Answer:
The force exerted on the
is 
Explanation:
From the question we are told that
The area is 
The magnitude of charge placed on them is 
The charge placed between the plate is 
The electric field generated around the plate is mathematically represented as

Substituting values


The force exerted the charge
is mathematically represented as

Substituting values


Answer:
Earth: 22.246 N
Moon: 3.71 N
Jupiter: 58.72 N
Explanation:
The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.
To calculate anything, we need to convert to standard measurements.
5.00 lbs = 2.27 kg
On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N
Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N
Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N
Answer:

Explanation:
Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

Where Vf is the final velocity of the object, (in our case 80 m/s)
Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)
and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.
Therefore we have:

Notice that the units of acceleration in the SI system are
(meters divided square seconds)