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:
e = 1.21 mV
Explanation:
given,
length of rod = 10 m
height of drop = 4.89 m
Earth’s magnetic field = 12.4 µT
acceleration of gravity = 9.8 m/s²
velocity of the beam
v = 9.79 m/s
emf of the beam
e = B l v
e = 12.4 x 10⁻⁶ x 9.79 x 10
e = 1.21 x 10⁻³ V
e = 1.21 mV
Answer:
k = 44000 N/m
Explanation:
Given the following data;
Maximum gravitational potential energy = 770 J
Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J
Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m
To find the spring constant, k;
The elastic potential energy of an object is given by the formula;
Substituting into the equation, we have;
Cross-multiplying, we have;
k = 44000 N/m