Answer:
work done by the force, W = -147.97 J
Given:
displacement of the box, s = 4.73 m
angle between force and displacement, 
Force, F = 39.7 N
Solution:
Work done by the force can be expressed as the dot product of Force, F and displacement 's' is given by:
W = Fscos
W = 39.7
W = -147.97 J
where negative sign indicates that work is done on the block
Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
Answer:
27,000 m
450 m/s
Explanation:
Assuming the initial velocity is 0 m/s:
v₀ = 0 m/s
a = 15 m/s²
t = 60 s
A) Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²
Δy = 27,000 m
B) Find: v_avg
v_avg = Δy / t
v_avg = 27,000 m / 60 s
v_avg = 450 m/s
I think the answer is c but I’m not sure
Answer:
F = m a = m v / t where v is the change in velocity in time t
F = p / t since m v is equal to p
F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N
Or you can use the impulse equation
