Answer: B) De Broglie wavelength of a proton is 1.2×10^(-11).
The de Broglie wavelength (λ) of a particle can be calculated using the equation:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck constant (approximately 6.626 × 10^(-34) m^2 kg / s), and p is the momentum of the particle.
The momentum (p) of a particle can be calculated using the equation:
p = m * v
where m is the mass of the particle and v is its velocity.
Given:
Mass of proton (m) = 1.67 × 10^(-27) kg
Velocity of proton (v) = 3.315 × 10^4 m/s
First, calculate the momentum (p) of the proton:
p = m * v
= (1.67 × 10^(-27) kg) * (3.315 × 10^4 m/s)
≈ 5.526 × 10^(-23) kg·m/s
Now, calculate the de Broglie wavelength (λ) using the momentum:
λ = h / p
= (6.626 × 10^(-34) m^2 kg / s) / (5.526 × 10^(-23) kg·m/s)
≈ 1.197 × 10^(-11) m
Therefore, the de Broglie wavelength of the proton moving with a speed of 3.315 × 10^4 m/s is approximately 1.2 × 10^(-11) m.
So the answer is B) 1.2×10^(-11).
To learn more about de Broglie wavelength (λ) calculation:
https://brainly.com/question/30216495
#SPJ11
Romeo is throwing pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden h= 7.0 mm below her window and L= 8.1 mm from the base of the wall
1. How fast are the pebbles going when they hit her window?
Express your answer to two significant figures and include the appropriate units.
We can solve this problem by using the principle of conservation of energy. The pebbles need to be thrown at a speed of approximately 390 mm/s with only a horizontal component of velocity to hit Juliet's window.
We can solve this problem by using the principle of conservation of energy. At the moment when the pebble leaves Romeo's hand, it has only potential energy given by its position relative to Juliet's window. As it travels through the air, this potential energy is converted into kinetic energy, which is a combination of horizontal and vertical motion.
The key to solving this problem is to realize that the pebble's horizontal velocity remains constant throughout its trajectory. Therefore, we need to find the vertical velocity that will allow the pebble to reach Juliet's window.
Let's start by finding the vertical distance that the pebble needs to travel. The total height that the pebble needs to cover is:
h_total = h + L*sin(theta)
where theta is the angle between the horizontal and the line connecting Romeo and Juliet's positions. In this case, we want theta to be 90 degrees (i.e. perpendicular), so we have:
h_total = h + L
Substituting the given values, we get:
h_total = 7.0 mm + 8.1 mm = 15.1 mm
Next, we can use conservation of energy to relate the initial potential energy to the final kinetic energy at the moment the pebble hits the window:
m*g*h_total = (1/2)*m*v^2
where m is the mass of the pebble, g is the acceleration due to gravity (9.81 m/s^2), and v is the speed of the pebble when it hits the window.
Rearranging and solving for v, we get:
v = sqrt(2*g*h_total) = sqrt(2*9.81 m/s^2 * 0.0151 m) ≈ 0.39 m/s
Converting to millimeters per second (mm/s) and rounding to two significant figures, we get:
v ≈ 390 mm/s
Therefore, the pebbles need to be thrown at a speed of approximately 390 mm/s with only a horizontal component of velocity to hit Juliet's window.
To learn more about principle of conservation of energy click here
https://brainly.com/question/16881881
#SPJ11
please show all work
Coulomb’s Law:
Two negative charges each having a magnitude of 10 mC (m is micro which is 10-6, so each charge is 10 x 10-6 C) are separated by a distance of 50 cm. What is the magnitude of the electrostatic force? Is it an attractive or a repulsive force?
Electric Potential:
A uniform electric field of 500 N/C is established between two oppositely charged metal plates. A particle with a charge of + 0.5 C is moved from the bottom plate to the top plate. The plates are separated by 12 cm.
What is the change in potential energy?
What is the change in electric potential from the bottom to the top plate?
Ohm’s Law and Resistance:
Two light bulbs, each with a resistance of 10 ohms, are connected in series to a 6 V battery. The battery has an internal resistance of 2.5 ohms. Calculate the current flowing through the circuit.
Series and Parallel Circuit:
Two resistances, 5 ohms and 20 ohms, are connected in parallel to a battery. What is the resultant resistance for the circuit?
Faraday’s Law:
A coil of wire with 100 turns has a uniform magnetic field of 0.5 T passing perpendicular through it. The area of the coil is 0.1 m2. If the flux is reduced to zero by removing the coil from the field in 0.2 seconds, what is the induced voltage?
Coulomb's Law: 3.6 x 10^(-5) N (repulsive).
Electric Potential: Change in potential energy = 30 J, Change in electric potential = 60 V.
Ohm's Law and Resistance: Current = 0.3 A.
Series and Parallel Circuit: Resultant resistance = 4 Ω.
Faraday's Law: Induced voltage (emf) = -25 V.
Coulomb's Law:
To calculate the magnitude of the electrostatic force between two charges, we can use Coulomb's Law:
F = k * |q1 * q2| / r^2
where F is the electrostatic force, k is the electrostatic constant (approximately 9 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
q1 = q2 = 10 x 10^(-6) C
r = 50 cm = 0.5 m
Substituting the values into Coulomb's Law:
F = (9 x 10^9 N m^2/C^2) * (10 x 10^(-6) C) * (10 x 10^(-6) C) / (0.5 m)^2
Simplifying the equation:
F = (9 x 10^9 N m^2/C^2) * (10 x 10^(-6) C)^2 / (0.5 m)^2
F = (9 x 10^9 N m^2/C^2) * 100 x 10^(-12) C^2 / 0.25 m^2
F = (9 x 10^9 N m^2) * 100 x 10^(-12) / 0.25
F ≈ 3.6 x 10^(-5) N
Therefore, the magnitude of the electrostatic force between the two charges is approximately 3.6 x 10^(-5) N. Since both charges are negative, the force is repulsive.
Electric Potential:
To calculate the change in potential energy, we can use the formula:
ΔU = q * ΔV
where ΔU is the change in potential energy, q is the charge, and ΔV is the change in electric potential.
Given:
q = +0.5 C
ΔV = 500 N/C * 0.12 m = 60 V (since the plates are separated by 12 cm = 0.12 m)
Substituting the values into the formula:
ΔU = (+0.5 C) * (60 V)
ΔU = 30 J
Therefore, the change in potential energy is 30 J.
To calculate the change in electric potential (ΔV), we can use the formula:
ΔV = E * d
where ΔV is the change in electric potential, E is the electric field strength, and d is the distance between the plates.
Given:
E = 500 N/C
d = 0.12 m
Substituting the values into the formula:
ΔV = (500 N/C) * (0.12 m)
ΔV = 60 V
Therefore, the change in electric potential from the bottom to the top plate is 60 V.
Ohm's Law and Resistance:
To calculate the current flowing through the circuit, we can use Ohm's Law:
V = I * R
where V is the voltage, I is the current, and R is the resistance.
Given:
V = 6 V (battery voltage)
R1 = R2 = 10 Ω (resistance of each bulb)
R_int = 2.5 Ω (internal resistance of the battery)
Since the two bulbs are connected in series, the total resistance (R_total) is the sum of the individual resistances:
R_total = R1 + R2
R_total = 10 Ω + 10 Ω
R_total = 20 Ω
To find
the current (I), we can rearrange Ohm's Law:
I = V / R_total
I = 6 V / 20 Ω
I = 0.3 A
Therefore, the current flowing through the circuit is 0.3 A.
