Answer:State the question so I can answer
Explanation:
Answer:
The magnitude of the net force is √2F.
Explanation:
Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

Then, it means that the net force acting on the test charge has a magnitude of √2F.
Answer: current I = 0.96 Ampere
Explanation:
Given that the
Resistance R = 60 Ω
Power = 55 W
Power is the product of current and voltage. That is
P = IV ...... (1)
But voltage V = IR. From ohms law.
Substitutes V in equation (1) power is now
P = I^2R
Substitute the above parameters into the formula to get current I
55 = 60 × I^2
Make I^2 the subject of formula
I^2 = 55/60
I^2 = 0.92
I = sqr(0.92)
I = 0.957 A
Therefore, 0.96 A current must be applied.
Answer:
The ball would hit the floor approximately
after leaving the table.
The ball would travel approximately
horizontally after leaving the table.
(Assumption:
.)
Explanation:
Let
denote the change to the height of the ball. Let
denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let
denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:
.
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was
.
The height of the table was
. Therefore, after hitting the floor, the ball would be
below where it was before leaving the table. Hence,
.
The equation becomes:
.
Solve for
:
.
In other words, it would take approximately
for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at
) until the ball hits the floor.
The ball was in the air for approximately
and would have travelled approximately
horizontally during the flight.