To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
[tex]\omega = \frac{v}{R}[/tex]
Where,
[tex]\omega =[/tex]Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
[tex]\alpha = \frac{\omega}{t}[/tex]
Where
[tex]\alpha =[/tex]Angular acceleration
[tex]\omega =[/tex] Angular velocity
t = Time
Our values are
[tex]v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]
[tex]v = 16.67m/s[/tex]
[tex]r = 0.25m[/tex]
[tex]t=6s[/tex]
Replacing at the previous equation we have that the angular velocity is
[tex]\omega = \frac{v}{R}[/tex]
[tex]\omega = \frac{ 16.67}{0.25}[/tex]
[tex]\omega = 66.67rad/s[/tex]
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
[tex]\alpha = \frac{\omega}{t}[/tex]
[tex]\alpha = \frac{66.67}{6}[/tex]
[tex]\alpha = 11.11rad/s^2[/tex]
Therefore the angular acceleration of a point on the outer edge of the tires is [tex]11.11rad/s^2[/tex]
Which is true about the inductance of a solenoid?
O It depends on the number of turns per unit length but not on the area of each turn.
O It depends on the area of each turn but not on the number of turns per unit length.
O It depends on the number of turns per unit length and the area of each turn.
Answer:
It depends on the number of turns per unit length and the area of each turn.Explanation:
The inductance of a long solenoid can be approximated by:
[tex]L=\mu \frac{N^2A}{L}[/tex]
Where:
[tex]N=Number\hspace{3}of\hspace{3}turns\\A=Cross-sectional\hspace{3}area\\L=Length\hspace{3}of\hspace{3}the\hspace{3}coil\\\mu=Magnetic\hspace{3}permeability[/tex]
Therefore, according to this, we can conclude that the inductance of a solenoid depends on the number of turns per unit length and the area of each turn.
A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
a. v = 9.8 m/s up, a = 0.
b. v = 0, a = 9.8 m/s2 down.
c. v = 0, a = 9.8 m/s2 up.
d. v = 9.8 m/s down, a = 0.
e. v = 0, a = 0.
Answer:
b. v = 0, a = 9.8 m/s² down.
Explanation:
Hi there!
The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?
Let´s take a look at the height function:
h(t) = h0 + v0 · t + 1/2 g · t²
Where
h0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.
Another way to see it (without calculus):
When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.
A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.At the instant when d = 30 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?
The torque exerted about the center of the disk is 1.76 Nm.
Explanation:To determine the torque exerted about the center of the disk, we need to calculate the perpendicular component of the force applied by the string. Since the force is applied tangentially, it can be resolved into two components: the perpendicular component and the parallel component. The perpendicular component creates the torque.
The perpendicular component of the force is equal to the tension in the string multiplied by the sine of the angle between the force and the radius. In this case, the angle is 90 degrees, so sin(90) = 1. Therefore, the torque is equal to the tension in the string multiplied by the radius of the disk.
Given that the force applied by the string is 16 N and the radius of the disk is 11 cm (0.11 m), the torque exerted about the center of the disk is 16 N * 0.11 m = 1.76 Nm.
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The torque exerted on the disk is calculated using the formula Torque = Force x Distance, where the distance is the radius of the disk. The torque in this situation is 1.76 N.m.
Explanation:In your question, we're being asked to find the torque exerted on a disk with radius R that's being pulled along a frictionless surface by a force F. The torque exerted on an object can be calculated using the formula Torque = Force x Distance, where distance refers to the perpendicular distance from the axis of rotation to the line of action of the force. In this case, the force is being applied at the edge of the disk, and the radius of the disk serves as the lever arm distance. Therefore, the torque exerted about the center of the disk is Torque = F * R = 16 N * 0.11 m = 1.76 N.m. Please note that the radius was converted from cm to m in order to keep the units consistent.
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A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at and it leaves the bat traveling to the left at an angle of above horizontal with a speed of If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.
Answer
given,
mass of the baseball = 0.145 Kg
Assuming the horizontal velocity of ball = V x = 50 m/s
ball left at an angle of = 30°
At the speed of 65 m/s
time of contact = 1.75 ms
velocity of time along horizontal direction
v_{horizontal} = -65 cos 30° - 50
= -106.29 m/s
Impulse = force x time
impulse is equal to change in momentum
now,
force x time = m v
[tex]F = \dfrac{mv}{t}[/tex]
[tex]F = \dfrac{0.145 \times (-106.29)}{1.75 \times 10^{-3}}[/tex]
F_{horizontal} = - 8.806 kN
now velocity in vertical component
v_{vertical} = 65 sin 30°
= 32.5 m/s
[tex]F = \dfrac{mv}{t}[/tex]
[tex]F = \dfrac{0.145 \times (32.5)}{1.75 \times 10^{-3}}[/tex]
F_{vertical} = 2.692 kN
(a) The horizontal component of the average force on the ball is 7,633.63 N.
(b) The vertical component of the average force on the ball is 2,929 N.