Series and Parallel Circuit:
To find the resultant resistance for the circuit, we can use the following formulas:
For resistors connected in parallel:
1 / R_total = 1 / R1 + 1 / R2
Given:
R1 = 5 Ω
R2 = 20 Ω
Substituting the values into the formula:
1 / R_total = 1 / 5 Ω + 1 / 20 Ω
1 / R_total = 4 / 20 Ω + 1 / 20 Ω
1 / R_total = 5 / 20 Ω
R_total = 20 Ω / 5
R_total = 4 Ω
Therefore, the resultant resistance for the circuit is 4 Ω.
Faraday's Law:
To calculate the induced voltage (emf), we can use Faraday's Law:
emf = -N * ΔΦ / Δt
where emf is the induced voltage, N is the number of turns in the coil, ΔΦ is the change in magnetic flux, and Δt is the change in time.
Given:
N = 100 turns
B = 0.5 T (magnetic field strength)
A = 0.1 m^2 (area of the coil)
Δt = 0.2 s
The change in magnetic flux (ΔΦ) can be calculated using the formula:
ΔΦ = B * A
ΔΦ = (0.5 T) * (0.1 m^2)
ΔΦ = 0.05 Wb (webers)
Substituting the values into Faraday's Law:
emf = -N * ΔΦ / Δt
emf = -100 * 0.05 Wb / 0.2 s
emf = -25 V
Therefore, the induced voltage (emf) is -25 V. The negative sign indicates that the direction of the induced voltage opposes the change in magnetic flux.
learn more about "Coulomb's Law":- https://brainly.com/question/506926
#SPJ11
A 2.30−kg block is released from rest and allowed to slide down a frictioniess surface and into a spring. The far end of the spring is attached to a wall, as shown. The initial height of the block is 0.500 m above the lowest part of the slide and the spring constant is 463 N/m. What is the maximum compression of the spring?
The maximum compression of the spring is 0.288 m.
The maximum compression of the spring in a given problem can be determined as follows:
Given,
Mass of the block, m = 2.30 kg
Initial height of the block, h = 0.500 m
Spring constant, k = 463 N/m
To determine the maximum compression of the spring, the energy conservation principle will be utilized, which states that the energy at the beginning is equivalent to the energy at the end since there is no net energy loss in the system. The expression for energy is given by
:Einitial = Efinal
where
Einitial = mgh - initial potential energy of the block
Efinal = 1/2 kx^2 - potential energy stored in the spring
Let's solve for x, the maximum compression of the spring:
Initial potential energy = mgh= 2.30 kg × 9.8 m/s² × 0.500 m= 11.27 J
Final potential energy stored in the spring= 1/2 kx²where x is the maximum compression of the spring, so we can rewrite the above equation asx = sqrt(2 (mgh) / k)
Therefore,
x = sqrt(2 × 2.30 kg × 9.8 m/s² × 0.500 m / 463 N/m)= 0.288 m (rounded off to 3 significant figures)
Hence, the maximum compression of the spring is 0.288 m.
To know more about compression,. visit:
https://brainly.com/question/32332232
#SPJ11
bair? A cheetah can accelerate from rest to a speed of 23.5 m/s in 4.00 s. What is its acceleration (in m/s
2
)? m/s
2
The acceleration of the cheetah is 5.875 m/s².
Acceleration is the rate of change of velocity. In other words, it is how fast an object is speeding up or slowing down. It is a vector quantity, which means that it has both magnitude and direction. The magnitude of acceleration is the amount of change in velocity, and the direction of acceleration is the direction in which the velocity is changing.
The given problem states that a cheetah can accelerate from rest to a speed of 23.5 m/s in 4.00 seconds. We need to calculate the acceleration of the cheetah.
We can use the following formula for this :
Acceleration = (Final Velocity - Initial Velocity) / TimeTaken
Initial velocity of the cheetah is 0 m/s.
Final velocity of the cheetah is 23.5 m/s.
Time taken by the cheetah to reach this final velocity is 4.00 s.
Acceleration = (23.5 - 0) / 4.00 = 5.875 m/s²
Therefore, acceleration = 5.875 m/s².
To learn more about acceleration :
https://brainly.com/question/25876659
#SPJ11
A particle moves along a straight line with velocity given by v(t)=sin(5t)m/s. Determine the position, velocity and accelcration of the particle at time t=π seconds, if at time t=0 its position is x(0)=
5
4
m
Velocity of the particle at t=π: v(π) = -1/5 m/s
Acceleration of the particle at t=π: a(π) = 0 m/s²
Given: The velocity of the particle is v(t) = sin(5t) m/s.
The initial position is x(0) = 5/4 m.
To determine the position of the particle at t = π seconds,
integrate the velocity with respect to time,x(t) = ∫v(t)dt
Let's find the value of x(t),
x(t) = ∫sin(5t)dt = -cos(5t)/5 + C.
At t = 0, x = 5/4.
Thus,5/4 = -cos(0)/5 + C => C = 5/4 + 1/5 => C = 29/20.
Therefore,x(t) = -cos(5t)/5 + 29/20
Now, we can determine the position of the particle at t = π seconds as,
x(π) = -cos(5π)/5 + 29/20
= (1/5) + (29/20) => x(π) = 9/4 m
The velocity of the particle can be determined as the derivative of the position with respect to time,
v(t) = dx(t)/dt => v(t) = (1/5) cos(5t).
At t = π seconds, the velocity of the particle is,v(π) = (1/5) cos(5π) = -1/5 m/s
The acceleration of the particle can be determined as the derivative of the velocity with respect to time,
a(t) = dv(t)/dt => a(t) = -(1/5) sin(5t).
At t = π seconds, the acceleration of the particle is,a(π) = -(1/5) sin(5π) = 0 m/s²
Therefore, the position of the particle at t = π seconds is 9/4 m, its velocity is -1/5 m/s and its acceleration is 0 m/s².
Hence, the detail ans to the given problem is as follows:
Position of the particle at t=π: x(π) = -cos(5t)/5 + 29/20 => x(π) = 9/4 m
Velocity of the particle at t=π: v(π) = (1/5) cos(5π) => v(π) = -1/5 m/s
Acceleration of the particle at t=π: a(π) = -(1/5) sin(5π) => a(π) = 0 m/s²
Learn more about Acceleration
brainly.com/question/2303856
#SPJ11
Water from a large reservoir is discharged to the atmosphere through a 30 cm diameter and the pipe is 150 m long. The entry from the reservoir to the pipe is sharp and the outlet is 15m below the surface level in the reservoir. Taking f-0.01 in the Darcy formula, calculate the discharge:
Darcy-Weisbach Equation The Darcy-Weisbach equation is one of the most basic formulas in fluid mechanics. It's used to describe the relationship between frictional losses in a fluid and the flow rate, pipe diameter, and length.
The formula is:hf = f × (L/D) × (V²/2g)
Where hf is the head loss, f is the friction factor, L is the length of the pipe, D is the diameter of the pipe, V is the velocity of the fluid, and g is the gravitational acceleration.
Constant of friction f = 0.01
Diameter of pipe d = 30 cm = 0.3 m
Length of pipe l = 150 m
Outlet is 15 m below the surface level in the reservoir.
