If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final
1 answer:
5m/s
Explanation:
Given parameters:
Mass of ball = 0.1kg
Force on the ball = 5N
time taken = 0.1s
Unknown:
final speed of the ball = ?
Solution:
According to newton's second law "the net force on a body is the product of its mass and acceleration".
Force = mass x acceleration equation 1
Acceleration =
V is the final velocity
U is the initial velocity
T is the time taken
U = O since it is a stationary body;
a =
Input "a" into equation 1
F = m x
5 = 0.1 x
V = 5m/s
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<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>
Explanation:
a) Given S(t) = 76 + 128t − 16t²
s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
Displacement in 4 seconds = 332 - 76 = 256 ft
Time = 4 - 0 = 4 s
Average velocity in first 4 seconds is 64 ft/s upward
a) Given S(t) = 76 + 128t − 16t²
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft
Displacement in 4 seconds = 78 - 332 = -254 ft
Time = 4 - 0 = 4 s
Average velocity in second 4 seconds is 63.5 ft/s downward