Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
Answer: a) 19.21m b) 3.92secs
Explanation:
a) Maximum height reached by the object is the height reached by an object before falling freely under gravity.
Maximum height = U²/2g
U is the initial velocity = 19.6m/s
g is acceleration due to gravity = 10m/s²
Maximum Height = 19.6²/2(10)
H = 19.21m
b) The time elapsed before the stone hits the ground is the time of flight T= 2U/g
T= 2(19.6)/10
T = 39.2/10
Time elapsed is 3.92secs
Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
Hence, this is the required solution.
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