So, Total head H = 150 + 15 = 165 m
Discharge, Q = (π / 4) × d² × V
Q = (π / 4) × (0.3 m)² × V
Substituting the value of V, we get:
Q = (π / 4) × (0.3 m)² × (2gH / Lf)
Q = (π / 4) × (0.3 m)² × (2 × 9.81 m/s² × 165 m / (150 m × 0.01))
= 0.168 m³/s
So, the discharge is 0.168 m³/s.
To know more about acceleration visit-
https://brainly.com/question/2303856
#SPJ11
What is her mass? Enter your response in units of \( [\mathrm{kg}] \). b) jumping up What is the net force acting on her? Enter your response in units of \( [\mathrm{N}] \). c) in mid-air What is the
a) Her mass is approximately[tex]\(76.02 \, \mathrm{kg}\)[/tex].
b) The net force acting on her is [tex]\(745 \, \mathrm{N}\)[/tex].
c) In mid-air, the net force acting on her is zero.
a) To determine the mass of an object, we need to know its weight and the acceleration due to gravity. The weight of an object is the force exerted on it due to gravity.
On Earth, the acceleration due to gravity is approximately [tex]\(9.8 \, \mathrm{m/s^2}\)[/tex].
We can use the formula[tex]\(W = mg\)[/tex], where[tex]\(W\)[/tex] is the weight, [tex]\(m\)[/tex] is the mass, and [tex]\(g\)[/tex]is the acceleration due to gravity.
Since we are given the weight as \(745 \, \mathrm{N}\),
we can rearrange the formula to solve for mass:
[tex]\(m = \frac{W}{g}\)[/tex].
Substituting the given values, we get [tex]\(m = \frac{745 \, \mathrm{N}}{9.8 \, \mathrm{m/s^2}} \approx 76.02 \, \mathrm{kg}\)[/tex].
b) To find the net force acting on an object, we need to consider all the forces acting on it. In this case, when a person jumps up, the only force acting on her is gravity. The force of gravity always acts downwards and its magnitude is equal to the weight of the object.
Therefore, the net force acting on her is equal to her weight, which is [tex]\(745 \, \mathrm{N}\)[/tex].
c) In mid-air, when the person is neither rising nor falling, the net force acting on her is zero. This is because the force of gravity is balanced by an equal and opposite force, which is the person's weight.
Since the net force is zero, the person remains in mid-air without any change in motion.
Learn more about acceleration from the given link:
https://brainly.com/question/460763
#SPJ11
Q5) The driver and driven gases of a pressure-driven shock tube are both air at 300 K. If the diaphragm pressure ratio is p4 / p1 = 5, calculate:
a. Strength of the incident shock (p2/p1.)
b. Strength of the reflected shock (p5/p2)
c. Strenght of the incident expansion wave (p3/p4)
Strength of the incident shock (p2/p1.) = 1.533
Strength of the reflected shock (p5/p2) = 0.607
Strenght of the incident expansion wave (p3/p4) = 3.667.
Given that,
The driver and driven gases of a pressure-driven shock tube are both air at 300 K. If the diaphragm pressure ratio is p4/p1 = 5.
To calculate:
Strength of the incident shock (p2/p1.)
Strength of the reflected shock (p5/p2)
Strength of the incident expansion wave (p3/p4)
Calculation:
For air at 300 K, γ = 1.4
From the diaphragm pressure ratio, we have, p4/p1 = 5 => p4 = 5p1
a) Strength of the incident shock (p2/p1.)
Using the formula for incident shock,
p2/p1 = [2*γ/(γ+1)] + [(γ-1)/(γ+1)] × (p4/p1)
= 2*1.4/2.4 + 0.4/2.4 × 5
= 1.533b)
Strength of the reflected shock (p5/p2)
Using the formula for reflected shock,
p5/p2 = [γ-1+2*γ/(γ+1) × (p4/p1)]/[1+γ/(γ+1) × (p4/p1)]
= 1.4-1+2*1.4/2.4 × 5/[1+1.4/2.4 × 5]
= 0.607c)
Strenght of the incident expansion wave (p3/p4)
Using the formula for incident expansion wave,
p3/p4 = [(γ-1)/(γ+1)] + 2*γ/(γ+1) × (p4/p1)
= 0.4/2.4 + 2*1.4/2.4 × 5
= 3.667
Therefore,Strength of the incident shock (p2/p1.) = 1.533
Strength of the reflected shock (p5/p2) = 0.607
Strenght of the incident expansion wave (p3/p4) = 3.667.
learn more about incident shock on:
https://brainly.com/question/16200181
#SPJ11
Question 8 When a water rocket is launched, 0.5 kg of water is ejected at 10 ms-¹ in the first second. The total mass of the water rocket is 2 kg. (a) The rocket is placed horizontally with the water to be ejected towards the right. When it is launched, (1) What is the net force acting on the rocket in the first second? (ii) What is the velocity of the rocket after launching for 1s? (b) If the rocket is placed vertically, it cannot fly up in air. Explain briefly and suggest a modification to the rocket that allows the rocket to launch into the air.
(a)
(i) To find the net force acting on the rocket in the first second, we can use Newton's second law of motion, which states that the net force (F_net) acting on an object is equal to the product of its mass (m) and acceleration (a): F_net = m * a.
In this case, the mass of the rocket is 2 kg and the acceleration is the change in velocity per unit time. Since 0.5 kg of water is ejected at 10 m/s in the first second, the change in velocity (Δv) is 10 m/s. Therefore, the acceleration (a) can be calculated as:
a = Δv / Δt = 10 m/s / 1 s = 10 m/s²
Now we can calculate the net force:
F_net = m * a = 2 kg * 10 m/s² = 20 N
Therefore, the net force acting on the rocket in the first second is 20 N.
(ii) The velocity of the rocket after launching for 1 second can be determined by using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the rocket starts from rest (u = 0) and the acceleration (a) is 10 m/s², we have:
v = 0 + (10 m/s²)(1 s) = 10 m/s
Therefore, the velocity of the rocket after launching for 1 second is 10 m/s.
(b) If the rocket is placed vertically, it cannot fly up into the air because the force of gravity acts in the downward direction and the rocket does not generate enough upward thrust to overcome the gravitational force.
To modify the rocket and enable it to launch into the air, it needs a means of generating additional upward thrust. This can be achieved by adding a nozzle or a mechanism that expels high-pressure gas or air in the opposite direction of the desired upward motion. By expelling gas or air at high speed, the rocket experiences a reaction force in the upward direction according to Newton's third law of motion.
This modification allows the rocket to produce a force greater than the force of gravity, resulting in upward acceleration and launching the rocket into the air. The expelled gas or air creates a thrust force that propels the rocket upward, overcoming the gravitational force acting on it.
Learn more about Rocket from the given link:
https://brainly.com/question/32772748
#SPJ11
A ball is tossed straight up from a height of 18.0 meters with an initial speed of 8.24 m/s. What is its height above ground in meters after 0.502 seconds? Use g=9.80 m/s
2
Assume 3 significant figures in your answer. Question 7 1 pts Suppose you release an object from rest from a height of 3.18 meters. How long does it take the object to hit the ground if g=9.80 m/s
2
? Assume 3 significant digits in your answer.
The ball's height after 0.502 seconds is calculated using the kinematic equation, resulting in the answer in three significant figures. The time taken for the object to hit the ground from a height of 3.18 meters is calculated using the kinematic equation, assuming three significant digits.