The given parameters;
mass of the bat, m₁ = 0.145 kgtime, t = 1.75 ms initial velocity of the ball to right, v₀ = 50 m/sfinal velocity of the ball to left v = 55 m/s at 40⁰The horizontal component of the ball's acceleration is calculated as follows;
[tex]-v_f_x = v_0_x - a_xt\\\\a_xt = v_0_x + v_f_x\\\\a_x = \frac{v_0_x + v_f_x}{t} \\\\a_x = \frac{50 \ + \ 55cos(40)}{0.00175} \\\\a_x = 52,645.7 \ m/s^2[/tex]
The horizontal component of the average force on the ball is calculated as follows;
[tex]F_x = ma_x\\\\F_x = 0.145 \times 52,645.7 \\\\F_x = 7,633.63 \ N[/tex]
The vertical component of the ball's acceleration is calculated as follows;
[tex]v_f_y = v_0y + a_yt\\\\v_f_y = 0+ a_yt\\\\a_y = \frac{v_f_y}{t} \\\\a_y = \frac{55 \times sin(40)}{0.00175} \\\\a_y = 20,200 \ m/s^2[/tex]
The vertical component of the average force on the ball is calculated as follows;
[tex]F_y = ma_y\\\\F_y = 0.145 \times 20,200\\\\F_y = 2,929 \ N[/tex]
"Your question is not complete, it seems to be missing the following information";
A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s , and it leaves the bat traveling to the left at an angle of 40 ∘ above horizontal with a speed of 55.0 m/s . The ball and bat are in contact for 1.75 ms .
find the horizontal and vertical components of the average force on the ball.
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A 2.3 g spider is hanging at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every 3 seconds.What is the spring constant of the silk thread?
Answer:[tex]k=10.091\times 10^{-3} N/m[/tex]
Explanation:
Given
mass of spider [tex]m=2.3 gm[/tex]
Largest amplitude can be obtained by Tapping after every 3 second
i.e. Time period of oscillation is [tex]T= 3 s[/tex]
considering spider to execute Simple harmonic motion
time of oscillation is given by
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]
substituting values
[tex]3=2\pi \sqrt{\frac{2.3\times 10^{-3}}{k}}[/tex]
[tex]\frac{3}{2\pi }=\sqrt{\frac{2.3\times 10^{-3}}{k}}[/tex]
[tex]0.477=\sqrt{\frac{2.3\times 10^{-3}}{k}}[/tex]
[tex](0.477)^2=\frac{2.3\times 10^{-3}}{k}[/tex]
[tex]\frac{2.3\times 10^{-3}}{k}=0.228[/tex]
[tex]k=10.091\times 10^{-3} N/m[/tex]
In a mass spectrometer, ions are directed into a magnetic field, where they curve and strike a detector. If a variety of singly ionized atoms travel at the same speed through the magnetic field, would you expect them all to be deflected by the same amount, or would different ions be bent to different amounts? Defend your answer.
Answer:
the trajectory depends on the velocity and mass of the ions, therefore, they should be in different quantities
Explanation:
A mass spectrometer works based on the electrical force to accelerate the ions to a certain speed and the magnetic force to give a circular trajectory, let's use Newton's second law
F = m a
q v B = m v2 / r
r = (q / m) B / v
Here we can see that the radius of curvature of the ions depends on the velocity, but also on the mass of each of them. Therefore, even when they enter with the same speed they deviate in different trajectories depending on their mass.
The heavier ions will have smaller radii than the lighter ions
Consequently, the trajectory depends on the velocity and mass of the ions, therefore, they should be in different quantities
A satellite beams microwave radiation with a power of 13 toward the Earth's surface, 550 away. When thebeam strikes Earth, its circular diameter is about 1000. Find the rms electric field strength of the beam at the surface of the Earth.
