Answer:
82.8 J
Explanation:
The work done to raise the crate is ...
PE = Mgh = (3 kg)(9.8 m/s^2)(2 m) = 58.8 J
The kinetic energy added to send the box flying is ...
KE = (1/2)Mv^2 = (1/2)(3 kg)(4 m/s)^2 = 24 J
So, the total work involved in this activity is ...
58.8 J +24 J = 82.8 J
Answer:
Kf= 36 J
W(net) = 32 J
Explanation:
Given that
m = 2 kg
F= 4 N
t= 2 s
Initial velocity ,u= 2 m/s
We know that rate of change of linear momentum is called force.
F= dP/dt
F.t = ΔP
ΔP = Pf - Pi
ΔP = m v - m u
v= Final velocity
By putting the values
4 x 2 = 2 ( v - 2)
8 = 2 ( v - 2)
4 = v - 2
v= 6 m/s
The final kinetic energy Kf
Kf= 1/2 m v²
Kf= 0.5 x 2 x 6²
Kf= 36 J
Initial kinetic energy Ki
Ki = 1/2 m u²
Ki= 0.5 x 2 x 2²
Ki = 4 J
We know that net work is equal to the change in kinetic energy
W(net) = Kf - Ki
W(net) = 36 - 4
W(net) = 32 J
Answer:
a =45 m/s2
t = 2 seconds
Explanation:
Hi, to answer this question we have to apply the next formula:
v^2 = u^2 +2 a d
Where:
v = final velocity = 90 m/s
u = initial velocity = 0 m/s (shots from rest)
a = acceleration (m/s2)
d = distance = 90m
90^2 = 0^2 + 2a(90)
Solving for a:
8,100= 180 a
8,100/180 = a
a = 45 m/s^2
For time:
v = u + at
90 = 0 + 45t
90/45=t
t =2 seconds
Answer:
Low satellite has high orbital velocity
Explanation:
let v be the orbital speed of the satellite orbiting at a height h is given by

where, M be the mass of planet, r be the radius of planet and h be the height of planet from the surface of planet.
here we observe that more be the height lesser be the orbital velocity.
So, a satellite which is at low height has high orbital velocity.
Answer:
see below
Explanation:
acceleration = Δv /Δt
for this situation 60 / 10 = 6 m/s^2
B) vf = vo + at
vf = 0 + 6(3) =<u> 18 m/s after 3 seconds </u>
<u />
C) vf = at
60 = 6 ( t) t = 10 seconds ( actually, this was given)
d = 1/2 a t^2
= 1/2 (6) (10)^2 = <u>300 m </u>
<u />