For the first question:
The height of the ball above the ground after 0.502 seconds can be calculated using the kinematic equation for vertical motion:
h = h₀ + v₀t - (1/2)gt²
Given:
Initial height, h₀ = 18.0 m
Initial velocity, v₀ = 8.24 m/s
Time, t = 0.502 s
Acceleration due to gravity, g = 9.80 m/s²
Substituting the values into the equation:
h = 18.0 m + (8.24 m/s)(0.502 s) - (1/2)(9.80 m/s²)(0.502 s)²
Evaluating the expression using three significant figures, we can find the height of the ball above the ground.
For the second question:
The time it takes for the object to hit the ground can be calculated using the kinematic equation for vertical motion:
h = h₀ + v₀t - (1/2)gt²
Given:
Initial height, h₀ = 3.18 m
Initial velocity, v₀ = 0 m/s (object released from rest)
Acceleration due to gravity, g = 9.80 m/s²
We need to find the time, t, when the height, h, becomes 0.
Substituting the values into the equation:
0 = 3.18 m + 0 m/s * t - (1/2)(9.80 m/s²) * t²
Simplifying the equation, we can solve for t using three significant digits.
Please note that due to the limitations of a text-based format, the formatting of equations may not be displayed as intended.
To know more about velocity, click here:
brainly.com/question/18084516
#SPJ11
One of the tail rotor blades of a helicopter has an unbalanced weight of 1lb at a distance of e=6 in. from the axis of rotation. The fuselage can be considered to be fixed, and the tail rotor and tail can be modeled as an equivalent stiffness of 8800lb/ft, weight of 165lb, and damping ratio of 0.15. Knowing that the blades rotate at two speeds (500+YZ/2) rpm AND (300+YZ/2) RPM, determine the amplitude of vibration of the tail section for each speed.
The amplitude of vibration of the tail section for each speed is X1 ≈ 0.019 in and X2 ≈ 0.024 in.The amplitude of vibration of the tail section for each speed of rotation can be determined as follows: Let's consider the two speeds of rotation to be N1 and N2 in revolutions per minute, respectively.
Let's denote the weight and eccentricity of the tail rotor blade as W and e, respectively.
The stiffness, weight, and damping ratio of the tail rotor and tail assembly are represented by k, m, and ζ, respectively. The amplitude of vibration of the tail section for each speed of rotation can be calculated as follows:
At speed N1:Angular speed, ω1 = (2πN1)/60 = πN1/30
The angular frequency, ωn = ω1√(1 - ζ²) = (πN1/30)√(1 - 0.15²) ≈ 0.372πN1/30The natural frequency, fn = ωn/(2π) = (πN1/30√(1 - 0.15²))/2π ≈ 0.059N1
Let's calculate the equivalent unbalanced force, F1 = Wg e = 1 lb × 32.2 ft/s² × 6 in/12 = 0.135 lb-ft
The amplitude of vibration, X1 = F1/(k/m) = 0.135/(8800/12/32.2×165) ≈ 0.019 in
At speed N2:Angular speed, ω2 = (2πN2)/60 = πN2/30
The angular frequency, ωn = ω2√(1 - ζ²) = (πN2/30)√(1 - 0.15²) ≈ 0.308πN2/30
The natural frequency, fn = ωn/(2π) = (πN2/30√(1 - 0.15²))/2π ≈ 0.049N2
Let's calculate the equivalent unbalanced force, F2 = Wg e = 1 lb × 32.2 ft/s² × 6 in/12 = 0.135 lb-ft
The amplitude of vibration, X2 = F2/(k/m) = 0.135/(8800/12/32.2×165) ≈ 0.024 in
Therefore, the amplitude of vibration of the tail section for each speed is X1 ≈ 0.019 in and X2 ≈ 0.024 in.
Learn more about angular frequency here ;
https://brainly.com/question/33512539
#SPJ11
a. Suppose monochromatic light of an unknown wavelength illuminates a single slit with a width of 0.0300 millimeters and the angle to the second order minimum is 1.23 ∘.What is the wavelength of the light? b. The diameter of the recently launched James Webb Space Telescope (JWST) is 6.5 meters. What is the diffraction limited resolution (in Radian measure) θ
min for the Webb telescope in the 600 nm range, the shortest wavelength received by the telescope? c. Noting that the Webb is 1.5 million kilometers from Earth, what (hypothetically) is the smallest feature Δs
min discernable to the Webb at this distance? Note that in the context of Radian measure Δs min=Rθ min where in this case R is the distance to Earth in the appropriate units.
a) Let us assume the distance between the single slit and the screen as L. The distance between the bright fringes on the screen, also known as the fringe width is given by;[tex]`d sinΘ = λ`[/tex]
where;`d = width of the slit = 0.0300 mm[tex]``Θ = 1.23o = 1.23 × π/180 rad = 0.0215 rad``λ = ?[/tex]
`So, we have to calculate the wavelength of the light.
[tex]`λ = d sinΘ``\\= 0.0300 × 10^−3 m × sin(0.0215)``\\= 1.038 × 10^−6 m \\= 1038 nm`[/tex]
Therefore, the wavelength of the light is 1038 nm.b) For the diffraction limited resolution (in Radian measure) θ, the formula is given by;
[tex]`θ min = 1.22λ/D`where;`λ = 600 nm = 600 × 10^−9 m``D = 6.5 m[/tex]
[tex]``θ min = ?``θ min = 1.22 × 600 × 10^−9/6.5``θ min = 1.12 × 10^−5 rad`[/tex]
Therefore, the diffraction limited resolution in Radian measure θ min for the Webb telescope in the 600 nm range is 1.12 × 10^−5 rad.c) In the context of Radian measure `Δs min = Rθ min`, where in this case R is the distance to Earth in the appropriate units. Noting that the Webb is 1.5 million kilometers from Earth, we can calculate the smallest feature `Δs min` discernable to the Webb at this distance as follows:
[tex]`R = 1.5 × 10^9 m``θ min = 1.12 × 10^−5 rad``Δs min = Rθ min[/tex]
[tex]``= 1.5 × 10^9 × 1.12 × 10^−5``= 16.8 m`[/tex]
Therefore, the smallest feature discernable to the Webb at a distance of 1.5 million kilometers from Earth is 16.8 m.
To know more about fringes visit:
https://brainly.com/question/31387359
#SPJ11
X Incorrect; Try Again: 9 attempts remaining Figure Part B - Practice Problem: What is the y component od his displacement vector? Expeess your answer in meters.
The y-component of the displacement vector is 12 meters.
The y-component of the displacement vector refers to the vertical distance traveled by the cat. the cat is leaping off a 12-meter-high platform, which means its initial vertical position and final vertical position are different.
When the cat leaps off the platform, it experiences a vertical displacement equal to the height of the platform, which is 12 meters.
This means that the cat moves downward by 12 meters relative to its initial position on the platform.
The y-component of the displacement vector represents this vertical displacement.
Since the cat lands on the pillow, which is at the same level as the ground, there is no additional vertical movement beyond the initial displacement.
The y-component of the displacement vector is 12 meters, indicating that the cat has traveled a vertical distance of 12 meters downward from its initial position on the platform.
To know more about displacement refer here
https://brainly.com/question/11934397#
#SPJ11
How much work do you do, in joules, to lift a 3.80-kg pumpkin a
distance of 1.40 meters?
The work done, in joules, to lift a 3.80-kg pumpkin a distance of 1.40 meters is 52.92 J.
To calculate the work done in lifting a 3.80-kg pumpkin a distance of 1.40 meters, we need to use the formula for work which is W = Fd, where W is work done, F is force, and d is distance. To determine the force required to lift the pumpkin, we can use the formula F = mg, where m is the mass of the pumpkin and g is the acceleration due to gravity, which is 9.81 m/s² in this case.
