Ask question

Ask question

All categories

2 years ago

14

Examine the motion map. One animal is an antelope that is already running. The other is a cheetah that starts running after the

antelope passes it. Does A or B represent the motion of the cheetah?

1 answer:

weqwewe [10]2 years ago

6 0

Answer:

A although am not seeing the cheetah

You might be interested in

The change in momentum of an object is equal to the ____________ that acts on it.

meriva

Answer : The change in momentum of an object is equal to the impulse that acts on it.

Explanation :

Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.

The formula of change in momentum is,

$\Delta p=m\times \Delta v$

Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of an object.

$J=F\times \Delta t$

Proof :

$J=F\times \Delta t\\\\J=(m\times a)\times \Delta t\\\\J=m\times (a\times \Delta t)\\\\J=m\times \Delta v=\Delta p$

Hence, the change in momentum of an object is equal to the impulse that acts on it.

3 0

3 years ago

Read 2 more answers

Describe a situation when you might travel at a high velocity bit with low acceleration

iren [92.7K]

I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy. They start to close.
I try to stay awake, but it's no use.
I am so warm and comfortable and sleepy,
and I have just finished my dinner.
Finally I can't help it. Resistance is futile.
I give up, and fall deep asleep.
My head rests back against my soft, comfy seat.

My seat is in row 26 on the airplane I'm flying in
to visit my grandmother on the coast.
We are cruising at 560 miles an hour, bearing 280°,
at flight level 320 .
The temperature outside my window is -60°F .

5 0

3 years ago

What best describes a sound wave? transverse and electromagnetic longitudinal and mechanical longitudinal and electromagnetic tr

GalinKa [24]

Sound needs medium to travel and it can not travel without medium

so sound wave is a travelling wave

now we also know that sound wave propagate in form of rarefaction and compression.

So all medium particles travel in the direction of wave only

so it is a longitudinal wave also

so correct answer will be

<em>mechanical longitudinal </em>

7 0

3 years ago

Read 2 more answers

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

A car accelerate from a stop to 27 m/sec in 5 seconds. What was the car's acceleration?

Furkat [3]

The acceleration should 5.4 m/s^2

4 0

3 years ago