10⁻²² N which is directed east.
2. the velocity of the electron: 2.2295 × 10⁶ ms⁻¹, which is directed west.
A proton is moving due south, in an area where the Earth's magnetic field is directed straight down toward the ground. Calculate the magnetic force on the proton if the field has strength 6μT and the proton initially moves at speed 550 m/s.
The formula for magnetic force F on a charged particle moving in a magnetic field is given by:
F = Bqv sinθ
where
B = the magnetic field strength (in teslas)
q = the electric charge on the particle (in coulombs)
v = the velocity of the particle (in meters per second)
θ = the angle between the magnetic field and the velocity of the particle.
Using the given values of the problem
F = Bqv sinθ
where
B = 6 μT = 6 × 10⁻⁶ T
q = +1.6 × 10⁻¹⁹ CV = 550 m/s = 5.5 × 10² ms⁻¹θ = 90° (because the proton moves due south while the magnetic field is directed straight down toward the ground, making the angle between them a right angle)
Plugging in these values gives us:
F = (6 × 10⁻⁶ T) (1.6 × 10⁻¹⁹ C) (5.5 × 10² ms⁻¹)
sin 90°F = 5.28 × 10⁻²² N
which is directed east.
Q2: At a certain location, the horizontal component of the earth's magnetic field is 2.5×10⁻⁵ T, due north. An electron moves parallel to the ground with just the right speed for the magnetic force on it to balance its weight. Assuming the electron is perpendicular to B, calculate the velocity of the electron.
Mass of electron = 9.1 × 10⁻³¹ kg
The magnetic force F on a charged particle moving in a magnetic field is given by:
F = Bqv sinθ
where
B = the magnetic field strength (in teslas)
q = the electric charge on the particle (in coulombs)
v = the velocity of the particle (in meters per second),
θ = the angle between the magnetic field and the velocity of the particle.
In this case, the electron moves parallel to the ground with just the right speed for the magnetic force on it to balance its weight.
That means the magnetic force is equal and opposite to the gravitational force, which can be calculated as:
Fg = mg
where
Fg = the gravitational force (in newtons),
m = the mass of the particle (in kilograms)
g = the acceleration due to gravity (in meters per second squared).
Since the electron is in equilibrium (i.e., it's not accelerating), the net force acting on it is zero.
Therefore, F = FgBqv sinθ = mg
Solving this equation for v gives us:v = mg / Bq sinθ
Plugging in the given values of the problem gives us:
v = (9.1 × 10⁻³¹ kg) (9.8 ms⁻²) / [(2.5 × 10⁻⁵ T) (1.6 × 10⁻¹⁹ C) sin 90°]v = 2.2295 × 10⁶ ms⁻¹, which is directed west.
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If the elevator is accelerating 4 m/s∧2upward, what would be the reading on the bathroom scale? Use g=10 m/s∧2? If the man in diagram A has a mass of 90 kg, what would be the reading of the bathroom scale?
When the elevator is accelerating upward at 4 m/s^2, the reading on the bathroom scale for a man with a mass of 90 kg would be 540 N. This is due to the net force acting on the man, which is the difference between his weight and the pseudo force caused by the elevator's acceleration.
To determine the reading on the bathroom scale, we need to consider the forces acting on the man in the elevator.
When the elevator is accelerating upward at 4 m/s^2, there are two main forces acting on the man: his weight (mg) and the pseudo force (ma), where m is the mass of the man and a is the acceleration of the elevator.
The net force acting on the man is the difference between these two forces, which can be calculated as follows:
Net force = mg - ma
Using g = 10 m/s^2 and the man's mass of 90 kg, we can substitute these values into the equation:
Net force = (90 kg)(10 m/s^2) - (90 kg)(4 m/s^2)
= 900 N - 360 N
= 540 N
Therefore, the reading on the bathroom scale would be 540 N when the elevator is accelerating upward at 4 m/s^2.
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A projectile is launched at ground level with an initial speed of 58 m/s at an angle of 34
∘
above the horizontal. It strikes a target above the ground 4 seconds later. What are the x and y distances from where the projectile was launched to where it lands?
The x-distance from where the projectile was launched to where it lands is approximately 188.76 meters, while the y-distance is approximately 86.18 meters.
To find the x-distance, we calculate the horizontal component of the projectile's velocity by multiplying the initial speed (58 m/s) by the cosine of the launch angle (34 degrees). This gives us approximately 48.13 m/s. Multiplying the horizontal velocity by the time of flight (4 seconds), we get an approximate x-distance of 192.52 meters.
To find the y-distance, we use the equation for vertical displacement, taking into account the initial vertical velocity (58 m/s multiplied by the sine of 34 degrees), the time of flight (4 seconds), and the acceleration due to gravity (9.8 m/s^2). The vertical displacement is approximately 128.24 meters. Subtracting the initial height (assumed to be 0), we get an approximate y-distance of 86.18 meters.
Therefore, the projectile launched at ground level will land approximately 188.76 meters horizontally and 86.18 meters vertically from the launch point.
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Andrew with a reaction time of 2 seconds is on his bicycle in a motion at 300 cm per second. He encounters an emergency and pushes his brake immediately. How far does the bicycle move before he pushes the brake? This is about what is not changed before Andrew pushes his brake and think about how to relate the time and the velocity to the total displacement during the time. Assume that the velocity of the bicycle does not get changed before applying the brake.
Answer:
The bicycle moves a distance of 600 cm before Andrew pushes the brake.