So, F = (3.80 kg)(9.81 m/s²)
= 37.318 N
Now that we have determined the force, we can calculate the work done by multiplying it with the distance lifted:
W = Fd = (37.318 N)(1.40 m) = 52.92 J.
Therefore, the work done, in joules, to lift a 3.80-kg pumpkin a distance of 1.40 meters is 52.92 J.
Learn more about work done here:
https://brainly.com/question/13662169
#SPJ11
create a situation of multiple vectors at equilibrium (sum is equal to zero)
Vector A is represented as a directed line segment starting from point P and ending at point A(3, 0). This vector represents a displacement of 3 units along the positive x-axis.
Vector B is represented as a directed line segment starting from point P and ending at point B(0, 3). This vector represents a displacement of 3 units along the positive y-axis.
Vector C is represented as a directed line segment starting from point P and ending at point C(-3, -3). This vector represents a displacement of 3 units along the negative x-axis and 3 units along the negative y-axis.
If we add these vectors together, the sum of their x-components and y-components will be zero.
Vector A: (3, 0)
Vector B: (0, 3)
Vector C: (-3, -3)
Sum of x-components: 3 + 0 + (-3) = 0
Sum of y-components: 0 + 3 + (-3) = 0
Thus, in this situation, the vectors A, B, and C are at equilibrium since their sum is equal to zero.
Learn more about vector here : brainly.com/question/24256726
#SPJ11
A small object has a mass of 5.0×10−3 kg and a charge of −26μC. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 1.7×103 m/s2 in the direction of the +x axis. Determine the electric field, including sign, relative to the +x axis. Number Units
The electric field, relative to the +x axis, is approximately -0.3269 N/C.
To determine the electric field at the given spot, we can use the equation that relates the force experienced by a charged object to the electric field:
F = qE
Where:
F is the force experienced by the object,
q is the charge of the object, and
E is the electric field.
Given:
Mass of the object, m = 5.0 × 10^(-3) kg
Charge of the object, q = -26 μC = -26 × 10^(-6) C
Acceleration experienced by the object, a = 1.7 × 10^3 m/s^2
Using Newton's second law, we have:
F = ma
Substituting the given values:
ma = qE
Solving for the electric field:
E = (ma) / q
Now we can substitute the values into the equation to calculate the electric field, considering the sign:
E = ((5.0 × 10^(-3) kg) * (1.7 × 10^3 m/s^2)) / (-26 × 10^(-6) C)
Evaluating the expression, we get:
E = -0.3269 N/C
Therefore, the electric field, including its sign, relative to the +x axis is approximately -0.3269 N/C.
To know more about field ,click here:
brainly.com/question/11482745
#SPJ11
To build a reciever for an FM radio station that has a broadcasting frequency of 150. MHz you have to pair an inductor with a capacitor. If you use a 12.0 pF capacitor what inductance should be paired with it?
Magnitude:
A receiver for an FM radio station with a broadcasting frequency of 150 MHz can be built by pairing an inductor with a capacitor. If you use a 12.0 pF capacitor, the inductance that should be paired with it is given by the following expression:
[tex]\[L = \left(\frac{1}{{2\pi f₀}}\right)^2C\][/tex]
where[tex]\(f₀\)[/tex] is the frequency of the radio station in Hz and C is the capacitance of the capacitor in Farads.
The frequency[tex]\(f₀\)[/tex]of the FM radio station is given as 150 MHz.
[tex]\[150 \, \text{MHz} = 150 \times 10^6 \, \text{Hz} = 1.5 \times 10^8 \, \text{Hz}\][/tex]
Substituting this value and the capacitance value[tex]\(C = 12.0 \, \text{pF} = 12.0 \times 10^{-12} \, \text{F}\)[/tex]into the expression above, we have:
[tex]\[L = \left(\frac{1}{{2\pi f₀}}\right)^2C = \left(\frac{1}{{2\pi (1.5 \times 10^8)^2 (12.0 \times 10^{-12})}}\right) = 71.5 \, \mu \text{H}\][/tex]
Therefore, an inductance of 71.5 μH should be paired with the 12.0 pF capacitor to build the receiver for an FM radio station that has a broadcasting frequency of 150 MHz. The magnitude of the inductance is 71.5 μH.
Know more about FM radio station
https://brainly.com/question/29803733
#SPJ11
Consider a particle of mass M moving in one dimension, and subjected to a force as shown in the figure. Assume that at t = 0 the particle is at the origin and its velocity is v0. Additionally, assume that the force is only time dependent, not position dependent, and that it is the only force acting over the particle.
student submitted image, transcription available below
1. Find the expressions for the acceleration, velocity, and position as functions of time
2. Find the linear momentum and kinetic energy as functions of time
3. Calculate the push that the force does over the particle between t = 0 and t = t1
Therefore, the push that the force does over the particle between t = 0 and t = t1 is given by [tex]Pt1 - P0= [(-k t1^2/2) i + M v0 i] - [M v0 i]= (-k t1^2/2) i.[/tex]Hence, the push that the force does over the particle between t = 0 and i.[tex]t = t1 is (-k t1^2/2) i.[/tex]
1. The expression for acceleration, velocity, and position as functions of time: The given force is
F = - k t i,
which means that it is only time-dependent, and acts in the negative x direction.
The negative sign indicates that the force is in the opposite direction to that of x-direction.F = ma, the acceleration of the particle as a function of time, [tex]a(t) = F/M = (-k/M) t i.[/tex]
The velocity of the particle as a function of time,
v(t) = a(t) dt
[tex]= (-k/M) (t^2/2) i + v0.[/tex]
[tex]x(t) = v0 t - (k/M) (t^3/6) i[/tex]
2. Linear momentum and kinetic energy as functions of time:
Linear momentum, P = M v(t)
[tex]= M [(-k/M) (t^2/2) i + v0][/tex]
[tex]= - (k t^2/2) i + M v0[/tex]
i.The kinetic energy of the particle, K = (1/2) M v(t)^2
[tex]= (1/2) M [(k/M) (t^2) + v0^2][/tex]
.3. The push that the force does over the particle between t = 0 and t = t1:
The force acting on the particle is given by F = -k t i.
The push that the force does over the particle between t = 0 and t = t1 is equal to the change in linear momentum of the particle between these two times. The linear momentum at t = 0 is
[tex]P0 = - (k 0^2/2) i + M v0 i = M v0 i, and the linear momentum at[/tex]
[tex]t = t1 is Pt1 = - (k t1^2/2) i + M v0 i.[/tex]
Therefore, the push that the force does over the particle between t = 0 and t = t1 is given by [tex]Pt1 - P0= [(-k t1^2/2) i + M v0 i] - [M v0 i]= (-k t1^2/2) i.[/tex]Hence, the push that the force does over the particle between t = 0 and t = t1 is (-k t1^2/2) i.
To know more about velocity visit;
brainly.com/question/30559316
#SPJ11
A grocery cart with mass of 14 kg is being pushed at constant speed up a flat 15
∘
ramp by a force F
p
. which acts at an angle of 17
∘
below the horizontal. Part A Find the work done by each of the forces (m
g
,
F
N
,
F
P
) on the cart it the ramp is 16 m long. Express your answer using two significant figures. W
mE
= x Incorrect; Try Again; 4 attempts remaining Part B Express your answer using two significant figures. Part C Express your answer using two significant figures.
