Answer:
Maximum torque on the wire is 
Explanation:
It is given that,
Diameter of the wire, d = 11.1 cm = 0.111 m
Radius of wire, r = 0.0555 m
Magnetic field, 
Current, I = 5 A
We need to find the maximum torque on the wire. Torque is given by :

Torque is maximum when, 



or

So, the maximum toque on the wire is
. Hence, this is the required solution.
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
The gravitational potential energy is given by :
P = mgh
The kinetic energy of an object is given by :

As the ball reaches the bottom of the ramp, its potential energy decreases and kinetic energy increases.
It imply that, when the ball at the top most height, its gravitational potential energy is maximum and zero kinetic energy and when ball reaches the bottom of the ramp, it will have maximum kinetic energy and zero potential energy.
The AU ... Astronomical Unit ... used to be defined as the average distance between the Sun and Earth during the year.
Now it's defined as 149,597,870,700 meters exactly.
Answer:
Explanation:
Given
initial velocity 
At maximum height velocity of ball is zero
using

where v=final velocity
u=initial velocity
a=acceleration
s=displacement



time taken by the ball to reach the maximum height




At
height of ball is




i.e. ball is moving downward
height at 