Answer:
[tex]E_{rms} = 2.27 V/m[/tex]
Explanation:
Power of microwave radiation = 13 kW = 13 x 10³ W
Diameter = 1000 m
Radius = R = 1000/2 = 500 m
distance, r = 500 Km
r = 500 x 10³ m
Electric strength of beam of light is given as
[tex]S = \dfrac{P}{A} = c\epsilon_0 E_{rms}^2[/tex]
Now,
[tex]E_{rms} =\sqrt{\dfrac{P}{Ac\epsilon_0}}[/tex]
c is speed of light
ε₀ is permittivity of free space
A is the area of circle = π r² = π x (500 x 10³)² = 7.853 x 10¹¹ m²
[tex]E_{rms} =\sqrt{\dfrac{13\times 10^3}{7.853 \times 10^{11}\times 3 \times 10^8 \times 8.85\times 10^{-12}}}[/tex]
[tex]E_{rms} =\sqrt{\dfrac{13\times 10^3}{7.853 \times 10^{11}\times 3 \times 10^8 \times 8.85\times 10^{-12}}}[/tex]
[tex]E_{rms} = 2.27 V/m[/tex]
Electricity is distributed from electrical substations to neighborhoods at 1.6×104 V . This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house. Part A How many turns does the primary coil on the transformer have if the secondary coil has 140 turns? Express your answer using two significant figures. n prim n p r i m = nothing turns SubmitRequest Answer Part B No energy is lost in an ideal transformer, so the output power P out from the secondary coil equals the input power P in to the primary coil. Suppose a neighborhood transformer delivers 230 A at 120 V . What is the current in the 1.6×104 V line from the substation? Express your answer using two significant figures. I i I i = nothing A SubmitRequest Answer Provide Feedback Next
Answer:
(A) 18667 turns
(B) 1.7 A
Solution:
As per the question:
Voltage at which the electricity is distributed, [tex]V_{p} = 1.6\times 10^{4}\ Hz[/tex]
Frequency of the oscillating voltage, f = 60 Hz
Step down voltage, [tex]V_{s} = 120\ V[/tex]
No. of turns in the secondary coil, [tex]N_{s} = 140\ turns[/tex]
Current in the secondary coil, [tex]I_{s} = 230\ A[/tex]
Now,
(A) To calculate the primary no. of turns, we use the relation:
[tex]\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}[/tex]
[tex]N_{p} = \frac{V_{p}}{V_{s}}\times N_{s}[/tex]
[tex]N_{p} = \frac{1.6\times 10^{4}}{120}\times 140 = 18,667\ turns[/tex]
(B) To calculate the current in the primary coil, [tex]I_{p}[/tex], we use the relation:
[tex]\frac{V_{p}}{V_{s}} = \frac{I_{s}}{I_{p}}[/tex]
[tex]I_{p} = \frac{V_{s}}{V_{p}} \times {I_{s}}[/tex]
[tex]I_{p} = \frac{120}{1.6\times 10^{4}} \times 230 = 1.7\ A[/tex]
You have a flashlight that uses 0.75 watts of power and requires 1.5 volt battery.How many joules of energy are used by the flashlight in 15 minutes? Hint: Joules are watts and seconds and you are computing the difference between the electric energy at two points A and B.What the total charge q0? Hint: the statement of the problem provides delta V.How many particles, each having a charge of 1.60 x 10^-6, are needed to produce the total charge at q0?
Answer:
(a) Energy will be 675 J
(B) charge will be 450 C
(C) Total number of particles will be [tex]281.25\times 10^6[/tex]
Explanation:
We have given that a flashlight uses 0.75 watts of power
So power P = 0.75 watt
Voltage is given as V = 1.5 volt
Time is given as t = 15 minutes
We know that 1 minute = 60 sec
So 15 minutes = [tex]15\times 60=900sec[/tex]
(A) We know that energy is given by [tex]E=P\times T=0.75\times 900=675j[/tex]
(b) We know that energy is also given by [tex]E=QV[/tex]
So [tex]675=Q\times 1.5[/tex]
[tex]Q=450C[/tex]
Now we have given charge on each particle [tex]=1.6\times 10^{-6}C[/tex]
So number of charge particle [tex]n=\frac{450}{1.6\times 10^{-6}}=281.25\times 10^6[/tex]
Romeo (85.0 kg) entertains Juliet (59.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. (a) How far (in m) does the 75.0 kg boat move toward the shore it is facing? m
Answer:
0.7274 m
Explanation:
X denotes the distance from the reference point
M denotes the mass of the point
The center of mass of a system is given by
[tex]X_{CM}=\sum_{n=1}^n\frac{X_n\times M_n}{M_n}[/tex]
Taking the position of Juliet as the reference point
[tex]X_{CM}=\frac{59\times 0+85\times 2.7+75\times 1.35}{59+85+75}\\\Rightarrow X_{CM}=1.51027\ m[/tex]
After the boat has moved the distance between the center of mass of the boat and the shore remains the same. The change is observed relative to the rear of the boat
[tex]X_{CM}=\frac{75\times 1.35+(85+59)\times 2.7}{59+85+75}\\\Rightarrow X_{CM}=2.23767\ m[/tex]
The displacement of the boat towards the shore is
[tex]2.23767-1.51027=0.7274\ m[/tex]
By applying the conservation of momentum for a system with no external forces, the distance the boat moves when Juliet goes to the rear to kiss Romeo is 2.124 meters toward the shore.
The question involves the concept of conservation of momentum, particularly in systems with no external forces. Here, we consider a scenario where Juliet transitions from the front to the rear of the boat, causing the boat to move toward shore. To solve for the movement of the boat toward the shore, we need to realize that the total momentum of the system (Juliet, Romeo, and the boat) before and after Juliet's movement must remain constant, as there is no external force in the horizontal direction.
Initially, both Juliet and the boat are at rest, so the total initial momentum is zero. When Juliet, having mass 59.0 kg, moves a distance of 2.70 m towards Romeo, the boat will move in the opposite direction to conserve momentum.
To find the distance the boat moves, we use the conservation of momentum equation: (mass of Juliet) x (distance Juliet moves) = (mass of boat) x (distance boat moves). Rearranging gives us the distance the boat moves: distance boat moves = (mass of Juliet x distance Juliet moves) / mass of boat.