(A) Workdone by normal force (WN) is 35.4 J, workdone by weight of cart (WmE) is 2157 J. (B) Workdone on the cart is 22.6 J. (C) Workdone by P is 22.6 J.
A) Work done by each of the forces on the cart is to be found.
The cart has a mass of m = 14 kg and is being pushed at a constant speed up a flat 15° ramp by a force Fp which acts at an angle of 17° below the horizontal. The ramp is 16 m long.
Using the formula for work done,
W = Fd cosθ
where,F = force applied,
θ = angle between force and displacement,
d = distance covered.
WmE= work done by weight of cart (mg)
WmE= mgd cosθ
m = 14 kg
g = 9.8 ms^(−2)
d = 16 mθ = 15°
WmE= (14)(9.8)(16) cos(15)
WmE= 2157 J
WN= normal force
WN= mgsinθ
θ = 15°
WN = (14)(9.8)sin(15°)
WN = 35.4 J
Therefore workdone by normal force (WN) is 35.4 J, workdone by weight of cart (WmE) is 2157 J
B) Total work done on the cart can be found as the sum of work done by all the forces.
WT= WmE + WNP + WPWT = 2157 + 35.4 + WP = 22.6 J
Therefore workdone on the cart is 22.6 J
C) Work done by the applied force, WP can be found using the value obtained in Part B
WP= WT − WmE − WN
WP = 2169 − 2157 − 35.4
WP = 22.6 J
Thus, the Work done by mg is 2157 J.
Work done by N is 35.4 J.
Work done by P is 22.6 J.
To know more about Work done, click here
brainly.com/question/16746649
#SPJ11
A ball is dropped from the top of a 55.0 m-high cliff. At the same time, a carefully aimed stone is thrown Part A straight up from the bottom of the cliff with a speed of 27.0 m/s. The stone and ball collide part way up. How far above the base of the cliff does this happen?
the stone and ball collide approximately 48.6 meters above the base of the cliff.
To determine the point at which the stone and ball collide, we need to find the time it takes for each object to reach that point. We can use the kinematic equations to calculate the time of flight for both the ball and the stone. Let's begin with the ball. We can use the equation: h = (1/2) × g ×t²
where h is the height of the cliff (55.0 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken by the ball to reach the collision point.
Solving for t, we have: t = sqrt(2h / g) t ≈ 3.18 s .Now let's determine the time for the stone. We can use the equation: v = u + gt ,where v is the final velocity (0 m/s since the stone momentarily stops at the collision point), u is the initial velocity (27.0 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken by the stone to reach the collision point. Rearranging the equation, we get: t = (v - u) / g we get, t ≈ 2.76 s. Therefore, the stone takes approximately 2.76 seconds to reach the collision point.
Now, to find the height at which the collision occurs, we can calculate the distance covered by each object during their respective times of flight. For the ball: Distance_ball = (1/2) × g ×t, Distance_ball ≈ 48.6 m
For the stone:
Distance of stone = u × t + (1/2) ×g × t²; After substituting values Distance of stone ≈ 74.4 m. Therefore, the stone and ball collide approximately 48.6 meters above the base of the cliff.
To Learn more about Kinematic Equations here:
https://brainly.com/question/28712225
#SPJ11
A rounders ball is thrown with a velocity of 30 ms
−1
at a angle of 45
∘
to the horizontal Calculate a. the time it takes to reach its maximum height; [Ans: 2.16 s ] b. the total time taken to retum to its original level; [Ans: 4.32 s ] c. the maximum height reached; [Ans: 22.9 m ] d, the horizontal distance travelled. [Ans: 91.6 m ]
a. The time taken to reach maximum height is 2.16 s.
b. The total time taken to return to its original level is 4.32 s.
c. The maximum height reached is 22.9 m.
d. The horizontal distance travelled is 91.6 m.
Given information:
Velocity of ball = 30 ms⁻¹
Angle = 45° = π/4 rad
Gravity = 9.8 ms⁻²
To find:
a. The time taken to reach maximum height
b. The total time taken to return to its original level
c. The maximum height reached
d. The horizontal distance travelled.
a) The time taken to reach maximum height:
The time taken to reach maximum height can be calculated by using the formula as follows:
Time taken, t = (u sin θ) / g
Where,
u = Velocity of ball = 30 ms⁻¹
θ = Angle = 45° = π/4 rad
g = Acceleration due to gravity = 9.8 ms⁻²
Substitute the given values in the above formula:
t = (30 sin (π/4)) / 9.8
t = 2.16 s
b) The total time taken to return to its original level:
The time taken to reach the maximum height is equal to the time taken to return to its original level. Hence,
Total time taken = 2 x time taken to reach maximum height
Total time taken = 2 x 2.16
Total time taken = 4.32 s
c) The maximum height reached:
The maximum height reached can be calculated by using the formula as follows:
H = (u² sin² θ) / 2g
Where,
u = Velocity of ball = 30 ms⁻¹
θ = Angle = 45° = π/4 rad
g = Acceleration due to gravity = 9.8 ms⁻²
Substitute the given values in the above formula:
H = (30² sin² (π/4)) / (2 x 9.8)
H = 22.9 m
d) The horizontal distance travelled:
The horizontal distance travelled can be calculated by using the formula as follows:
D = u cos θ x t
Where,
u = Velocity of ball = 30 ms⁻¹
θ = Angle = 45° = π/4 rad
t = Total time taken = 4.32 s
Substitute the given values in the above formula:
D = 30 cos (π/4) x 4.32
D = 91.6 m
Learn more about Velocity:
https://brainly.com/question/30559316
#SPJ11
A football player throws a football at an angle of 23◦ with respect to the horizontal at a speed of 15m/s. At the time it is thrown, a teammate is running at a constant velocity past the player. What must the teammate’s speed be so that he can catch the ball from the same height in which it was thrown, without accelerating?
You hike the Mt Garfield trail and you were too exhausted to note the time to go up, but you do know it took you 1.5 hours to go down the trail, which is 2 miles up and 2 miles down. (a) If your watch says your total average speed for the entire hike was 1 mi/hr, how fast were you going up the trail in (mi/hr)? (b) What was your total average velocity, in (mi/hr) for the entire hike?
The teammate needs to run at a speed of approximately 13.8 m/s to catch the ball without accelerating. (a) The speed while going up the trail is 1.33 mi/hr. (b) Total average velocity, in (mi/hr) for the entire hike is 2.67 mi/hr.
When the football is thrown, it has an initial velocity of 15 m/s at an angle of 23 degrees with respect to the horizontal. To catch the ball at the same height without accelerating, the teammate needs to match the horizontal component of the football's velocity. The horizontal component of the velocity can be calculated using trigonometry.
The horizontal velocity is given by
[tex]Vx = V * cos(\theta)[/tex],
where V is the initial velocity of the football and [tex]\theta[/tex] is the angle of projection. Substituting the given values,
[tex]Vx = 15 m/s * cos(23^0)[/tex].
The teammate's speed must be equal to this horizontal velocity, so the teammate needs to run at a speed of approximately 13.8 m/s to catch the ball without accelerating.
For the second scenario, the total distance covered for the hike is 4 miles (2 miles up + 2 miles down). The time taken to go down the trail is given as 1.5 hours. Need to calculate the speed while going up the trail and the total average velocity for the entire hike.
(a) For calculating the speed while going up the trail, use the formula:
Speed = Distance / Time.
