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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

3 years ago

Read 2 more answers

Which statement correctly defines power?

Anna007 [38]

Answer:

$Power=V\, I$ which corresponds to the second option shown: "voltage times amperage"

Explanation:

The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

$Power=\frac{V\,Q}{t} =V\,\frac{Q}{t} = V\, I$

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.

This corresponds to the second option shown in the question: "Voltage times amperage".

6 0

3 years ago

Gasoline burns inside a car’s engine. how does this fuel enable a car to move?

Svetllana [295]

The purpose of a gasoline car engine is to convert gasoline into motion so that your car can move. Currently the easiest way to create motion from gasoline is to burn the gasoline inside an engine.
Therefore, a car engine is an internal combustion engine -- combustion takes place internally.
There is such a thing as an external combustion engine. A steam engine in old-fashioned trains and steam boats is the best example of an external combustion engine. The fuel (coal, wood, oil, whatever) in a steam engine burns outside the engine to create steam, and the steam creates motion inside the engine. Internal combustion is a lot more efficient (takes less fuel per mile) than external combustion, plus an internal combustion engine is a lot smaller than an equivalent external combustion engine. This explains why we don't see any cars using steam engines.

To understand the basic idea behind how a reciprocating internal combustion engine works, it is helpful to have a good mental image of how "internal combustion" works.

One good example is an old Revolutionary War cannon. You have probably seen these in movies, where the soldiers load the cannon with gun powder and a cannon ball and light it. That is internal combustion, but it is hard to imagine that having anything to do with engines.

A potato cannon uses the basic principle behind any reciprocating internal combustion engine: If you put a tiny amount of high-energy fuel (like gasoline) in a small, enclosed space and ignite it, an incredible amount of energy is released in the form of expanding gas. You can use that energy to propel a potato 500 feet. In this case, the energy is translated into potato motion. You can also use it for more interesting purposes. For example, if you can create a cycle that allows you to set off explosions like this hundreds of times per minute, and if you can harness that energy in a useful way, what you have is the core of a car engine!

8 0

3 years ago

Humans and fish brains are similar because they both have a?

velikii [3]

Answer:

it is because they can store and retrieve informatiion just like human beings

4 0

3 years ago

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the we

Alex_Xolod [135]

Answer:

w = 2w₀ the angular velocity of man doubles

Explanation:

In this exercise, releasing the weights reduces the moment of inertia

I= I₀ / 2

Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved

L₀ = L

I₀ w₀ = I w

I₀ w₀ = I₀ / 2 w

w = 2w₀

therefore the angular velocity of man doubles

7 0

3 years ago

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