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Blababa [14]
3 years ago
14

3. Why is static electricity not useful as a power

Physics
1 answer:
shtirl [24]3 years ago
3 0
The answer is B, because all energy is released at once in static electricity.

There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.
You might be interested in
M+2-4=3 solve and get 50 pts
Katen [24]
Hey!

First, let's write the problem.
M+2-4=3
Subtract the numbers, we would do the following operation, 2-4=-2
M-2=3
Add 2 to both sides.
M-2+2=3+2

This tells us that our final answer would be,
M=5

Thanks!
-TetraFish
3 0
3 years ago
Read 2 more answers
At the beginning of a basketball game, the referee tosses the ball straight up with a speed of 4.6m/s. A player cannot touch the
maria [59]

Answer:

t=0.47s

Explanation:

the ball has uniformly accelerated movement due to gravity

Vo=initial speed=4.6m/s

g=gravity=-9.8m/s^2

Vf=final speed=0, the player must wait for the ball to stop. so the final speed will be 0

we can use the following ecuation

T=(Vf-Vo)/g

T=(0-4.6)/-9.8m/s^2

T=0.47s

8 0
3 years ago
a pelican flying along a horizontal path drops a fish from a height of 5.4m. the fish travels 8.0m horizontally before it hits t
oksian1 [2.3K]

Answer:

7.0 m/s

Explanation:

6 0
3 years ago
A pitcher exerts a force on a baseball that is 30 times the balls weight. How fast is the pitcher accelerating the ball?
iVinArrow [24]

Answer:

 Pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

Explanation:

  Force applied on baseball = 30 times weight of the ball.

   Weight of ball = mg, where m is the mass of ball and g is acceleration due to gravity value.

  We have force applied is also equal to product of mass and acceleration.

                            F = ma = 30 x mg

                                 a = 30g

   So, pitcher is accelerating the ball at 30 times of acceleration due to gravity = 294 m/s²

8 0
3 years ago
a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. th
Savatey [412]

Answer:

15.7 m

Explanation:

The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

v_x = v cos \theta

where

v = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle between the direction of v and the horizontal

Substituting,

v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s

We know that the projectile hits the ground in a time of

t = 2.08 s

so the horizontal distance covered is

d = v_x t = (7.55 m/s)(2.08 s)=15.7 m

8 0
3 years ago
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