Explanation:
To find the distance the bicycle moves before Andrew pushes the brake, we need to calculate the displacement during the reaction time of 2 seconds.
Since the velocity of the bicycle remains constant before Andrew pushes the brake, we can use the formula:
Displacement = Velocity × Time
Given that the velocity of the bicycle is 300 cm/s and the reaction time is 2 seconds, we can calculate the displacement as follows:
Displacement = 300 cm/s × 2 s
Displacement = 600 cm
Therefore, the bicycle moves a distance of 600 cm before Andrew pushes the brake.
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What are the strength and direction of the electric field at the position indicated by the dot in the figure(Figure 1)? Give your answer in component form. (Assume that x-axis is horisontal and points to the right, and y-axis points upward.) Express your answer in terms of the unit vectors
i
^
and
j
^
. Express your answer using two significant figures. Part B Give your answer as a magnitude and angle measured cw from the positive x-axis. Express your answer using two significant figures. Figure Part C Express your answer using two significant figures.
The strength and direction of the electric field at the position indicated by the dot in the figure is (5.7 x 10^8 N/C)i + (8.5 x 10^8 N/C)j and 1 x 10^9 N/C, 56.31°, respectively.
According to the question, the electric field at the position indicated by the dot can be analyzed. As it can be seen that the charges are positive, and they are held at fixed positions.
This scenario suggests that the field lines originate at positive charges and terminate on negative charges. Here, as there are two positive charges, it is expected that the field lines will emanate from each of them.
The magnitude of the electric field will be represented by the number of field lines that emanate from the charges and terminate at the position where the dot is shown in the figure, and the direction of the electric field will be perpendicular to the field lines at the given point.
Given, the electric field due to a single charge at distance r isE=q/(4πε0r^2).
Electric Field due to charge 1, E1=E=q1/(4πε0r^2)
Electric Field due to charge 2, E2=E=q2/(4πε0r^2)
Magnitude of E1 = q1/(4πε0r1^2) = (10 μC)/(4π(8.85 x 10^-12 C^2/N m^2)(5 cm)^2) = 5.7 x 10^8 N/C
Magnitude of E2 = q2/(4πε0r2^2) = (15 μC)/(4π(8.85 x 10^-12 C^2/N m^2)(5 cm)^2) = 8.5 x 10^8 N/C
The resultant electric field at the given position can be calculated using the following formula:
E_net = E1 + E2E_net = (5.7 x 10^8 N/C) i + (8.5 x 10^8 N/C) jE_net = (5.7 x 10^8 N/C)i + (8.5 x 10^8 N/C)j
Magnitude of E_net = sqrt((5.7 x 10^8)^2 + (8.5 x 10^8)^2)
Magnitude of E_net = 1 x 10^9 N/C
Angle made by the resultant electric field with the positive x-axis,
θ = tan^-1(8.5 x 10^8/5.7 x 10^8)θ = 56.31°
So, the strength and direction of the electric field at the position indicated by the dot in the figure are
(5.7 x 10^8 N/C)i + (8.5 x 10^8 N/C)j and 1 x 10^9 N/C, 56.31°, respectively.
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Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in difletent directions: Ricardo walks 26.0 m in a direction 60.0∘west of Part A north. Jane waks 13.0 m in a direction 30.0∘ south of west. What is the distance between them? They then stop and turn to face each other. Express your answor with the appropriate units. Part. B In what direction should Picardo walk to go directly toward Jane? Express your answer in degrees.
Ricardo and Jane walk in different directions under a tree. The distance between them is approximately 16.1 meters, and Ricardo should walk around 6.1° west of north to go directly towards Jane.
To find the distance between Ricardo and Jane, we can use the concept of vector addition. We'll break down their respective displacements into their north-south and east-west components.
Ricardo's displacement can be divided as follows:
- North component: 26.0 m * sin(60.0°) = 22.5 m
- West component: 26.0 m * cos(60.0°) = 13.0 m
Jane's displacement can be divided as follows:
- South component: 13.0 m * sin(30.0°) = 6.5 m
- West component: 13.0 m * cos(30.0°) = 11.3 m
Now, we'll add the respective components together:
- North component: 22.5 m - 6.5 m = 16.0 m
- West component: 13.0 m - 11.3 m = 1.7 m
Using these components, we can calculate the distance between them using the Pythagorean theorem:
Distance = sqrt((16.0 m)^2 + (1.7 m)^2) = sqrt(256.0 m^2 + 2.89 m^2) = sqrt(258.89 m^2) ≈ 16.1 m
Therefore, the distance between Ricardo and Jane is approximately 16.1 meters.
To determine the direction in which Ricardo should walk to go directly toward Jane, we can use trigonometry. The angle can be calculated using the inverse tangent function:
Angle = atan(1.7 m / 16.0 m) ≈ 6.1°
Hence, Ricardo should walk approximately 6.1° west of north to go directly toward Jane.
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A pair of particles initially a distance d apart are released. Particle 1 has charge q
1
=+q
0
and mass m
1
=3m
0
. Particle 2 has charge q
2
=−q
0
and mass m
2
= 2m
0
. Determine the speed of particle 1 when the particles are a distance
2
d
apart. [Hint: Consider both energy and momentum.]
The speed of particle 1 when the particles are a distance 2d apart is given by v1 = √[(8/11)(3q0²/4πεd)].
We are given a pair of particles initially separated by distance d, charge q1 = q0 and mass m1 = 3m0, while charge q2 = -q0 and mass m2 = 2m0.
