If Planet Y orbits the Sun but does not rotate on its axis at all, how long would a day be on Planet Y?

A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)

castortr0y [4]

Answer: D.) 39,200 J

Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height. In this case m = 400 kg, g = 9.8, h = 10 m thus:

$P.E.=(400kg)(9.8\frac{m}{s^2} )(10m)=39,200 J$

P.E.= 39,200 Joules

7 0

3 years ago

Read 2 more answers

Air in the atmosphere is heated by the ground this warm air then rises and cooler air falls this is an example of what type of p

Alona [7]

The answer for both is ‘B’

8 0

3 years ago

(100 pts) Why can't you use distance divide by time to calculate the instantaneous speed?

Fiesta28 [93]

Answer:

Instantaneous speed means speed at any instant

that means Speed is changing with time

You know speed is distance/time

So that means distance is also changing with time

So we take infinitesimal small distance per infinitesimal small time As we assume speed is constant in infinitesimal small time dt

So, we take speed = ds/dt

ds = infinitesimal small distance

dt = infinitesimal small time

As its ratio is equal to speed at any instant

Note : We are taking infinitesimal small distance

But :) we are taking infinitesimal small time also

As you know if denominator is small fraction is large So fraction always give large value

So it's not O ( this makes confuse to most of students)

So, thanks

Good question

Keep thinking like this :)

4 0

4 years ago

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Which of Newton's laws of motion best illustrates the principle of inertia?

andrew-mc [135]

Newton's first law of motion best illustrates the principle of inertia

5 0

3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$