Answer:
h = 250 m
Explanation:
Given that,
The helicopter exerts 6480 N of force on the water while rising high enough to fly over a mountain.
The helicopter does 
of work on the water lifting it.
We need to find how high does it lift the water. Work done is given by :
So, it can lift the water to a height of 250 m.
Refer to the figure shown below.
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
Answer:
A homopolar motor is a direct current electric motor with two magnetic poles, the conductors of which always cut unidirectional lines of magnetic flux by rotating a conductor around a fixed axis so that the conductor is at right angles to a static magnetic field.
Explanation:
Answer:
Explanation:
We could use the following suvat equation:
where
s is the vertical displacement of the coin
v is its final velocity, when it hits the water
t is the time
g is the acceleration of gravity
Taking upward as positive direction, in this problem we have:
s = -1.2 m
And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:
where the negative sign means the direction is downward.