By substituting the given values, we get:
distance boat moves = (59.0 kg x 2.70 m) / 75.0 kg
distance boat moves = 2.124 m
Therefore, the boat moves 2.124 meters toward the shore when Juliet moves to the rear to kiss Romeo.
5.37 At steady state, a power cycle develops a power output of 10 kW while receiving energy by heat transfer at the rate of 10 kJ per cycle of operation from a source at temperature T. The cycle rejects energy by heat transfer to cooling water at a lower temperature of 300 K. If there are 100 cycles per minute, what is the minimum theoretical value for T, in K?
Answer: minimum theoretical value of T = 750k
Explanation:
Assuming the the cycle is reversible and is ideal then
Wnet/Qh = Nmin .... equa 1
Equation 1 can be rewritten as
(Th -TL)/ Th ...equation 2
Th= temp of hot reservoir
TL= temp of low reservoir= 300
Wnet = power generated=10kw
He = energy transfer=10kj per cycle
Qhe = power transfer = (100/60)*10Kj = 16.67kw
Sub into equat 1
Nmin = 10/16.67 = 0.6
Sub Nmin into equation 2
Th = -TL/(Nmin -1) = -300k/(0.6 - 1)
Th =750k
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approaches a long embankment inclined upward at 16° above the horizontal. As she slides up the embankment, she encounters a coefficient of friction of 0.128. Determine the height to which she will travel before coming to rest.
Final answer:
To find the height Sheila will ascend before stopping, we use energy conservation, calculating her initial kinetic energy and the work done against friction to solve for the final potential energy, which includes the height she reaches on the incline.
Explanation:
To determine the height Sheila will travel up the embankment before coming to rest, we need to apply the principles of energy conservation and include the work done against friction. Sheila's initial kinetic energy (KE) at the bottom of the hill is converted into gravitational potential energy (PE) and the work done against friction as she moves up the incline.
We can calculate the initial kinetic energy using the formula KE = 1/2 * m * v2, where m is her mass and v is her velocity. Then, we find the work done against friction, which is equal to the force of friction times the distance traveled (Wfriction = Ffriction * d). The force of friction is calculated as the coefficient of friction multiplied by the normal force, which, on an incline, is the component of the gravitational force perpendicular to the slope.
Using these concepts, we can set up the equation: KEinitial = PEfinal + Wfriction. Solving for the final potential energy gives us PEfinal = m * g * h, where h is the height she will reach. The height can be determined by rearranging this equation after calculating the work done against friction and knowing the initial kinetic energy.
It is important to note that since no values are given for distances or the length of the incline and we are assuming a constant coefficient of friction, we find the height as a function of distance she travels up the slope until she comes to rest.
A transverse harmonic wave travels on a rope according to the following expression:
y(x,t) = 0.14sin(2.1x + 17.7t)
The mass density of the rope is μ = 0.104 kg/m. x and y are measured in meters and t in seconds.
1)
What is the amplitude of the wave?
m
Your submissions:
2)
What is the frequency of oscillation of the wave?
Hz
Your submissions:
3)
What is the wavelength of the wave?
m
Your submissions:
4)
What is the speed of the wave?
m/s
Your submissions:
5)
What is the tension in the rope?
N
Your submissions:
6)
At x = 3.4 m and t = 0.48 s, what is the velocity of the rope? (watch your sign)
m/s
Your submissions:
7)
At x = 3.4 m and t = 0.48 s, what is the acceleration of the rope? (watch your sign)
m/s2
Your submissions:
8)
What is the average speed of the rope during one complete oscillation of the rope?
m/s
Your submissions:
9)
In what direction is the wave traveling?
+x direction
-x direction
+y direction
-y direction
+z direction
-z direction
Your submissions:
10)
On the same rope, how would increasing the wavelength of the wave change the period of oscillation?
the period would increase
the period would decrease
the period would not change
This answer clarifies various aspects related to a transverse harmonic wave, such as amplitude, frequency, and tension in the rope. The wave is traveling in the -x direction. And, the period of oscillation will increase if the wavelength increases.
Explanation:To answer such questions, we need to understand the equation of the wave given by y(x,t) = A sin (kx + ωt). Here, 'A' represents the amplitude of the wave, 'k' is the wave number, and 'ω' is the angular frequency. We get these values by comparing the given equation with the standard wave equation.
