Answer:
F = 69.5 [N]
Explanation:
We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

where:
N = normal force [N]
miu = friction coefficient
f = friction force = 22 [N]
Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

where:
F = force exerted [N]
f = friction force [N]
m = mass = 95 [kg]
a = acceleration = 0.5 [m/s²]
Now replacing:
![F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]](https://tex.z-dn.net/?f=F%20-%2022%20%3D%2095%2A0.5%5C%5CF%20%3D%2047.5%20%2B%2022%5C%5CF%20%3D%2069.5%20%5BN%5D)
Answer:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings
Answer:
r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.
Explanation:
From a sum of forces:
where Ff = μ * N and 
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
(eq2)
Solving for r:

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E

Therefore, the maximum speed is 0.20 m/s