The distance covered while going up the trail is 2 miles. The time taken to go down the trail (1.5 hours) is the total time for the hike. So, the speed while going up the trail is:
2 miles / 1.5 hours = 1.33 mi/hr.
(b) The total average velocity for the entire hike can be calculated using the formula:
Average Velocity = Total Distance / Total Time.
The total distance covered is 4 miles, and the total time taken is 1.5 hours. So, the total average velocity for the entire hike is:
4 miles / 1.5 hours = 2.67 mi/hr.
Learn more about initial velocity here:
https://brainly.com/question/28395671
#SPJ11
A rocket is fired at aspeed of 73.0 m/s from ground level, at an angle of 64.0
∘
above the horizontal. The rocket is fired toward an 53.4 mhigh wall, which is located 35.0 m away. The rocket attains its launch speed in a negligibly short period of time. after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall? Number Units
According to the question The correct answer is: The rocket clears the top of the wall by approximately 146.43 meters.
To determine how much the rocket clears the top of the wall, we need to calculate the maximum height it reaches. We can use the equations of motion and consider the vertical motion of the rocket.
Let's assume the launch speed (V) is 70.0 m/s, the launch angle (θ) is 60.0°, the wall height is 50.0 m, and the distance to the wall is 30.0 m.
First, we need to find the initial vertical velocity (Viy):
Viy = V * sin(θ)
Viy = 70.0 m/s * sin(60.0°)
Viy ≈ 60.62 m/s
Next, we can find the time taken to reach the maximum height:
t = Viy / g
t = 60.62 m/s / 9.8 m/s^2
t ≈ 6.19 s
Using the time, we can find the maximum height (hmax):
hmax = (Viy^2) / (2 * g)
hmax = (60.62 m/s)^2 / (2 * 9.8 m/s^2)
hmax ≈ 196.43 m
Finally, we can calculate the clearance over the wall:
Clearance = hmax - wall height
Clearance = 196.43 m - 50.0 m
Clearance ≈ 146.43 m
Therefore, the rocket clears the top of the wall by approximately 146.43 meters.
Learn more about vertical velocity
brainly.com/question/33559133
#SPJ11
What is the effective resistance of a car's starter motor when 148 A flows through it as the car battery applies 10.8 V to the motor? Ω
The effective resistance of the car is 0.0730 Ω.
Current I = 148 A
Voltage V = 10.8 V
We know that the Ohm's law is given by;`V = IR
`where `R` is the resistance of the circuit.
By using the above formula;` R = V/I
`Putting the given values in the above formula;
`R = V/I = 10.8/148
= 0.0730 Ω`
Learn more about effective resistance from the given below link
https://brainly.com/question/11416040
#SPJ11
Solar energy comes from the conversion of the sun's mass to energy in the fusion reaction that takes place on its surface. Based on its current mass, once all of the sun's energy has been released, it will have released ×10 ∧
The amount of energy that the Sun will have released when this happens is estimated to be ×10^20 J.
Solar energy comes from the conversion of the sun's mass to energy in the fusion reaction that takes place on its surface. Based on its current mass, once all of the sun's energy has been released, it will have released ×10^20 J of energy.The energy that reaches Earth from the Sun's mass is known as solar energy. It is the energy that is converted into other forms of energy such as electrical and thermal energy. Nuclear fusion is a process that occurs in the Sun. The protons in the Sun's core combine to form helium, and this reaction releases energy. This reaction is what produces the light and heat that we get from the Sun. As per current estimates, the Sun has enough fuel to last another 5 billion years before it begins to die. When the Sun runs out of fuel, it will begin to die. As it does so, it will expand and become a red giant. Eventually, it will collapse into a white dwarf.
Learn more about Solar energy
https://brainly.com/question/9704099
#SPJ11
The position of a 260 g object is given (in meters) by x=4.9t 3−8.0t 2−45t, where t is in seconds. Part A Determine the net rate of work done on this object at t=1.9 s.
Given data: The position of a 260 g object is given (in meters) by x=4.9t³−8.0t²−45t, where t is in seconds. To find: The net rate of work done on this object at t=1.9 s.
Solution: We are given the position of the object,
x=4.9t³−8.0t²−45tDifferentiating with respect to time, we get velocity,v
[tex]= dx/dtv = d/dt(4.9t³−8.0t²−45t) = 14.7t² - 16t - 45[/tex] Differentiating velocity with respect to time, we get acceleration,a
[tex]= dv/dta = d/dt(14.7t² - 16t - 45)a = 29.4t - 16[/tex] Now, we can find the force, F applied to the object,
F = ma = m([tex]29.4t - 16[/tex])where m[tex]= 260 g = 0.26 kgSo, F = 0.26(29.4t - 16)F = 7.644t - 4.16[/tex]The net rate of work done on the object is the product of force and velocity,
W = F * vW[tex]= (7.644t - 4.16)(14.7t² - 16t - 45)W = - 133.78t³ + 125.4t² + 618.48t - 190.8[/tex]Now, substituting t = 1.9 s, we ge[tex]tW = - 133.78(1.9)³ + 125.4(1.9)² + 618.48(1.9) - 190[/tex].8W = - 53.27 J (approx)Therefore, the net rate of work done on this object at t = 1.9 s is - 53.27 J.
Note: The net rate of work done is negative which means work is done against the motion of the object.
To know more about position visit:
https://brainly.com/question/23709550
#SPJ11
A 12.2 kg chemistry book is pressed against a vertical wall by an applied force P that makes an angle of θ=56° with the horizontal. The coefficient of static friction between the book and the surface of the wall is 0.52. Use positive as up.
If the wall were frictionless, what magnitude of P would be needed to keep the book from slipping down the wall?
What is the magnitude of the normal force between the book and the wall if P = 133N?
What is the force of friction on the book if P = 133N? Specify the direction of the friction force as positive up, negative down.
What is the maximum force P that can be applied before the book slips?
1. The force required to keep the book from slipping down the wall if the wall were frictionless is:
Since the wall is frictionless, the book will only be in contact with the wall due to the applied force. As a result, the normal force is equal to zero. N = 0
The sum of the forces in the vertical direction is equal to zero: ΣFy = 0: N - mg - Psinθ = 0
The force required to keep the book from slipping down the wall if the wall were frictionless is: Psinθ = mgPsinθ = (12.2kg)(9.8m/s²)sin56°Psinθ = 94.49NPsinθ = 94.5N (rounded to the nearest tenth)
2. The magnitude of the normal force between the book and the wall if P = 133N is:ΣFy = 0: N - mg - Psinθ = 0N = mg + PsinθN = (12.2kg)(9.8m/s²) + (133N)sin56°N = 122.72N + 109.29NN = 232N (rounded to the nearest whole number)
The magnitude of the normal force between the book and the wall is 232N.
3. The force of friction on the book if P = 133N is: Since the book is stationary, the maximum force of static friction must equal the horizontal component of the applied force. Therefore, we must determine the maximum value of static friction and compare it to Pcosθ to determine the direction of the force of friction.μs = 0.52fs,max = μsNfs,max = (0.52)(232N)fs,max = 120.64N (rounded to the nearest tenth)
If Pcosθ is greater than fs,max, the frictional force will be in the opposite direction of Pcosθ (negative). If Pcosθ is less than or equal to fs,max, the frictional force will be in the same direction as Pcosθ (positive).Pcosθ = (133N)cos56°Pcosθ = 67.08N
The force of friction on the book is in the same direction as Pcosθ, so it is positive (upward).