The amplitude of the wave is given by A and here A=0.14 m.The frequency of the wave is given by f=ω/2π=17.7/2π ≈ 2.82 Hz.The wavelength of the wave λ = 2π/k = 2π/2.1 ≈ 2.99 m.The speed of the wave is given by the relation v=λf ≈ 2.99 m * 2.82 Hz ≈ 8.43 m/s.The tension in the rope is given by T = μv^2 = 0.104 kg/m * (8.43 m/s)^2 ≈ 7.4 N.The velocity and acceleration at given values of x and t can be obtained from the time derivative of your wave equation. They require more advanced calculations and depend on the given values of x and t.The average speed of the rope during one complete oscillation is 0, as every point on the rope returns to its initial position during one oscillation.The wave is travelling in the -x direction since the sign before 't' in the wave equation is positive.If the wavelength of the wave is increased, the period of oscillation will increase as the period T=1/f and frequency f=v/λ and v is constant for a given medium.Learn more about Transverse harmonic wave here:https://brainly.com/question/14014748
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A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on a straight road. The hospital is 70 km down that road from that nearest point. If the vehicle can drive at 30 kph over the terrain and at 130 kph on the road, how far down the road should the vehicle aim to reach the road to minimize the time it takes to reach the hospital? (Round your answer to two decimal places.)
Final answer:
To minimize the time it takes to reach the hospital, the vehicle should aim to reach the road 30 km from the starting point.
Explanation:
To minimize the time it takes to reach the hospital, the four-wheel-drive vehicle should aim to reach the road at a distance that allows it to travel at its maximum speed on the road. The vehicle can drive at 30 kph over the terrain and at 130 kph on the road.
So, if the vehicle aims to reach the road after traveling 30 km over the terrain at 30 kph, it will take 1 hour. Then, on the road, it will travel at 130 kph for the remaining 40 km, which will take approximately 18.46 minutes. Therefore, the vehicle should aim to reach the road at a distance of 30 km from the starting point to minimize the time it takes to reach the hospital.
The escape velocity on earth is 11.2 km/s. What fraction of the escape velocity is the rms speed of H2 at a temperature of 31.0 degrees Celsius on the earth? Note that virtually all the molecules will have escaped the earth's atmosphere if this fraction exceeds 0.15.
To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as
[tex]v_{rms} = \sqrt{\frac{3RT}{n}}[/tex]
Where,
T = Temperature
R = Gas ideal constant
n = Number of moles in grams.
Our values are given as
[tex]v_e =11.2km/s = 11200m/s[/tex]
The temperature is
[tex]T = 30\°C = 30+273 = 303K[/tex]
Therefore the root mean square velocity would be
[tex]v_{rms} = \sqrt{\frac{3(8.314)(303)}{0.002}}[/tex]
[tex]v_{rms} = 1943.9m/s[/tex]
The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity
[tex]\alpha = \frac{v_{rms}}{v_e}[/tex]
[tex]\alpha = \frac{1943.9}{11200}[/tex]
[tex]\alpha = 0.1736[/tex]
Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736
A rod of mass M = 4 kg, length L = 1.8 meters, and moment of inertia ML2/12 is free to move on a frictionless surface. The rod is at rest when a puck of mass m = 0.4 kg approaches with a speed vi = 20 m/s perpendicular to the rod's length and strikes the rod at a point d = 0.3 meters from its lower end. After the collision, the puck moves backward with a speed vf = 10 m/s. The rod's center of mass moves forward with speed v and rotates with angular speed ω. (a) What is the speed v of the rod's center of mass after the collision?
Answer:
The speed of the rod's center of mass after the collision is 6 m/s.
Explanation:
Given that,
Mass of rod = 4 kg
Length l = 1.8 m
Moment of inertia [tex]I=\dfrac{ML^2}{12}[/tex]
Mass of puck = 0.4 kg
Initial speed= 20 m/s
Distance = 0.3 m
Final speed = 10 m/s
(a). We need to calculate the speed v of the rod's center of mass after the collision
As there is no external force acting on the system so, linear and angular momentum of the system will be conserved.
Using conservation of momentum
[tex]m_{i}v_{i}=m_{f}v_{f}+Mv[/tex]
Put the value into the formula
[tex]0.4\times20=-0.4\times10+2v[/tex]
[tex]v=\dfrac{8+4}{2}[/tex]
[tex]v=6\ m/s[/tex]
Hence, The speed of the rod's center of mass after the collision is 6 m/s.
An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in a time interval of length 4.30 s.
A.) Find the angular acceleration in revolutions per second per second.
= -1.43 rev/s^2
B.) Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A.
= _________? rev
C.) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A?
= ____________? s
The angular acceleration is -1.43 rev/s². The number of revolutions made by the fan blades is -1870.5 rev. It will take an additional 125.87 seconds for the fan to come to rest.
Explanation:To find the angular acceleration, we can use the formula:
Angular acceleration = (final angular velocity - initial angular velocity) / time
Substituting the given values:
Angular acceleration = (180 rev/min - 550 rev/min) / 4.30 s = -1.43 rev/s²
For part B, the number of revolutions made by the fan blades can be found using the formula:
Number of revolutions = (final angular velocity - initial angular velocity) * time
Substituting the given values:
Number of revolutions = (180 rev/min - 550 rev/min) * 4.30 s = -1870.5 rev
For part C, the time required for the fan to come to rest can be found using the formula:
Time = (final angular velocity - initial angular velocity) / angular acceleration
Substituting the given values:
Time = (0 rev/min - 180 rev/min) / -1.43 rev/s² = 125.87 s
A uniform, thin rod of length h and mass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. (a) What is the speed of its center of mass just before it hits the horizontal surface? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
The speed of the center of mass just before it hits the horizontal surface can be found using the principle of conservation of energy.
