a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Answer:
The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.
Explanation:
Given that,
Velocity = 270 m/s
Acceleration = 9.0g s
We need to calculate the altitude
Using formula of centripetal acceleration
Where, v = velocity
r = altitude
a = acceleration
Put the value into the formula
Hence, The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.
Answer:
112.06 kg - Thats heavy !
Explanation:
Let's do force balance here. Let the object of our interest be George. The forces acting on him are the tension in the upward direction, his weight in the downward direction and the centrifugal force in the downward direction. Considering the upward and downward directions on the y-axis and f=given the fact that George doesn't move up or down, the forces are balanced along the y-axis. Hence doing force balance:
magnitude of forces upward =magnitude of forces downward
i.e., Tension(T) = Weight(mg) + Centrifugal force (mv²/r)
where: 'm' is the mass of George, g is the acceleration due to gravity (9.8 m/s²). v is the speed with which George moves (14.1 m/s) and r is the radius of the circle in which he's moving at the instant (Here since he's swinging on the rope, he moves in a circle with radius as the length of the rope and hence r=7.3m).
therefore, T = m (9.8 + (14.1)²/7.3) = 4150 N
Therefore, m = 112.06 kg
The answer is....
214.28
which would be 214.3 or 214
Answer:
F = 2553.6 N
Explanation:
v² = u² + 2as
0² = 16² + 2a4.0
a = -32 m/s²
F = ma = 79.8(32) = 2553.6 N
