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Scrat [10]
2 years ago
5

9. Suppose you ride your Segway to the library traveling at 0.5 km/min. It takes you 25

Physics
1 answer:
3241004551 [841]2 years ago
8 0

Answer:

12.5 km

Explanation:

0.5 x 25 = 12.5

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Atmospheric pressure is greater at the base of a mountain than at
aliina [53]

At the top of the mountain, when he tightens the cap onto the bottole, there is some water and some air inside the bottle.  Then he brings the bottle down to the base of the mountain.

The pressure on the outside of the bottle is greater  than it was when he put the cap on.  If anything could get out of the bottlde, it would. But it can't . . . the cap is on too tight. So all the water and all the air has to stay inside, and anything that can get squished into a smaller space has to get squished into a smaller space.

The water is pretty much unsquishable.

Biut the air in there can be <em>COMPRESSED</em>.  The air gets squished into a smaller space, and the bottle wrinkles in slightly.

8 0
2 years ago
Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass
LuckyWell [14K]
The answer is b.) the momentum before the collision is greater than the momentum after the collision
8 0
2 years ago
Read 2 more answers
What is energy converted to if it is not used to do work?
Romashka [77]

Answer:

the human body isn't very efficient at converting food into useful work. The human body is less than 5% efficient most of the time. The rest of the energy is converted to heat, which may or may not be useful, depending on how cool or warm a person wants to be.

Explanation:

3 0
2 years ago
What's the momentum (in kg-m/s) of a cannonball with a mass of 21 moving with a velocity of 25 m/s?
lbvjy [14]
P=mv
P=21*25=525 kg*m/s
8 0
3 years ago
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

t = 4 s

4 0
3 years ago
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