Answer:
The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.
We know V=IR (Ohm's law).
We are given R=180Ω and I=0.1A, then V=(0.1AΩ)(180Ω). Therefore
V=18V
Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:


The problem tells us the radius of the planet is twice that of the radius on earth, so:

If we substituted that into the gravity of the planet equation we would end up with the following formula:

Which yields:

So we can now compare the two gravities:

When simplifying the ratio we end up with:

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton,
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E

where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd




Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
I think Option B, i’m sorry if that’s wrong.