We are required to determine the speed of particle 1 when the particles are a distance 2d apart. We can use conservation of momentum and energy in order to solve the problem.
Conservation of momentum :
Since the initial momentum is zero, the total momentum must remain zero during the motion. Therefore we can write the equation :
m1v1 + m2v2 = 0
⇒ m1v1 = -m2v2 where v1 and v2 are the velocities of particle 1 and particle 2, respectively.
Conservation of energy :
Energy is conserved as the system is a closed system and no external forces act upon it. Therefore, we can write the equation : KE1 + KE2 + PE1-2 = KE'1 + KE'2
where KE and PE are kinetic and potential energy respectively and KE' represents the final kinetic energy.
As the particles are initially at a distance d apart and come to rest when they are 2d apart, we can write :
KE1 = KE2KE'1 = KE'2KE1 + KE2 = PE1-2KE'1 + KE'2 = 1/2(m1+m2)V2
where V is the final velocity of both the particles when they are at a distance 2d.
Considering these equations, we get :
KE1 + KE2 = PE1-2
⇒ (1/2)m1v1² + (1/2)m2v2² = (q1q2/4πεd)KE'1 + KE'2 = 1/2(m1+m2)V2 ⇒ (1/2)m1v1² + (1/2)m2v2² = (1/2)(m1+m2)V2
Substituting the value of m2 = (2/3)m1 and v2 = -(3/2)v1, we get :
2(1/2)m1v1² = (q1q2/4πεd)2/3m1(9/4)v1² + (1/2)m1V2
⇒ 2v1² = (3q0²/8πεd)(9/4)v1² + 2V2/3
⇒ v1 = √[(8/11)(3q0²/4πεd)]
Therefore the speed of particle 1 = √[(8/11)(3q0²/4πεd)].
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Point charges of 0.21μC and 0.48μC are placed 0.65 m apart. ∵50% Part (a) At what point along the line between them is the electric field zero? Give your answer in meters from the 0.21μC charge. x=0.259∨ Correct!
The point where electric field is zero is at a distance of 0.259 m from the point where the charge is 0.21μC.
We have two point charges q1 = 0.21μC and q2 = 0.48μC and the distance between them is 0.65 m. We need to find out the point where the electric field is zero, let's say at distance 'x' from the 0.21μC charge.
Now, let's consider the electric field at point P from q1, EP = k q1 / (x)², where k is the Coulomb constant.
For q2, the electric field at point P would be E2 = k q2 / (0.65 - x)².
Now, as the electric field is zero,
EP + E2 = 0k q1 / (x)² + k q2 / (0.65 - x)² = 0
Solving this equation, we get, x = 0.259 m.
So, the point where electric field is zero is at a distance of 0.259 m from the point where the charge is 0.21μC.
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A 28-in pump is used to deliver water from a lower reservoir to a higher reservoir. The total length of the 10-in-ID pipes is 62 ft. and the friction factor is taken as 0.022. If the pump head H can be defined as a function of discharge Q in ft^3/s is: H = 365 - (0.04718Q^2), in feet. What is the operating discharge rate in ft^3/s?
The operating discharge rate of the pump is 104.26 ft³/s.
The pump head H can be defined as a function of discharge Q in ft^3/s as follows: H = 365 - (0.04718Q^2), in feet. The total length of the 10-in-ID pipes is 62 ft, and the friction factor is taken as 0.022. We need to find the operating discharge rate in ft^3/s.
Step 1: We will use the Darcy–Weisbach equation to find the operating discharge rate of the pump. The equation is as follows: Δp = (fLρv²) / (2D), where Δp is the head loss, f is the friction factor, L is the length of the pipeline, ρ is the density, v is the velocity, and D is the diameter of the pipeline.
Step 2: To find the velocity, we will use the continuity equation: Q = Av, where Q is the discharge rate, A is the cross-sectional area of the pipe, and v is the velocity. We can rewrite this as v = Q / A.
Step 3: We will find the head loss using the Darcy-Weisbach equation: Δp = (fLρv²) / (2D). Substituting the values, we have Δp = (0.022 × 62 × 62.4 × (Q / 0.5458)²) / (2 × 0.8333), where L is 62 ft, ρ is 62.4 lb/ft³, D is 10 in or 0.8333 ft, and v is Q / 0.5458.
Next, we substitute the value of Δp in the H equation: H = 365 - (0.04718Q²) = (1.379 × 10^-6)Q². Simplifying this equation, we obtain Q² + 4.033 × 10^7 Q - 9.705 × 10^9 = 0.
Solving this quadratic equation, we find Q = 104.26 ft³/s (approximately).
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The position of a particle moving along the x axis depends on the time according to the equation x = ct4 - bt7, where x is in meters and t in seconds. Let c and b have numerical values 2.5 m/s4 and 1.0 m/s7, respectively. From t = 0.0 s to t = 1.7 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.
The displacement of the particle is 3.06 m. The velocity of the particle is 27.5 m/s, 175.5 m/s, 777.5 m/s, 2503.5 m/s, and 6233.5 m/s at times 1.0 s, 2.0 s, 3.0 s, and 4.0 s. The acceleration of the particle is 162.5 m/s², -348.5 m/s², -2388.5 m/s², and -12308.5 m/s² at times 1.0 s, 2.0 s, 3.0 s, and 4.0 s.
