Answer:
60000 J
Explanation:
Assuming the force is applied parallel to the displacement of the elephant, the work done to move it across the floor is

where
F = 2000 N is the force applied
d = 30 m is the displacement of the elephant
Substituting the numbers into the formula, we find

Answer:
a. 
b. 
c. 
Explanation:
First, look at the picture to understand the problem before to solve it.
a. d1 = 1.1 mm
Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:
To solve the equations we have to convert all units to those of the international system. (mm→m)

μ0 is the constant of proportionality
μ0=4πX10^-7 N*s2/c^2
b. d2=3.6 mm
Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:
J: current density
c: outer radius
b: inner radius
The cilinder's current is negative, as it goes on opposite direction than the wire's current.




c. d3=7.4 mm
Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.
Answer: static stretching
Explanation:
e.g rubberband
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass
kg, charge +e =
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed
m/s. The proton comes momentarily to rest at a distance
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are
m apart?
Explanation:
The given data is as follows.
Mass of proton =
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
=

where, 
U = 
Putting the given values into the above formula as follows.
U = 
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is
.