The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?
1 answer:
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Ok so we know:
The time (t) is 18seconds
The acceleration (a) is 2.2m/s2
The displacement (r) is 660
Using the equation
With 'u' being the initial velocity we want, we get:
So:
So:
So the original/initial velocity was 16.8666 or 16.87 m/s
Hope this helped
The time (t) is 18seconds
The acceleration (a) is 2.2m/s2
The displacement (r) is 660
Using the equation
With 'u' being the initial velocity we want, we get:
So:
So:
So the original/initial velocity was 16.8666 or 16.87 m/s
Hope this helped
Answer:
the location of the center of gravity for the entire body is 1.08 m
Explanation:
Given the data in the question;
w1 = 458 N, y1 = 1.34 m
w2 = 120 N, y2 = 0.766 m
w3 = 89.8 N, y2 = 0.204 m
The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.
so,
= (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )
so we substitute in our values
= (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )
= 723.9592 / 667.8
= 1.08 m
Therefore, the location of the center of gravity for the entire body is 1.08 m
The answer is 41.25 mph.
The average speed (v) can be expressed as:
v = D/T
where
D is total distance: D = d1 + d2 + d3 + d4
T is total time: T = t1 + t2 + t3 + t4
Now let's calculate distances using the formula: d = v * t:
d1 = v1 * t1 = 20.0 * 0.500 = 10 mi
d2 = v2 * t2 = 40.0 * 1.00 = 40 mi
d3 = v3 * t3 = 30.0 * 0.500 = 15 mi
d4 = v4 * t4 = 50.0 * 2.00 = 100 mi
Thus, the average speed is:
The average speed (v) can be expressed as:
v = D/T
where
D is total distance: D = d1 + d2 + d3 + d4
T is total time: T = t1 + t2 + t3 + t4
Now let's calculate distances using the formula: d = v * t:
d1 = v1 * t1 = 20.0 * 0.500 = 10 mi
d2 = v2 * t2 = 40.0 * 1.00 = 40 mi
d3 = v3 * t3 = 30.0 * 0.500 = 15 mi
d4 = v4 * t4 = 50.0 * 2.00 = 100 mi
Thus, the average speed is: