Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.


W is the work done by the friction.







v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.
Answer:
3.62 V
Explanation:
L = 80 cm = 0.8 m
f = 15 rps
B = 60 m T = 0.060 T
ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s
v = r ω
here, r be the radius of circular path. Here r = length of rod = L
v = 0.80 x 94.2 = 75.36 m/s
The motional emf is given by
e = B v L = 0.060 x 75.36 x 0.8 = 3.62 V
Thank you for posting your question here at brainly. I hope the answer will help. Below are the choices that can be found elsewhere:
<span>A. 1.5 * 10^3 Watts
B. 7.3 * 10^2 Watts
C. 3.5 * 10^2 Watts
D. 2.5 * 10^2 Watts
</span>
<span>Work = force*displacement = 10^2*87 = 8,700 joule
Power = work/time = 8,700/6 = 1.45*10^3 (rounded up to 1.5 kw). The answer is A. </span>
Explanation:
It is given that,
Speed of a wave, v = 251 m/s
Wavelength of the wave, λ = 5.1 cm = 0.051 m
(1) The frequency of the wave is given by :



(2) Angular frequency of the wave is given by :



(3) The period of oscillation is given by T as :


T = 0.000203 seconds
or
T = 0.203 milliseconds
Hence, this is the required solution.
Answer:
W = 172.5 J
Explanation:
given,
mass of the fruit crate = 14.5 kg
initial velocity to lift = 0.500 m/s
increase in the tension = 150 N
lift of crate = 1.15 m
work done by the tension = ?
work done = force x displacement
W = F s cos θ
θ = 0°
W = F s x cos 0
W = 150 x 1.15 x 1
W = 172.5 J
Work done on the crate by the tension force = W = 172.5 J