I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.
Answer:
This can be translated to:
"find the electrical charge of a body that has 1 million of particles".
First, it will depend on the charge of the particles.
If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C
Then the total charge of the body will be:
Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C
If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)
Q = 1.6*10^-13 C
But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.
Answer:
hello your question is not properly arranged attached below is the arranged table and solution
answer : attached table below
Explanation:
Given data:
02 molecules size = 10^-10m
smoke particles size = 0.3 mm
cloud droplets size = 20 mm
Rain droplets size = 3 mm
Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths
Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.
and For Mie scattering the wavelength is the same as the wavelength.
