3 years ago

If you push on a railroad boxcar with a force of 200 N and it doesn’t move, you can conclude that

Physics

1 answer:

Nitella [24]3 years ago

4 0

You can conclude that the force of static friction acting on the boxcar is equal to 200N while you're pushing on it.

You might be interested in

Why is sustainable use important?

marusya05 [52]

To make sure that we have enough resourses to sustain ourselves. You see, if we use up, say, all water on earth, then obviously we won't have any more and we'll die.

4 0

3 years ago

Read 2 more answers

Two condition required for work done

mixas84 [53]

1) There must be a force
2) There must be displacement

5 0

2 years ago

Which type of radiation travels at the speed of light and penetrates matter easily?

Alina [70]

I think the answer is Gamma Rays

5 0

3 years ago

Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

8 0

3 years ago

How fr do you think you would go on a car while sneezing for 2.5 seconds ?

arsen [322]

depends on the speed limit

4 0

3 years ago

Read 2 more answers