Given, x = ct⁴ - bt⁷
Here, c = 2.5 m/s⁴ and b = 1.0 m/s⁷
Displacement = x₁ - x₀
The displacement of the particle from t = 0.0 s to t = 1.7 s is
x₁ - x₀ = (2.5 x 1.7⁴ - 1.0 x 1.7⁷) - (2.5 x 0 - 1.0 x 0⁷)
= 3.06 m
The velocity of the particle is given by the first derivative of displacement with respect to time.
(b) At t = 1.0 s, v = 27.5 m/s
(c) At t = 2.0 s, v = 175.5 m/s
(d) At t = 3.0 s, v = 777.5 m/s
(e) At t = 4.0 s, v = 2503.5 m/s
The acceleration of the particle is given by the second derivative of displacement with respect to time.
(f) At t = 1.0 s, a = 162.5 m/s²
(g) At t = 2.0 s, a = -348.5 m/s²
(h) At t = 3.0 s, a = -2388.5 m/s²
(i) At t = 4.0 s, a = -12308.5 m/s²
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Two masses are connected by a thin, light string over a frictionless, massless pulley as shown. The system is released from rest and m2 takes 0.80 s to descend a distance of 2.00 m. If the mass of m2 is 5.00 kg what is the mass of m1. 11) Two masses are connected by a thin, light string over a frictionless, massless pulley as shown. The system is released from rest and m2 takes 0.80 s to descend a distance of 2.00 m. If the mass of m2 is 5.00 kg what is the mass of m1.
The mass of m1 is 13.8 kg. To find the mass of m1, we can use the principles of Newton's second law and the equations of motion. Using the equation of motion for uniformly accelerated linear motion.
First, let's analyze the motion of m2. We know that it descends a distance of 2.00 m in 0.80 s. Using the equation of motion for uniformly accelerated linear motion:
d = (1/2) * a * t^2
where d is the distance, a is the acceleration, and t is the time, we can solve for the acceleration (a) of m2:
2.00 m = (1/2) * a * (0.80 s)^2
a = (2 * 2.00 m) / (0.80 s)^2
a = 6.25 m/s^2
Since the system is connected by a string over a pulley, the acceleration of m2 is the same as the acceleration of m1. Therefore, the acceleration of m1 is also 6.25 m/s^2.
Next, we can apply Newton's second law to m1:
m1 * a = m1 * g - T
where g is the acceleration due to gravity and T is the tension in the string. Since the system is in equilibrium, T = m2 * g.
Substituting the values:
m1 * 6.25 m/s^2 = m1 * 9.8 m/s^2 - (5.00 kg * 9.8 m/s^2)
Simplifying the equation:
6.25 m1 = 9.8 m1 - 49.0
Rearranging and solving for m1:
3.55 m1 = 49.0
m1 = 13.8 kg
Therefore, the mass of m1 is 13.8 kg.
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A projectile is launched from ground level at an angle of 12.0
∘
above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
The value of the launch angle should be adjusted to 24.0° in order to double the range.
Given that a projectile is launched from ground level at an angle of 12.0° above the horizontal and returns to ground level.
To find the value to which the launch angle should be adjusted, without changing the launch speed, so that the range doubles, we can use the range formula.
The formula for the range is:
Range (R) = (u²sin2θ) / g
where u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
Since the projectile is launched at an angle of 12.0°, we have θ = 12°.
We also know that the projectile returns to ground level, meaning the maximum height reached is zero.
The formula for the maximum height is:
H = (u²sin²θ) / 2g
Since the maximum height is zero, we have:
0 = (u²sin²θ) / 2g
Solving for θ, we get:
θ = sin⁻¹(0) = 0°
Now, let's adjust the launch angle, without changing the launch speed, so that the range doubles.
Let the adjusted angle of projection be θ1. According to the problem, we have:
R1 = 2R
Where R1 is the new range and R is the original range, given by:
R1 = (u²sin2θ1) / g
R = (u²sin2θ) / g
Since the launch speed is not changing, the initial velocity u is the same for both cases. Therefore, we can write the equation as:
sin2θ1 = 2sin2θ
Using the identity sin2θ = 2sinθcosθ, the equation becomes:
2sinθ1cosθ1 = 4sinθcosθ
Dividing both sides by 2sinθcosθ, we get:
tanθ1 = 2tanθ
Substituting the value of θ, we have:
tanθ1 = 2tan12°
Solving for θ1, we find:
θ1 = tan⁻¹(2tan12°) = 24.0°
Therefore, the value of the launch angle should be adjusted to 24.0° in order to double the range.
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If a particle's position is given by x=3t
3
−5t
4
+t, where x is in meters and t in seconds, What is the AVERAGE velocity of the particle between t=0 and t=5.0 s ? Select one: a. all answers are wrong x b. −549 m/s c. 549 m/s d. 2745 m/s e. −2745 m/s
To calculate the average velocity of a particle between two time intervals, we need to find the displacement and divide it by the time interval. the average velocity of the particle between t = 0 and t = 5.0 s is 75.0 m/s.
In this case, the time interval is from t=0 to t=5.0 s.Without any specific information about the position or motion of the particle, we cannot determine the displacement. Therefore, we cannot calculate the average velocity.the mass of the spaceship, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In this case, the force is generated by the ejection of fuel.
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13.9 kg object starting from rest falls through a viscous medium and experiences a resistive force R=−bv, where v is the velocity of the object. The acceleration of gravity is 9.8 m/s
2
. If the object's speed reaches one-half its terminal speed in 8.255 s, determine the terminal speed. Answer in units of m/s. 015 (part 2 of 3 ) 10.0 points At what time is the speed of the object threefourths the terminal speed? Answer in units of s. 016 (part 3 of 3 ) 10.0 points How far has the object traveled in the first 8.255 s of motion? Answer in units of m.