Explanation:To find the speed of the center of mass just before it hits the horizontal surface, we can use the principle of conservation of energy. When the rod falls freely, it gains gravitational potential energy which is converted into kinetic energy. At the lowest point, where the center of mass hits the horizontal surface, all the gravitational potential energy is converted into kinetic energy. Therefore, we can equate the gravitational potential energy at the top to the kinetic energy at the bottom:
mgh = (1/2)mv^2
Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, and v is the speed of the center of mass. Solving for v:
v^2 = 2gh
v = sqrt(2gh)
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A 0.200-kg mass is attached to the end of a spring with a spring constant of 11 N/m. The mass is first examined (t = 0) when the mass is 17.0 cm from equilibrium in the positive x-direction, and is traveling at 2.0 m/s in the positive x-direction.
a) Write an equation x(t) that describes the position of this mass as a function of time. Express this function in terms of numerical values, trigonometric functions and the time variable "t".b) Repeat for v(t), the speed of the mass as a function of time.c) Repeat for a(t), the acceleration of the mass as a function of time.
Answer:
a) x (t) = 0.3187 cos (7.416 t + 1.008) , b) v = -2,363 sin (7,416 t + 1,008)
c) a = - 17.52 cos (7.416t + 1.008)
Explanation:
The spring mass system creates a harmonic oscillator that is described by the equation
x = Acos (wt + φ)
Where is the amplitude, w the angular velocity and fi the phase
a) Let's reduce the SI system
x = 17.0 cm (1 m / 100 cm) = 0.170 m
The angular velocity is given by
w = √ (k / m)
w = √ 11 / 0.200
w = 7.416 rad / s
Let's look for the terms of the equation with the data for time zero (t = 0 s)
0.170 = A cos φ
Body speed can be obtained by derivatives
v = dx / dt
v = -A w sin (wt + φ)
2.0 = -A 7.416 sin φ
Let's write the two equations
0.170 = A cos φ
2.0 / 7.416 = -A sin φ
Let's divide those equations
tan φ= 2.0 / (7.416 0.170)
φ= tan⁻¹ (1,586)
φ= 1.008 rad
We calculate A
A = 0.170 / cos φ
A = 0.170 / cos 1.008
A = 0.3187 m
With these values we write the equation of motion
x (t) = 0.3187 cos (7.416 t + 1.008)
b) the speed can be found by derivatives
v = dx / dt
v = - 0.3187 7.416 sin (7.416 t +1.008)
v = -2,363 sin (7,416 t + 1,008)
c) the acceleration we look for conserved
a = dv / dt
a = -2,363 7,416 cos (7,416 t + 1,008)
a = - 17.52 cos (7.416t + 1.008)
A stone with heat capacity C = 1.2 J/K is left outside on a cold day to reach a temperature of 273.15 K. The stone is then brought inside where the air temperature is 293.15 K. The stone is used as the cold side of a reversible engine. (The air is the hot side.) What is the maximum work that can be accomplished ?
To solve the problem it is necessary to apply the concepts related to Helmholtz free energy. By definition in a thermodynamic system the Helmholtz energy is defined as
[tex]\Delta F = \Delta U - T\Delta S[/tex]
Where,
[tex]\Delta U[/tex] is the internal energy equivalent to
[tex]\Delta U = C \Delta T[/tex]
And [tex]\Delta S[/tex] means the change in entropy represented as
[tex]\Delta S = C ln \frac{T_2}{T_1}[/tex]
Note: C means heat capacity.
Replacing in the general equation we have to
[tex]\Delta F = C \Delta T - T C ln \frac{T_2}{T_1}[/tex]
The work done of a thermodynamic system is related by Helmholtz free energy as,
[tex]W = - \Delta F[/tex]
[tex]W = -(C \Delta T - T C ln \frac{T_2}{T_1})[/tex]
[tex]W = T C ln \frac{\T_2}{T_1}-C \Delta T[/tex]
Replacing with our values we have,
[tex]W = (293.15)(1.2)ln(\frac{293.15}{273.15})-(1.2)(20)[/tex]
[tex]W = 0.858 J[/tex]
Therefore the maximum work that can be accomplished is 0.858J
The maximum work that can be accomplished by the stone is 24 J.
Explanation:To find the maximum work that can be accomplished by the stone as the cold side of a reversible engine, we can use the formula for the maximum work of a Carnot engine:
Wmax = Qh - Qc
where Qh is the heat transferred from the hot reservoir and Qc is the heat transferred to the cold reservoir.
