Answer:
The charge carried by the droplet is 
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude
points straight down and if the mass of the droplet is 
We need to calculate the acceleration
Using equation of motion

Put the value into the formula



We need to calculate the charge carried by the droplet
Using formula of electric filed


Put the value into the formula


Hence, The charge carried by the droplet is 
Answer:
B. Acceleration in the direction of motion speeds you up
Explanation:
Acceleration is defined as when something gains speed. For example, when a car speeds up. Deceleration is when the car slows down, and looses speed. When defining these terms, think of a car going faster, then slowing down at a red light.
The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.
When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.
The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.
Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.
If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

Therefore,

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.
Answer:
(θ) = 60°
Explanation:
Given:
Speed of canoe Vc = 2 m/s
Speed of River Vr = 1 m/s
Computation:
Vc (Cosθ) = Vr
2 (Cosθ) = 1
(Cosθ) = 1 / 2
(Cosθ) = (Cos60)
(θ) = 60°
Complete Question
A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Answer:

Explanation:
From the question we are told that:
Angular velocity 
Mass 
Diameter d 
Off Time 
Oscillation at Power off 
Generally the equation for Angular displacement is mathematically given by




Generally the equation for Time to come to rest is mathematically given by



Therefore Angular displacement is

