This is EXACTLY the same scenario as the skydiver jumping
out of the airplane, except the whole thing is turned on its side.
==> The skydiver leaves the airplane.
The force of gravity on him (his weight) makes him accelerate down.
But the air resists his downward motion.
The faster he falls, the more UPWARD force the air exerts on him.
The more upward force the air exerts, the less he accelerates down.
When his falling speed is great enough, he stops accelerating, and
falls with a constant speed. He calls that speed his 'terminal velocity'.
==> The submarine turns on its engines, at maximum power.
The force of the engines makes the sub accelerate forward.
But the water resists its forward motion.
The faster it moves, the more BACKWARD force the water exerts on it.
The more backward force the water exerts, the less it accelerates forward.
When the forward speed is great enough, it stops accelerating, and moves
with a constant speed. I don't know if they use the same term in submarines,
but you might say that speed is the 'terminal velocity' in water.
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Calculating the average speed is simple using the formula <span>speed = distance/time</span>
Answer:
97 s
Explanation:
Given:
Δx = 9600 m
v₀ = 198 m/s
v = 0 m/s
Find: t
Δx = ½ (v + v₀) t
9600 m = ½ (0 m/s + 198 m/s) t
t = 97 s
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen
Neglect air resistance.
The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J
The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J
Answer: d. It has decreased by 200 J
