Answer: F = mg(1 + 4m / (½M + m))
Explanation:
"At this point seems" unclear. If the particle is at the top of the disc and angular velocity is negligible, then the force would equal the weight of the particle. F = mg
The more interesting question would be what force is needed to keep the particle attached when significant angular rotation has been achieved. The maximum point would be diametrically opposed to the starting point.
I will analyze it there
The potential energy will convert to kinetic energy
mgh = ½Iω²
mg(2R) = ½(½MR² + mR²)ω²
4mgR = R²(½M + m)ω²
ω² = 4mg / (R(½M + m))
With m at the lowest position, the force of attachment must support the weight of m and provide for the needed centripetal acceleration
F = m(g + ω²R)
F = m(g + 4mg / (R(½M + m))R)
F = mg(1 + 4m / (½M + m))
This is called a subduction zone. The two plates converge, but one is forced under the other.
Answer:
A. 
B. 
Explanation:
The frequency has an inversely proportional relationship with the concept of wavelength, the greater the wavelength, the lower the frequency. For electromagnetic waves, the frequency is equal to the speed of light, divided by the wavelength.
A.
B.
Answer:
The power dissipated is reduced by a factor of 2
Explanation:
The power dissipated by a resistor is given by:
where
I is the current
R is the resistance
by using Ohm's law, 
, we can rewrite the previous equation in terms of the voltage applied across the resistor (V):
In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have 
while V remains the same; substituting into the formula, we have:
so, the power dissipated is reduced by a factor 2.
Answer: C
the A4 V star is farther from the sun than F8 V
Explanation:
An A-type main-sequence star (A V) or A dwarf star is a main-sequence (hydrogen- burning) star of spectral type A and luminosity class stars of spectral type, the hot stars, have spectra characterized by very few spectral lines the A4 V star is farther from the sun than F8 V
