Am not really sure but what i see is D
Gravity
The moon doesn't smash into the earth because the gravity from the earth keeps the moon in orbit around it.
A) 1.05 N
The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:

where
P = 4.20 W is the power
F is the magnitude of the pulling force
v = 4.0 m/s is the speed of the wire
Solving the equation for F, we find

B) 3.03 T
The electromotive force induced in the circuit is:
(1)
where
B is the strength of the magnetic field
v = 4.0 m/s is the speed of the wire
L = 10.0 cm = 0.10 m is the length of the wire
We also know that the power dissipated is
(2)
where
is the resistance of the wire
Subsituting (1) into (2), we get

And solving it for B, we find the strength of the magnetic field:

Answer:
a) 4.0 rad/s2
Explanation:
- For rigid bodies, Newton's 2nd law becomes :
τ = I * α (1)
where τ is the net external torque applied, I is the rotational inertia
of the body with respect to the axis of rotation, and α is the angular
acceleration caused by the torque.
- At the same time, we can apply the definition of torque to the left side of (1), as follows:

where τ = external net torque applied by Fnet, r is the distance
between the axis of rotation and the line of Fnet, and θ is the
angle between both vectors.
In this particular case, as Fnet is applied tangentially to the disk, Fnet
and r are perpendicular each other.
- Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:
(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air (
)
(b) The plates can be close together to 1.7 mm with this applied voltage
<u>Explanation:</u>
Given data:
Dielectric strength of air = 
Distance between the plates = 2.00 mm = 
Potential difference, V = 
We need to find
a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not
b) the minimum distance at which the plates can be close together with this applied voltage.
The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

From the above, concluding that The electric field strength between two parallel conducting plates (
) does not exceed the breakdown strength for air (
)
b) To find how close together can the plates be with this applied voltage:
The formula would be,

Apply all known values, we get