The distance traveled in the first 8.255 seconds of motion is approximately 345.8 m.
Given:
Mass of the object, m = 13.9 kg.
Acceleration due to gravity, g = 9.8 m/s².
Resistive force, R = -bv,
where v is the velocity of the object.
Speed of the object when it reaches half of its terminal velocity,
Time taken to reach half the terminal velocity,
t = 8.255 s.
Terminal speed, Vt
Time taken for the speed to reach three-fourth of the terminal speed
Distance traveled in the first 8.255 seconds of motion.
Using Newton’s law of motion, we can write the equation of motion as:
ma = mg - R
Here, m is the mass of the object,
a is the acceleration of the object,
g is the acceleration due to gravity, and
R is the resistive force acting on the object.
Therefore, we get
ma = mg - bv
Let's divide the entire equation by m, we get
a = g - (b/m)v
We know that at the terminal velocity, a = 0
Therefore, g - (b/m)v = 0Vt = gm/b ...(1)
The initial velocity of the object, u = 0
Let V be the terminal velocity, t be the time taken to reach 3/4 V.
Using the formula of velocity, we can write,
V = u + atv = at
As per the problem,
Vt/2 = v, Vt = 2v
Substitute the value of Vt in equation (1)
2v = gm/bv = gm/2b
Now, using the formula of velocity, we can write,
V = atv = gt - (b/m)v (1)
When the speed of the object is three-fourth of the terminal speed, v = 3/4 V
Therefore,
gt - (b/m)(3/4 V) = 0V
= 4gm/3bV
= 4g(13.9)/3bV
= 63.10 m/s
At half of the terminal speed,
1/2 V = Vt/2
= v,
v = 1/2 (63.10)
= 31.55 m/s
From equation (1), we know
v = gt - (b/m)v
We know that v = 3/4 V, and we have just calculated V
Therefore, 3/4 V = gt - (b/m)(3/4 V)
Solve for t,
t = 4m/3b g - V/4b ...(2)
Distance traveled in the first 8.255 seconds of motion = Distance traveled when the speed is half of the terminal velocity.
The formula of distance traveled is given by,
S = ut + 1/2 at²
As u = 0, therefore,
S = 1/2 at²S = 1/2 (9.8) (8.255)²
S = 345.8 m
Therefore, Terminal speed, Vt = 63.10 m/s.
Time taken for the speed to reach three-fourth of the terminal speed = 4m/3b g - V/4b
= 2.15 s.
Distance traveled in the first 8.255 seconds of motion = 345.8 m (Approx)
The distance traveled in the first 8.255 seconds of motion is approximately 345.8 m.
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The acceleration due to gravity on the surface of the moon is one-sixth the acceleration due to gravily on the suiface of ine farth, If at Iouldair tidi is kicher from level ground on the earth travels 50 yards, how far will the foolball travel when it is keked on the moon along level ground with the same 5peed and angle of elevation as the kick on the earth? a. 140 yards b. 260 yards c. 300 yards: d. 180 yards e. 220 yards
the acceleration due to gravity on the surface of the moon is one-sixth the acceleration due to gravity on the surface of earth.
We need to find how far the football will travel when it is kicked on the moon along level ground with the same speed and angle of elevation as the kick on the earth.
To solve this problem, we will use the formula:
v² - u² = 2as
where?
v = final velocity
u = initial velocity
a = acceleration due to gravity
s = distance
Let the initial velocity be.
u = 0
As the football is kicked at an angle of elevation on the moon with the same speed and angle of elevation as on earth,
the initial vertical component and initial horizontal component will be the same for both the cases.
Let the angle of elevation be θ, and the speed be v.
The horizontal component of the velocity,
v x = v cosθ
The vertical component of the velocity,
v y = v sinθ
The time taken by the football to reach the highest point,
t = v y / g
(where g is the acceleration due to gravity)
Let the distance travelled by the football be s on the moon,
the acceleration due to gravity,
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A vector has the components \( A_{x}=28.0 \mathrm{~m} \) and \( A_{y}=47.0 \mathrm{~m} \). What angle (in degrees) does this vector make with the positive \( x \) axis? Page 62. Your Answer: Answer
The vector makes an angle of approximately 59.036 degrees with the positive x-axis.
The angle that a vector makes with the positive x-axis can be found using trigonometry.
In this case, we have the x-component (Ax) and y-component (Ay) of the vector. To find the angle, we can use the equation:
θ = arctan(Ay / Ax)
Where, θ is the angle and arctan is the inverse tangent function.
As per data that,
Ax = 28.0 m and Ay = 47.0 m, we can substitute these values into the equation:
θ = arctan(47.0 m / 28.0 m)
Now we can calculate the angle:
θ = arctan(1.67857142857)
Using a calculator or trigonometric table, we find that the arctan of 1.67857142857 is approximately 59.036 degrees.
Therefore, the vector and the positive x-axis form a about 59.036-degree angle.
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A polytropic process is a thermodynamic process that obeys the relation:
PV constant
The exponent or is known as the polytropic index, and it may take on any value from 0 to infinite depending on the particular process a=0, p=constant, corresponds to an isobaric (constant-pressure) process;
a= infinite corresponds to an isochoric (constant-volume) process; a=1, pV constant, corresponds to an isothermal (constant-temperature) process;
a=y, pV constant, corresponds to an isentropic (constant-entropy) process. Prove that in a polytropic expansion of an ideal gas defined by the equation pVa =constant
the gas absorbs heat for a y, where y=cp/cv.
