That relationship is as DIRECT as you can get.
The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.
Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)
N = the normal reaction, which is equal to
N = mg cos(12°)
R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)
F = the component of the weight acting down the incline. It is
F = mg sin(12°)
Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
Therefore, the static coefficient of friction is
μ = tan(12) = 0.213
Answer: μ = 0.21 (nearest tenth)
Answer:
Charge stored after insertion will be
Explanation:
We have given potential difference V =2.63 V
charge stored by the capacitor when without dielectric =
We know that , here is capacitance without dielectric
We have given
k = dielectric constant = 5.99
C = Capacitance with the dielectric = k = 3.55× =
Potential difference is due to the external charging source so it remains same
V = potential difference after insertion = 2.63 volts
New charge stored ,
Answer:
11.16 J
Explanation:
Elastic energy (E) stored in the tendon is given by 1/2ke^2
SPRINTERS
k = 31N/mm = 31N/mm × 1000mm/1m = 31,000N/m
e = 41mm = 41mm × 1m/1000mm = 0.041m
E = 1/2 × 31,000 × 0.041^2 = 26.06J
NON-ATHLETES
k = 31,000N/m
e = 31mm = 31mm × 1m/1000 = 0.031m
E = 1/2 × 31,000 × 0.031^2 = 14.90 J
Difference = 26.06 J - 14.90 J = 11.16 J
The magnitude will be =23.94 WS and the direction will be =98.48 m/s