Since the stone is the cold side, Qc is the negative of the heat capacity times the temperature difference:
Qc = -C(Tin - Tcold)
Substituting the given values, we have:
Qc = -(1.2 J/K)(293.15 K - 273.15 K) = -24 J
Since Qh is the negative of Qc, we have:
Qh = -Qc = -(-24 J) = 24 J
Therefore, the maximum work that can be accomplished by the stone is:
Wmax = 24 J
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A potter's wheel having a radius 0.47 m and a moment of inertia of 14.2 kg · m2 is rotating freely at 53 rev/min. The potter can stop the wheel in 8.0 s by pressing a wet rag against the rim and exerting a radially inward force of 68 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.
Answer:
the effective coefficient of kinetic friction between the wheel and the wet rag is 0.31
Explanation:
given information:
radius, r = 0.47 m
moment of inertia, I = 14.2 kg[tex]m^{2}[/tex]
angular velocity, ω0 = 53 rev/min = 53 x 2π/60 = 5.5 rad/s
time, t = 8.0 s
inward force, N = 68 N
τ = I α
F r = I α, F is friction force, F = μ N
μ N r = I α
μ = I α / N r
We have to find α
ωt = ω0 + αt. ωt = 0 because the wheel stop after 8 s
0 = 5.5 + α 8
α = -5.5/8 = 0.69 [tex]rad/s^{2}[/tex]
Now we can calcultae the coeffcient of kinetik friction
μ = I α / N r
= (14.2) (0.69) / (68) (0.47)
= 0.31
A truck is traveling horizontally to the right. When the truck starts to slow down, the crate on the (frictionless) truck bed starts to slide. In what direction could the net force be on the crate?
a. Straight up (the normal force).
b. Horizontal and to the left.
c. Straight down (because of gravity).
d. Horizontal and to the right.
e. No direction. The net force is zero.
To solve this problem it is necessary to apply Newton's first law which warns that every body remains in continuous motion or at rest until an external force acts on it.
From the statement it is said that there is no friction on the crate, then the sum of forces in the horizontal direction will be zero. Here as the truck is slowing down, there is not net horizontal force on the crate, it means that the crate is at rest.
The correct answer is e. No direction. The net force is zero.
When the truck slows down, the crate experiences a 'force' that is horizontal and to the left due to relative motion and the principle of inertia.
Explanation:The net force acting on the crate would be horizontal and to the left (option b). This is due to the principle of inertia. When the truck slows down, the crate tries to maintain its original speed and direction, which is horizontal and to the right. This will make the crate slide towards the back of the truck, giving a semblance of a force acting to the left on the crate. Note that this is a result of the relative motion between the crate and the truck and not an actual force acting on the crate.
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When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.
Answer:
P = 1.99 10⁸ Pa
Explanation:
The definition of the bulk module is
B = - P / (ΔV / V)
The negative sign is included for which balk module is positive, P is the pressure and V that volume
They tell us that the variation in volume is 9.05%, that is
ΔV / V = 9.0Δ5 / 100 = 0.0905
P = - B DV / V
P = 2.2 10⁹ (0.0905)
P = 1.99 10⁸ Pa
Final answer:
The force per unit area water exerts on a container when it freezes is calculated using the bulk modulus of water, resulting in a pressure of 1.99 × 10⁴ N/cm². This explains why freezing water can cause significant damage to rigid structures.
Explanation:
When water freezes, its volume increases by 9.05% (ΔV/V = 9.05 × 10⁻²). The force per unit area that water can exert on a container when it freezes can be calculated using the bulk modulus of water, which is given as B = 2.2 × 10⁹ N/m². The change in volume fraction (ΔV/V) is directly related to the change in pressure (ΔP) and the bulk modulus (B) by the formula ΔP = -B(ΔV/V), where the negative sign indicates a reduction in volume leads to an increase in pressure, but in this case, we're dealing with an expansion, so ΔV/V is positive.
To find the pressure in N/cm², we first calculate the pressure in Pascals (Pa) using the given values:
ΔP = B(ΔV/V)
= (2.2 × 10⁹ N/m²)(9.05 × 10⁻²)
= 1.99 × 10⁸ Pa.
Since 1 Pa equals 1 N/m² and 1 N/m² equals 0.0001 N/cm²,
the pressure in N/cm² is 1.99 × 10⁴ N/cm².
he drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 7 scale model. The tunnel operates with freshwater at 20 ˚C, whereas the prototype torpedo is to be used in seawater at 15.6 ˚C. To correctly simulate the behavior of the prototype moving with a velocity of 53 m/s, what velocity is required in the water tunnel?
Answer:20.03 m/s
Explanation:
Given
[tex]L_r=1:7[/tex]
velocity of Prototype [tex]v_p=53 m/s[/tex]
Taking Froude number same for both flow as it is a dimensionless number for different flow regimes in open Flow
[tex](\frac{v_m}{\sqrt{L_mg}})=(\frac{v_p}{\sqrt{L_pg}})[/tex]
[tex]v_m=v_p\times \sqrt{\frac{L_m}{L_p}}[/tex]
[tex]v_m=53\times \frac{1}{\sqrt{7}}[/tex]
[tex]v_m=20.03 m/s[/tex]
One at your summer lunar space camp activities is to launch n 1170-kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: II reaches an altitude of 211 km. What gain m gravitational potential memo does the launch accomplish? The mass and radios of the Moon are 7.36 x 10^22 kg and 1740 km, respectively.
To solve this problem it is necessary to apply the concepts related to gravitational potential energy.
The change in gravitational potential energy is given by,
[tex]\Delta PE = PE_f - PE_i[/tex]
Where,
[tex]PE = \frac{GMm}{R}[/tex]
Here,
G = Gravitational Universal Constant
M = Mass of Earth
m = Mass of Object
R = Radius
Replacing we have that
[tex]\Delta PE = \frac{GMm}{R+h} -\frac{GMm}{R}[/tex]
Note that h is the height for this object. Then replacing with our values we have,
[tex]\Delta PE = \frac{GMm}{R+h} -\frac{GMm}{R}[/tex]
[tex]\Delta PE = GMm(\frac{1}{R} -\frac{1}{R+h})[/tex]
[tex]\Delta PE = (6.65*10^{-11})(7.36*10^{22})(1170)(\frac{1}{1740*10^3} -\frac{1}{211*10^3+1740*10^3})[/tex]
[tex]\Delta PE = 57264.48*10^{11}(5.1255*10^{-7}-5.747*10^{-7})[/tex]
[tex]\Delta PE = 3.56*10^8J[/tex]
Therefore the gravitational potential is [tex]3.56*10^8J[/tex]
(a) What is the length of a simple pendulum that oscillates with a period of 3.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Length (Earth)=?
Length (Mars)=?
(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 3.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Mass (Earth)=?
Mass (Mars)=?
Answer:
Explanation:
The expression relating length and time period
T = 2π [tex]\sqrt{\frac{l}{g} }[/tex]
3.2 = [tex]2\pi \sqrt{\frac{l}{9.8} }[/tex]
l = 2.54 m
On Mars g = 3.7
[tex]3.2 = 2\pi \sqrt{\frac{L}{3.7} }[/tex]
L = .96 m
b )
Expression for elastic constant and time period is as follows
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
[tex]3.2=2\pi \sqrt{\frac{m}{20} }[/tex]
m = 5.19 N/s
Time period of oscillation due to spring is not dependent on g , so same time period will be found on Mars as that on the earth.
The Young�s modulus for steel is 20.7*1010 N/m2. How much will a 2.5 mm diameter wire 12 cm long, be strained when it supports a load of 450 N?
Answer:
ΔL = 53.14*10⁻⁶ m
Explanation:
Given
E = 20.7*10¹⁰ N/m²
D = 2.5 mm = 2.5*10⁻³ m
L = 12 cm = 0.12 m
P = 450 N
ΔL = ?
We can use the formula
ΔL = P*L / (A*E)
where A = π*D² / 4 = π*(2.5*10⁻³ m)² / 4
⇒ A = 4.908*10⁻⁶ m²
then
ΔL = (450 N)*(0.12 m) / (4.908*10⁻⁶ m²*20.7*10¹⁰ N/m²) = 53.14*10⁻⁶ m
The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????
Answer:
The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]
Explanation:
Given that,
Average energy [tex]E=2\times10^{-4}\ eV[/tex]
Photon = [tex]4\times10^{-5}\ eV[/tex]
We need to calculate the number of available energy states per unit volume
Using formula of energy
[tex]g(\epsilon)d\epsilon=\dfrac{8\pi E^2dE}{(hc)^3}[/tex]
Where, E = energy
h = Planck constant
c = speed of light
Put the value into the formula
[tex]g(\epsilon)d\epsilon=\dfrac{8\times\pi\times2\times10^{-4}\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]
[tex]g(\epsilon)d\epsilon=4.01\times10^{48}[/tex]
Hence, The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]
Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.84 Hz?
Answer:
m = 3.91 kg
Explanation:
Given that,
Mass of the object, m = 3.74 kg
Stretching in the spring, x = 0.0161 m
The frequency of vibration, f = 3.84 Hz
When the object is suspended, the gravitational force is balanced by the spring force as :
[tex]mg=kx[/tex]
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{3.74\times 9.8}{0.0161}[/tex]
k = 2276.52 N/m
The frequency of vibration is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]m=\dfrac{k}{4\pi^2f^2}[/tex]
[tex]m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}[/tex]
m = 3.91 kg
So, the mass of the object is 3.91 kg. Hence, this is the required solution.
A car speedometer that is supposed to read the linear speed of the car uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the car instead, how will that affect the speedometer reading? The speedometer
a. will still read the speed accurately.
b. will read low.
c. will read high.
A speedometer often known as aspeed meter do act as a boundary or gauge that helps us to take note, measures and displays the speed of a moving automobile.
The speedometer will read low.An increase in the radius of the tires does not increase the speed measured by a speedometer.
An increase in the size of the tires will cause the car to move faster than it would with smaller tires. This will leaf tothe speed of the car will be much more than what the speedometer is reading.
Conclusively we can say that option b is the best option that explains what the statement means
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