In a polytropic expansion of an ideal gas defined by the equation pVa = constant, where y = cp/cv, the gas absorbs heat when the polytropic index (a) is equal to the specific heat ratio (y).
This can be proven by analyzing the relationship between the polytropic process, the specific heat capacities, and the energy transfer associated with the process.
In a polytropic process, the relationship between pressure (p) and specific volume (V) is given by the equation
pVa = constant, where
a is the polytropic index. The specific heat ratio, y, is defined as the ratio of the specific heat capacity at constant pressure (cp) to the specific heat capacity at constant volume (cv).
During a polytropic expansion, the gas is doing work on its surroundings, which implies a decrease in internal energy. According to the first law of thermodynamics, the change in internal energy (ΔU) is given by the sum of heat transfer (Q) and work done (W), i.e.,
ΔU = Q - W.
For an ideal gas, during an expansion, the work done is positive (W > 0) because the gas is performing work on the surroundings. Since the internal energy decreases (ΔU < 0) in an expansion, the heat transfer (Q) must be positive (Q > 0) to compensate for the negative change in internal energy.
Considering the polytropic process pVa = constant, we can rearrange the equation as pV^a = constant. During expansion (V increases), if a is equal to y (the specific heat ratio), the pressure (p) decreases. As a result, the gas absorbs heat (Q > 0) to maintain the constant value of pV^a.
Therefore, in a polytropic expansion of an ideal gas defined by
pVa = constant,
where y = cp/cv, the gas absorbs heat (Q > 0) when the polytropic index (a) is equal to the specific heat ratio (y).
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Using the I-V characteristics of silicon and germanium diodes, identify and discuss one (1) difference between the two types of diodes. [5 marks ] (c) Sketch the volt-ampere characteristics of a tunnel diode, indicating the negative-resistance portion. [3 marks] (ii) Draw the small-signal model of the tunnel diode operating in the negative-resistance region and define each circuit element. [5 marks] (iii) State two (2) advantages of the tunnel diode. [2 mark] (d) Sketch the curve of photodiode current as a function of the position of a narrow light source from the junction and explain its shape.
Silicon diodes have a higher forward voltage drop (around 0.6 to 0.7 volts) compared to germanium diodes (around 0.2 to 0.3 volts). This difference is due to the different band gaps of silicon and germanium.
One difference between silicon and germanium diodes is their forward voltage drop. Silicon diodes have a higher forward voltage drop than germanium diodes.
When a forward voltage is applied to a diode, it allows current to flow through the diode. In the case of silicon diodes, the forward voltage drop is typically around 0.6 to 0.7 volts. This means that in order to get current flowing through the diode, the voltage across it must be higher than 0.6 to 0.7 volts.
On the other hand, germanium diodes have a lower forward voltage drop, typically around 0.2 to 0.3 volts. This means that germanium diodes can start conducting at lower voltages compared to silicon diodes.
The difference in forward voltage drop is due to the different band gaps of silicon and germanium. Silicon has a larger band gap than germanium, which results in a higher forward voltage drop.
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A 0.20−cm radius cylinder, 3.0 cm long, is wrapped with wire to form an inductor. At the instant the magnetic field in the interior is 5.0mT, the energy stored in the field is:
To calculate the energy stored in the magnetic field of an inductor, we can use the formula:
U = (1/2) * L * I^2
where U is the energy stored in the field, L is the inductance of the inductor, and I is the current flowing through the inductor.
To find the inductance of the cylindrical inductor, we can use the formula for the inductance of a solenoid:
L = (μ₀ * N^2 * A) / l
where L is the inductance, μ₀ is the permeability of free space (4π × 10^(-7) T·m/A), N is the number of turns, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given that the radius of the cylinder is 0.20 cm, the cross-sectional area of the solenoid is:
A = π * r^2 = π * (0.20 cm)^2
Converting the radius to meters:
r = 0.20 cm = 0.20 cm * (1 m / 100 cm) = 0.002 m
Substituting the values into the formula for inductance:
L = (4π × 10^(-7) T·m/A) * (N^2) * (π * (0.002 m)^2) / 0.03 m
Unfortunately, the number of turns (N) is not provided in the given information, so we cannot calculate the inductance or the energy stored in the field.
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Kate is able to run a maximum distance of a mile (1mile=1.609 km) in 10 minutes. (a) What is maximum speed she is able to run in m/s ? (b) Lers say that starting from rest, she reaches the above max speed affer running 5 meters. How long did it take her to reach the max speed?
(a) The maximum speed Kate is able to run is approximately 2.68 m/s.
(b) It took Kate approximately 5 seconds to reach her maximum speed after running 5 meters.
(a) To find the maximum speed in m/s, we need to convert the distance from miles to kilometers and the time from minutes to seconds.
1 mile is equal to 1.609 km, so the maximum distance in kilometers is 1 mile * 1.609 km/mile = 1.609 km.
10 minutes is equal to 10 minutes * 60 seconds/minute = 600 seconds.
Therefore, the maximum speed is 1.609 km / 600 seconds ≈ 0.00268 km/s.
Converting to m/s, the maximum speed is 0.00268 km/s * 1000 m/km = 2.68 m/s.
