Answer:
Approximately 
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant 
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case, is the same as the change in the ball's height: 
. By assumption, this ball was dropped with no initial velocity. As a result, 
. Since the ball is accelerating due to gravity, 
.
.
In this case, would be the velocity of the ball just before it hits the ground. Solve for
.
.
You can estimate one more digit past the smallest division on the measuring device. If you look at a 10mL graduated cylinder, for example, the smallest graduation is tenth of a milliliter (0.1mL). That means when you read the volume, you can estimate to the hundredths place (0.01mL).
Answer:
375 ms
Explanation:
the frequency of metronome , f = 160 beats per minute
f = 160 /60 beats per sec
f = 2.67 beats /s
the period of a single beat , T = 1/f
T = 1/2.67 s
T = 0.375 s = 375 ms
the period of a single beat is 375 ms
Answer:
The answer is below
Explanation:
Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:
The masses and distances for this question is in common units, Therefore the result would be in ratios
a) 4 MEarth / 2 MSolar / 3 AU
The force (F) = (4 * 3) / 3² = 4/3
b) 1 MEarth / 1 MSolar / 1 AU
The force (F) = (1 * 1) / 1² = 1
c) 1 MEarth / 2 MSolar / 2 AU
The force (F) = (1 * 2) / 2² = 1/2
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