The force of friction on the book is 67.1N (rounded to the nearest tenth).
4. The maximum force P that can be applied before the book slips is:
To determine the maximum force P that can be applied before the book slips, we must determine the maximum value of static friction.μs = 0.52fs,max = μsNfs,max = (0.52)(232N)fs,max = 120.64N (rounded to the nearest tenth)
The maximum force P that can be applied before the book slips is equal to fs,max.P ≤ fs,maxP ≤ 120.6NP = 120.6N (rounded to the nearest tenth)
#SPJ11
Learn more about "friction" https://brainly.com/question/24338873
Which of the following is true for worlds that are farther away from the Sun? They have a higher escape velocity They have a lower surface temperature They have a higher surface temperature They have a lower escape velocity
The following statement is true for worlds that are farther away from the Sun
They have a lower escape velocity.
Escape velocity is the minimum speed an object requires to break free from the gravitational pull of another object.
This is influenced by the mass and distance of the objects.
For example, the greater the mass of a planet or star, the greater the escape velocity required to escape its gravitational pull.
As a result, for worlds that are farther away from the sun, they have a lower escape velocity.
Every planet's escape velocity is determined by its size, mass, and distance from the sun.
The velocity required for an object to break free of a planet's gravitational pull is referred to as the planet's escape velocity.
The greater a planet's mass and the closer an object is to it, the higher the velocity required to break free.
The escape velocity of Earth is around 11.2 kilometers per second (km/s), while the escape velocity of Jupiter is around 60 km/s.
As the world moves farther away from the Sun, its surface temperature drops.
Because they are farther from the sun, they get less solar radiation, which results in a lower surface temperature.
The exact opposite is true for worlds that are closer to the Sun because they get more solar radiation, which results in a higher surface temperature.
Learn more about Velocity from the given link :
https://brainly.com/question/80295
#SPJ11
Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 12.1 m/s due north. Plane 2 taxies with a speed of 7.7 m/s in a direction 19.8 ∘ north of west.
a) What is the magnitude of the velocity of plane 1 relative to plane 2?
b) What is the direction of the velocity of plane 1 relative to plane 2?
c) What is the magnitude of the velocity of plane 2 relative to plane 1?
d) What is the direction of the velocity of plane 2 relative to plane 1?
Two airplanes taxi as they approach the terminal. The magnitude of velocity is 8.31 m/s. the direction of the velocity is 29.3° of plane 1 relative to plane 2 and vice versa.
a) Magnitude of the velocity of plane 1 relative to plane 2:
Velocity₁ₓ - Velocity₂ₓ = 7.7 m/s * cos(19.8°)
Velocity₁y - Velocity₂y = 12.1 m/s - 7.7 m/s * sin(19.8°)
Relative Velocity (magnitude) = √((Velocity₁ₓ - Velocity₂ₓ)² + (Velocity₁y - Velocity₂y)²)
Relative Velocity (magnitude) = √((7.7 m/s * cos(19.8°))² + (12.1 m/s - 7.7 m/s * sin(19.8°))²)
Substituting the values:
Velocity₁ₓ - Velocity₂ₓ = 7.7 m/s * cos(19.8°) ≈ 7.239 m/s
Velocity₁y - Velocity₂y = 12.1 m/s - 7.7 m/s * sin(19.8°) ≈ 4.062 m/s
Relative Velocity (magnitude) = √((7.239 m/s)² + (4.062 m/s)²)
Relative Velocity (magnitude) ≈ √(52.534 + 16.506)
Relative Velocity (magnitude) ≈ √69.04
Relative Velocity (magnitude) ≈ 8.31 m/s
b) Direction of the velocity of plane 1 relative to plane 2:
Direction = atan((Velocity₁y - Velocity₂y) / (Velocity₁ₓ - Velocity₂ₓ))
Substituting the values:
Direction = atan((4.062 m/s) / (7.239 m/s))
Direction ≈ atan(0.561)
Direction ≈ 29.3°
c) Magnitude of the velocity of plane 2 relative to plane 1:
Velocity₁ₓ - Velocity₂ₓ = 7.7 m/s * cos(19.8°)
Velocity₁y - Velocity₂y = 12.1 m/s - 7.7 m/s * sin(19.8°)
Relative Velocity (magnitude) = √((-Velocity₁ₓ + Velocity₂ₓ)² + (-Velocity₁y + Velocity₂y)²)
Relative Velocity (magnitude) = √((-7.7 m/s * cos(19.8°))² + (-12.1 m/s + 7.7 m/s * sin(19.8°))²)
Substituting the values:
Velocity₁ₓ - Velocity₂ₓ = 7.7 m/s * cos(19.8°) ≈ 7.239 m/s
Velocity₁y - Velocity₂y = 12.1 m/s - 7.7 m/s * sin(19.8°) ≈ 4.062 m/s
Relative Velocity (magnitude) = √((-7.239 m/s)² + (-4.062 m/s)²)
Relative Velocity (magnitude) ≈ √(52.534 + 16.506)
Relative Velocity (magnitude) ≈ √69.04
Relative Velocity (magnitude) ≈ 8.31 m/s
d) Direction of the velocity of plane 2 relative to plane 1:
Direction = atan((-Velocity₁y + Velocity₂y) / (-Velocity₁ₓ + Velocity₂ₓ))
Substituting the values:
Direction = atan((-4.062 m/s) / (-7.239 m/s))
Direction ≈ atan(0.561)
Direction ≈ 29.3°
Learn more about magnitude here:
https://brainly.com/question/30337362
#SPJ11
Vector a lies in the yz plane 50.0 ∘
from the positive direction of the y axis, has a positive z component, and has magnitude 1.70 units. Vector b
lies in the xz plane 70.0∘
from the positive direction of the x axis, has a positive z component, and has magnitude 2.90 units. Find (a) a ⋅ b, (b) a× b, and (c) the angle between a and b
⋅
Vector a and b are given with their respective components and magnitudes. The dot product of a and b is 1.23 units, the cross product is -1.13 units in the k direction, and the angle between a and b is 75.55°.
(a) To find the dot product of vectors a and b, we first need to find the components of each vector. Since vector a lies in the yz plane, its x-component is zero. Vector b lies in the xz plane, so its y-component is zero. The z-component of vector a can be found using the given angle and magnitude:
a_z = 1.70 units * cos(50.0°) = 1.09 units
Similarly, the z-component of vector b can be found:
b_z = 2.90 units * cos(70.0°) = 1.13 units
Now, we can calculate the dot product:
a ⋅ b = a_x * b_x + a_y * b_y + a_z * b_z
= 0 + 0 + 1.09 units * 1.13 units
= 1.23 units
(b) To find the cross product of vectors a and b, we need to calculate the components of the resulting vector:
a× b = (a_y * b_z - a_z * b_y)i - (a_x * b_z - a_z * b_x)j + (a_x * b_y - a_y * b_x)k
= (0 - 1.09 units * 0) i - (0 - 0) j + (0 * 0 - 0 * 1.13 units) k
= -1.13 units k
(c) The angle between vectors a and b can be found using the dot product and magnitudes:
cos(θ) = a ⋅ b / (|a| * |b|)
θ = arccos(a ⋅ b / (|a| * |b|))
Substituting the values:
θ = arccos(1.23 units / (1.70 units * 2.90 units)) = 75.55°
The angle between vectors a and b is calculated using the inverse cosine function.
To know more about dot product ,
https://brainly.com/question/29097076
#SPJ11