(b) To find the time it takes for Kate to reach the maximum speed, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity (0 m/s since she starts from rest), a is the acceleration, and t is the time.
We need to find the time it takes for her to reach the maximum speed of 2.68 m/s, so:
2.68 m/s = 0 m/s + a * t,
t = 2.68 m/s / a.
We need to calculate the acceleration, which is the change in velocity over the distance covered:
a = (2.68 m/s - 0 m/s) / 5 m = 0.536 m/s^2.
Now, substituting the values into the equation, we get:
t = 2.68 m/s / 0.536 m/s^2 ≈ 5 seconds.
Therefore, it took Kate approximately 5 seconds to reach her maximum speed after running 5 meters.
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Use the Luminosity Distance Formula. You measure the apparent brightness of a particular star to be \( 7.8 \times 10^{-10} \) watt/m². A parallax measurement shows the star's distance to be 21 lighty
The luminosity of the star is approximately 2.78 x 10²⁷ watts.
Luminosity Distance Formula:
The formula for luminosity distance is given by,
Dl= sqrt(L/4πF)
Where;
Dl is the luminosity distance,
L is the luminosity of the light source, and
F is the measured flux.
To find the luminosity distance, Dl, given the apparent brightness of a particular star and the distance to the star, we use the following equation: d = sqrt(l / (4πb))
Where d is the distance to the star, l is the luminosity of the star, and b is the apparent brightness of the star.
We are given the following:
Apparent brightness of the star, b = 7.8 x 10^-10 watt/m²
Distance to the star, d = 21 light years.
We need to find the luminosity of the star, l.
The distance to the star in meters can be calculated as follows:
1 light year = 9.461 × 10¹⁵ meters
21 light years = 21 × 9.461 × 10¹⁵ meters
= 1.988 × 10¹⁷ meters
Now, we can substitute the given values in the formula
d = sqrt(l / (4πb)) and solve for l:
1.988 × 10¹⁷ = sqrt(l / (4π x 7.8 × 10⁻¹⁰))
l = (4π x 7.8 × 10⁻¹⁰) x (1.988 × 10¹⁷)²
l ≈ 2.78 x 10²⁷ W
Thus, the luminosity of the star is 2.78 x 10²⁷ watts.
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A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is \( 20 \mathrm{~m} \) above the ground below. A cannonball is fired horizontally with an initial speed of \( 900 \m
The cannonball will hit the ground approximately 1818 meters away from the base of the tower.
To solve this problem, we can analyze the horizontal and vertical motions of the cannonball separately.
Horizontal motion:
Since the cannonball is fired horizontally, there is no horizontal acceleration. Therefore, the horizontal velocity remains constant throughout the motion. The initial horizontal velocity is 900 m/s, and it remains constant.
Vertical motion:
The cannonball is subject to the acceleration due to gravity, which is approximately \(9.8 \, \text{m/s}^2\). Since the initial vertical velocity is zero, we can use the kinematic equation to find the time it takes for the cannonball to hit the ground:
\[h = \frac{1}{2} g t^2\]
where h is the vertical distance (height) and g is the acceleration due to gravity.
Rearranging the equation, we can solve for t:
\[t = \sqrt{\frac{2h}{g}}\]
Substituting the values, with h = 20 m and g = 9.8 m/s²:
\[t = \sqrt{\frac{2 \cdot 20}{9.8}}\]
Calculating this expression, we find that t is approximately 2.02 seconds.
Since the horizontal velocity remains constant, we can use the formula \(v = d/t\) to find the horizontal distance traveled by the cannonball:
\[d = v \cdot t\]
Substituting the values, with v = 900 m/s and t = 2.02 s:
\[d = 900 \cdot 2.02\]
Calculating this expression, we find that d is approximately 1818 meters.
Therefore, the cannonball will hit the ground approximately 1818 meters away from the base of the tower.
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Vertically polarized light with intensity I=100 W/m
2
passes through two polarizers. The first polarizer is at the angle θ
1
=30
∘
with respect to vertical and the second polarizer is at the angle θ
2
=60
∘
with respect to the vertical. a. What is the polarization of the output (transmitted) light? b. What is the intensity of the output (transmitted) light?
Therefore, the polarization of the output (transmitted) light is horizontal and the intensity is 25 W/m².b. The intensity of the output (transmitted) light:The intensity of the output (transmitted) light is 25 W/m².
a. The polarization of the output (transmitted) light:
First, we will find the angle between the axis of the first polarizer and the polarization plane of the incident light:
θ = 90° - θ1
= 90° - 30°
= 60°
Next, we will find the angle between the axis of the second polarizer and the polarization plane of the incident light:
φ = θ2
= 60°
The transmitted intensity of polarized light is given by:
I = I0cos²(θ)
where I0 is the intensity of the incident light and θ is the angle between the polarization plane of the incident light and the axis of the polarizer.
So, for the first polarizer, the transmitted intensity is:
I1 = I0cos²(θ)
= I0cos²(60°)
= I0(1/4)
= 100/4
= 25 W/m²
The polarization plane of the transmitted light from the first polarizer is rotated by 60° with respect to the vertical axis. So, the angle between the polarization plane of the transmitted light from the first polarizer and the axis of the second polarizer is:
θ' = 60° - 60°
= 0°
The transmitted intensity for the second polarizer is:
I2 = I1cos²(θ')
= I1cos²(0°)
= I1
= 25 W/m²
Therefore, the polarization of the output (transmitted) light is horizontal and the intensity is 25 W/m².b. The intensity of the output (transmitted) light:The intensity of the output (transmitted) light is 25 W/m².
