Explanation:
Gravitational potential energy = mgh = (5)(9.81)(7) = 343.35J.
1) <span>A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.
2) </span><span>anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.
3) </span><span>during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA.
4) </span><span>a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.
5) </span><span>a partial lunar eclipse begins when the moon first touches earth's -->
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.
6) </span><span> a point at which the moon crosses earth's orbital plane is called a(n) -->
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".</span>
Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

Let d is the distance covered by car. Using second equation of motion as :

So, the car will cover a distance of 778.05 meters.
Answer:

Explanation:
Assuming the light takes essentially no time to reach you, the distance at which the lightning occurred can be calculated by multiplying the speed of sound by the time it takes to hear the thunder:

According to another source this is what I got
<span>0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m )
</span>Hope it helps