A dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.
s = vt - 1 / 2 at²
s = Displacement
v = Final velocity
t = Time
a = Acceleration
s = 5 m
t = 1 s
a = 10 m / s²
5 = ( v * 1 ) - ( 1 / 2 * 10 * 1 * 1 )
5 = v - 5
v = 10 m / s
The equation used to solve the given problem is an equation of motion. In a free fall motion, usually air resistance is not considered for easier calculation. If air resistance is considered acceleration cannot be constant throughout the entire motion.
Therefore, a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.
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Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
She could text/email,report it to them personally,post up flyers,tell one person to spread it around
To find average speed, we divide the distance of travel (in this case, 400 metres) by the time she took, 32 seconds. Therefore: 12.5 seconds is her average speed.