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A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.

Answers

Answer 1

Answer:

Answer:

The answers to the questions are;

(a) The velocity of the truck right after the collision is 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J

Explanation:

(a) From the principle of conservation of linear momentum, we have

m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄

Where:

m₁ = Mass of the car = 1225.0 kg

m₂ = Mass of the truck = 9700.0 kg

v₁ = Initial velocity of the car = 25.000 m/s

v₂ = Initial velocity of the truck = 20.000 m/s

v₃ = Final velocity of the car right after collision = 18.000 m/s

v₄ = Final velocity of the truck right after collision

Therefore

1225.0 kg × 25.000 m/s + 9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s + 9700.0 kg × v₄

That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄

Making v₄ the subject of the formula yields

v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s

The velocity of the truck right after the collision to five significant figures = 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision can be found by

The change in kinetic energy of the car truck system

Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy

That is ΔK.E. = ∑ Final K.E -∑ Initial K.E.

ΔK.E. = $((1)/(2) m_1v_3^(2)+(1)/(2) m_2v_4^(2)) - ((1)/(2) m_1v_1^(2) +(1)/(2) m_2v_2^(2) )$

= ( $(1)/(2)$ ·1225·18²+ $(1)/(2)$ ·9700·20.884²) - ( $(1)/(2)$ ·1225·25²+ $(1)/(2)$ ·9700·20²)

= 2313736.0616 kg·m²/s² - 2322812.5 kg·m²/s² = -9076.4384 kg·m²/s²

1 kg·m²/s² = 1 J ∴ -9076.4384 kg·m²/s² = -9076.4384 J

1) Frictional resistance between the tires and the road for the truck and car

2) Frictional resistance in the transmission system of the truck to increase its velocity

3) Sound energy, loud sound heard during the collision

4) Energy absorbed when the car and the truck outer frames are crushed

5) Heat energy in the form of raised temperatures at the collision points of the car and the truck.

6) Energy required to change the velocity of the car over a short distance.

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 48 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 9.0. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Answers

Answer:

Force that the output plunger applies to the car; F2 = 3888N

Explanation:

For a hydraulic device, the relationship between the force and the area using Pascal's principle is;

F1/A1 = F2/A2

Where;

F1 is force applied to the input piston

F2 is force that the output plunger applies to the car

A1 is Area of input piston

A2 is area of larger piston

We are given;

R2/R1 = 9

So,R2 = 9R1

F1 = 48N

Area of input piston;

A1 = π(R1)²

Area of output piston;

A2 = π(9R1)²

Since, (F1/A1) = (F2/A2)

Thus;

F1/(π(R1)²) = F2/(π(9R1)²)

If we simplify, π(R1)² will cancel out to give;

F1 = F2/9²

Thus;

F2 = 9² x F1

Plugging in 48N for F1, we have;

F2 = 9² x 48

F2 = 81 x 48

F2 = 3888N

Final answer:

Using the principle of Pascal's law and ignoring the height difference, the output force is found by the formula F2 = F1*(r2/r1)^2. Given F1 is 48N and r2/r1 is 9.0, the output force F2 equates to 3888N.

Explanation:

In the case of a hydraulic jack, the principle of Pascal's law is applied. According to this law, pressure applied at one point of the fluid is transmitted equally in all directions. Therefore, if we ignore the height difference between pistons, the pressure exerted on both pistons would be the same.

Pressure is equal to the force divided by the area, where area equals π times the radius squared (π*r^2). So, the pressure at the input piston (P1) is the force at the input piston (F1) divided by its area (A1): P1 = F1/A1, where A1 = π*(r1)^2.

For the output plunger(P2 = F2/A2), where F2 = force at the output plunger and A2 = π*(r2)^2. By equating the pressures (P1=P2) and simplifying, we find that F2 = F1*(r2/r1)^2, where r2/r1 is given as 9.0. So, the output force F2 would be 48N*(9.0)^2 = 3888N.

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Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 56 m long that spin at 13 rpm .A. At the tip of a blade, what is the speed?
B. At the tip of a blade, what is the centripetal acceleration?
C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .
D. What force is required to keep a 10 mg water droplet moving in this circular arc?
E. What is the ratio of this force to the weight of a droplet?

