Anaerobic metabolism:

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

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Why is an object's density expressed as a relationship between two units?

vfiekz [6]

Probably for kind of the same reason that speed is expressed as a
relationship between two units. You know, like miles per hour .

I guess the only reason is because no single unit has been invented
to describe density.

The rate of doing work or using energy would always be expressed
as a relationship between two units ... we would say that the rate of
work is "(so many) joules per second". But the "watt" was invented,
so we can say "(so many) watts" instead.

So I guess you're right. Density could be simpler to describe
if we only had a unit for it. Then we wouldn't have to say "(so many)
grams per cubic centimeter". We would just say "(so many) (new unit)".

Let's try it out:

"Uhhh, pardon me Professor . . . I've been working late in the lab,
and I believe I've identified a new substance, hitherto unknown to
the scientific community, and totally unexpected. In its pure form,
the substance appears to be pink, it smells like butterscotch, and
its density is approximately 27.4 Brianas. I think it's time we published
these findings ... with your name as lead investigator, of course."

I like it !

5 0

3 years ago

What type of fire can be put out safely with water?

nignag [31]

Flame of fire could get put out with water

3 0

3 years ago

Which careers typically require post-secondary education other than vocational school? Check all that apply. Boiler Operation Co

In-s [12.5K]

Answer: Construction management, Architecture, Civil Drafting.

Explanation: hope this helps :)

5 0

2 years ago

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Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo

choli [55]

Answer:

Yes, if the two carts are moving into opposite directions

Explanation:

The total momentum of the system of two carts is given by:

$p=m_1 v_1 + m_2 v_2$

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

Let's remind that v (the velocity) is a vector, so its sign depends on the direction in which the cart is moving.

We want to know if it is possible that the total momentum of the system can be zero, so it must be:

$p=0\\m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = -m_2 v_2$

From this equation, we see that this condition can only occur if v1 and v2 have opposite signs. Opposite signs mean opposite directions: therefore, the total momentum can be zero if the two carts are moving into opposite directions.

4 0

3 years ago