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3241004551 [841]
3 years ago
10

Q: Why are distances in space often measured in light years?

Physics
2 answers:
marin [14]3 years ago
6 0

Answer:

the answer is b

Explanation:

Hope it helps :)

nadezda [96]3 years ago
3 0

B) Distances in space are so great that a large unit is needed
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A cat is sleeping on the floor in the middle of a 2.8-m-wide room when a barking dog enters with a speed of 1.40 m/s. As the dog
marysya [2.9K]

Answer:

the cat is  0.4238 m in front of the dog as it leaps through the window

Explanation:

Given that;

acceleration a = 0.85 m/s²

speed v = 1.40 m/s

the cat is at rest, so initial velocity u = 0  

we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;

using the second equation equation of motion;

s = ut + 1/2 at²  

we substitute

2.8/2 = 0×t + 1/2 × 0.85 × t²

1.4 = 0.425t²

t = √( 1.4 / 0.425 )

t = 1.81497 sec

now, at acceleration 0.10 m/s²

the dog has to cover the distance;

s = ut + 1/2 at²  

s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²

s =  2.540958 - 0.1647

s = 2.3762 m  

The cant in front of the dog as it leaps through the window;

distance = 2.8 m - 2.3762 m

distance = 0.4238 m

Therefore, the cat is  0.4238 m in front of the dog as it leaps through the window

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3 years ago
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denis-greek [22]

Answer:

D

Explanation:

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3 years ago
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B)
Vlad1618 [11]

Answer:

a) 16m/s b) 192m

Explanation:

v1=32m/s a=-2m/s^2 t=8s v2=? d=??

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v2= 32m/s + -2m/s^2 * 8s

v2= 32m/s + -16m/s

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b) v2^2=v1^2 + 2ad

rearranging

v2^2-v1^2=2ad

v2^2-v1^2/2= a d

v2^2-v1^2/2a=d

16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d

d=192m

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2 years ago
You are pushing a heavy box across a rough floor. when you are initially pushing the box and it is accelerating, (a) you exert a
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The answer is C. If the box is accelerating, that means that the amount of force you are exerting is greater than the force of the box.

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