Answers

A) The speed of the tip of the blade is 76.2 m/s

B) The centripetal acceleration of the tip of the blade is $103.7 m/s^2$

D) The force required to keep the droplet moving in circular motion is 0.39 N

E) The ratio of the force to the weight of the droplet is 4.0

Explanation:

We know that the blade of the turbine is rotating at an angular speed of

$\omega = 13 rpm$

First, we have to convert this angular speed into radians per second. Keeping in mind that

$1 rev = 2 \pi$

$1 min = 60 s$

We get

$\omega = 13 rpm \cdot (2\pi rad/rev)/(60 s/min)=1.36 rad/s$

The linear speed of a point on the blade is given by:

$v=\omega r$

where

$\omega=1.36 rad/s$ is the angular speed

r is the distance of the point from the axis of rotation

For a point at the tip of the blade,

r = 56 m

Therefore, its speed is

$v=(1.36)(56)=76.2 m/s$

The centripetal acceleration of a point in uniform circular motion is given by

$a=(v^2)/(r)$

where

v is the linear speed

r is the distance of the point from the axis of rotation

In this problem, for the tip of the blade we have:

v = 76. 2 m/s is the speed

r = 56 m is the distance from the axis of rotation

Substituting, we find the centripetal acceleration:

$a=((76.2)^2)/(56)=103.7 m/s^2$

The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:

$F=m(v^2)/(r)$

where

m is the mass of the droplet

v is the linear speed

r is the distance from the centre of rotation

The data in this problem are

m = 10 mg = 0.010 kg is the mass of the droplet

v = 2.5 m/s is the linear speed

r = 16 cm = 0.16 m is the radius of the circular path

Substituting,

$F=(0.010)(2.5^2)/(0.16)=0.39 N$

The weight of the droplet is given by

$F_g = mg$

where

m = 10 mg = 0.010 kg is the mass of the droplet

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_g = (0.010)(9.8)=0.098 N$

The force that keeps the droplet in circular motion instead is

F = 0.39 N

Therefore, the ratio between the two forces is

$(F)/(F_g)=(0.39)/(0.098)=4.0$

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Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. Part A How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Answers

Final answer:

The total work done in moving an alpha particle from the center to the side of a square with electrons at its corners involves finding the potential energy change, which can be calculated using the charges, distances, and Coulomb's constant.

Explanation:

The question deals with the fundamental concepts of electrostatics and the energy associated with moving charges in an electric field. Given the aforementioned question, we are required to find the work done moving an alpha particle (a helium nucleus, having a charge of +2e) from the center of a square to one of its sides, with electrons (each having a charge of -e) being situated at its corners.

To determine the work done, we must consider the potential energy changes resulting from moving the alpha particle. The potential energy associated with two point charges is given by the formula: U = k*q1*q2/r, where q1 and q2 are charges, r is the distance between them, and k is Coulomb's constant.

First, we calculate the potential energy at the center due to all four electrons then find the potential energy at the midpoint of the side. The work done is the difference between these two potential energies. As the electrons are all at an equal distance from the alpha particle (in the center and on the side), the calculations would involve plugging in the values for the charge of an electron, the charge of an alpha particle, the given distance values, and Coulomb's constant into the aforementioned formula.

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Final answer:

The work required to move the alpha particle from the midpoint to the midpoint of one of the side of the square with four electrons at its corners would be zero as the net electric field at the midpoint due to the electrons is zero.

Explanation:

The subject of this question pertains to the concept of electrostatics and potential energy in physics. In this scenario, the alpha particle is initially at the midpoint of a square with four electrons at its corners. As per Coulomb's Law, the electrostatic force between two charges is inversely proportional to the square of the distance between them.

Since the alpha particle placed in the center of the square and four electrons at the corners form a symmetrical system, the net force and hence the net electric field at the midpoint due to the electrons is zero. Thus, no work would be required to move the alpha particle to the midpoint of one of the sides of the square as work done is calculated by the formula W = F x d x cos(θ), where F is force, d is the displacement, and θ is the angle between the force and displacement. Since F is equal to zero, the work done will also be zero.

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Monochromatic coherent light shines through a pair of slits. If the distance between these slits is decreased, which of the following statements are true of the resulting interference pattern?A) The distance between the maxima stays the same.B) The distance between the maxima decreases.
C) The distance between the minima stays the same.
D) The distance between the minima increases.
E) The distance between the maxima increases.

Answers

Answer:

The correct statements are D and E.

Explanation:

The fringe width is given by the following formula as :

$\beta =(\lambda D)/(d)$

Here,

$\lambda$ is wavelength of light

D is distance between slit and the screen

d is slit width.

If the between these slits is decreased, the fringe width increases. As a result, the distance between the minima increases and also the distance between the maxima increases.

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a wavelength of 3.5 cm?Express your answer in GHz and using two significant figures.
f = ________GHz
Microwave signals are beamed between two mountaintops 52 km apart. How long does it take a signal to travel from one mountaintop to the other?
Express your answer in ms and using two significant figures.
t = ________ms

Answers

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =(c)/(\lambda) =(3e8m/s)/(0.035m) = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =(d)/(t) (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = (d)/(c) = (52e3m)/(3e8m/s) = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

⇒ t = 0.2 ms (with two significative figures)

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