3 years ago

If a car is traveling at 35 mph how far will the car travel in 2.5 hours

Physics

1 answer:

Vanyuwa [196]3 years ago

4 0

Answer:

87.5 miles

Explanation:

d = s * t

d = 35 * 2.5

d =87.5

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What benefits does preforming this investigation in the physical world have over the computer simulation?

mezya [45]

Explanation:

Can be safer and cheaper than the real world. Able to test a product or system works before building it. Can use it to find unexpected problems. Can speed things up or slow them down to see changes over long or short periods of time.

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3 years ago

4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?

True [87]

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

509 J = 12x10xH

509 J = 120xH

H = 509/120

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Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.

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3 years ago

13. You push with 56 N on a 15-kg box, and there is a 23-N force of friction. How fast will the box accelerate?

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Answer:

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In the attached image, we can see the free body diagram. And using the second law of Newton it will be possible to find the acceleration of the box.

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4 years ago

What kind of energy do most electric inventions give off?

krek1111 [17]

Answer:

Electricity?

Explanation:

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3 years ago

A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th

vladimir1956 [14]

Answer:

Force, $F=2.27\times 10^4\ N$

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

$F.d=\dfrac{1}{2}m(v^2-u^2)$

Finally, it stops, v = 0

$F.d=-\dfrac{1}{2}m(u^2)$

$F=\dfrac{-mu^2}{2d}$

$F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}$

F = -22713.92 N

$F=2.27\times 10^4\ N$

So, the magnitude of the force that stops the bullet is $2.27\times 10^4\ N$

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3 years ago