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2 years ago

Paul is driving his car on a sunny afternoon. Which glasses should he ideally wear while driving?

1 answer:

8 0

$\alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha \alpha$ Answer:

Explanation:

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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0

2 years ago

Read 2 more answers

Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going

DerKrebs [107]

Answer:

The impact force will be same for both the cases.

Explanation:

The rate of change of momentum is known as the Impulse and is given by:

$I = \frac{\Delta p}{\Delta t}$

where

I = Impulse

$\Delta p = change in momentum$

$\Delta t = time interval$

Now,

In first case both the cars are identical and have same velocity and in the second case, the wall is stationary.

Also, in both the cases the car does not bounces off the things it hit.

Thus

$\Delta p = 0 - m\times v = - mv$

Thus

Impact force, $F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}$

Therefore, impact force is same for both the cases.

5 0

2 years ago

Read 2 more answers

I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =

Helen [10]

Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)

7 0

2 years ago

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.