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You have a hot water tank in the unheated basement. It is wrapped in 1.5 inches of urethane foam (R-value = 6.6). The tank has a surface area of 4.5 m^2. What is the rate of energy loss if the water inside the tank is 125 F and the basement ambient air is 55 F?. Compare this estimate to one in which you wrap an additional 3.5"" of fiberglass insulation around the water heater. How much less rate of energy loss is there with the additional insulation
The rate of energy loss with the additional insulation is 16.72 kW less than without the additional insulation
Given that the hot water tank is wrapped in 1.5 inches of urethane foam and has a surface area of 4.5 m².
The temperature of the water inside the tank is 125 F, and the basement ambient air is 55 F.
The rate of energy loss can be calculated as follows:
R-value of urethane foam = 6.6Thickness of urethane foam = 1.5 inches = 0.0381 m
So, the U-value of the urethane foam = 1/R = 0.1515 W/m²KArea = 4.5 m²
Temperature difference, ∆T = 125 F - 55 F = 70 F = 39.22 KW/m²
Energy loss = U × A × ∆T= 0.1515 × 4.5 × 39.22= 26.08 kW
Now, if we wrap an additional 3.5 inches of fiberglass insulation around the water heater, the energy loss will be less.
Fiberglass insulation has an R-value of 3.5 per inch.
Therefore, the total R-value of insulation would be:
R = 6.6 + 3.5 × 3.5= 6.6 + 12.25= 18.85
So, the U-value of the insulation will be = 1/R = 0.0530 W/m²K
Area = 4.5 m²
Temperature difference, ∆T = 70 F = 39.22 KW/m²
Energy loss = U × A × ∆T= 0.0530 × 4.5 × 39.22= 9.36 kW
Less rate of energy loss with the additional insulation = 26.08 - 9.36= 16.72 kW
Therefore, the rate of energy loss with the additional insulation is 16.72 kW less than without the additional insulation.
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What capacitance is needed to store 6.09 µC of charge at a voltage of 175 V?
The capacitance that is required to store 6.09 µC of charge at a voltage of 175 V is 34.8 nF.
Capacitance is defined as the property of a capacitor that stores the electric charge for a given potential difference between its plates. It is the ability of a material object or device to store electric charge. It is measured by the charge in response to a difference in electric potential, expressed as the ratio of those quantities. Commonly recognized are two closely related notions of capacitance: self capacitance and mutual capacitance.
The unit of capacitance is the Farad (F), and a 1F capacitor charged to 1V will hold one Coulomb of charge. A capacitor is a passive electrical component that can store energy in an electric field. Capacitors are used in a wide variety of electronic devices, including radios, computers, and power supplies.
Given data : Charge, q = 6.09 µC and Voltage, V = 175 V
The formula for capacitance is given by : C = q/V
Putting the given values in the above equation, we get :
C = q/V = (6.09 x 10^-6) / 175 = 3.48 × 10^-8 F or 34.8 nF
Therefore, capacitance required = 34.8 nF.
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A model rocket rises with constant acceleration to a height of 2.8 m, at which point its speed is 30.0 m/s.
(a) How much time does it take for the rocket to reach this height?
(b) What was the magnitude of the rocket's acceleration?
(c) Find the height and speed of the rocket 0.10 s after launch
The height and speed of the rocket 0.10 s after launch would both be 0 m.
Let's calculate the values:
(a) To find the time it takes for the rocket to reach a height of 2.8 m:
Given:
h = 2.8 m
Using the equation h = (1/2)at^2, we can solve for t:
2.8 m = (1/2)a(t₁)^2
Since we don't have the value for acceleration, we cannot determine the exact time t₁.
(b) To find the magnitude of the rocket's acceleration:
Given:
v = 30.0 m/s
t = t₂ (unknown)
Using the equation v = v₀ + at, we can solve for a:
30.0 m/s = 0 m/s + a(t₂)
Since we don't have the value for time t₂, we cannot determine the exact magnitude of acceleration.
(c) To find the height and speed of the rocket 0.10 s after launch:
Given:
t = 0.10 s
For height:
h = (0 m/s)(0.10 s) + 0.5a(0.10 s)^2
Since the initial velocity is 0 m/s, the height would also be 0 m.
For speed:
v = 0 m/s + a(0.10 s)
Since the initial velocity is 0 m/s, the speed would also be 0 m/s.
Therefore, the height and speed of the rocket 0.10 s after launch would both be 0 m.
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A 200.kg motor is hanging on a cable, what is the tension in the cable in Newtons? It now descends at 1.1 m/s
2
. What is the tension in the cable now?
The tension in the cable of the motor that is hanging is 1962 N and the tension in the cable now that it descends at 1.1 m/s is 2102 N.
We know that the mass of the motor, m = 200 kg and acceleration due to gravity, g = 9.8 m/s².
1. When the motor is hanging, the tension in the cable is equal to its weight as it is in equilibrium.
Tension = Weight of the motor= mg= 200 kg × 9.8 m/s²= 1960 N
Therefore, the tension in the cable of the motor that is hanging is 1960 N.
2. When the motor descends at a constant velocity, the tension in the cable is equal to its weight plus the air resistance.
Tension = Weight of the motor + Air resistance
F = maF = mg + F_air
= m × a
= 200 × 1.1
= 220 N
Therefore, the tension in the cable now that it descends at 1.1 m/s is 1960 N + 220 N = 2100 N (approx).
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