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.
Use conservation of momentum ;
m1u1 + m2u2 = m1v1 + m2v2
1200×15.6 + 0 = 2700v
v = 18720/2700
v = 6.933 or ~ 7 m/s
It is important to note that the electromagnetic spectrum has a variety of wavelength and frequency of light in it. Some lights we can see, while others are not visual to our naked eye. It is actually very important to determine the kind of light as different lights have different wavelengths and frequencies. some lights are of very high frequency like the gamma rays, while others are of far lower frequency. <span />
Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart
v is the velocity of jet
Q = A V
v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,
F n = 2295 N
now force along x direction
Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW
