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Brut [27]
3 years ago
6

The potential difference across two plates is 20.0 V. If the plates are 0.00500 m apart, what is the electric field between them

? Unit= N/C
Physics
2 answers:
Ksju [112]3 years ago
7 0

Answer:

4,000

Explanation:

Trust brother

GuDViN [60]3 years ago
7 0

Answer:

4,000

Explanation:

potential difference (V) = electric field (E) x separation between plates (d)

20=.00500x

x=4,000

You might be interested in
A 5.00 kg pendulum swings back and forth. At the top of its arc, it reaches a height of 0.36 m. What is the velocity of the pend
ella [17]
Tools we'll use:

--     Gravitational potential energy = (mass) x (gravity) x (height)

--     Kinetic energy (of a moving object) = (1/2) (mass) x (speed)²

When the pendulum is at the top of its swing,
its potential energy is

                             (mass) x (gravity) x (height)

                         =  (5 kg) x (9.8 m/s²) x (0.36 m)

                         =      (5 x 9.8 x 0.36)        joules

                         =            17.64 joules .

Energy is conserved ... it doesn't appear or disappear ...
so that number is exactly the kinetic energy the pendulum
has at the bottom of the swing, only now, it's kinetic energy:

                   17.64 joules  =  (1/2) x (mass) x (speed)²

                   17.64 joules  =  (1/2) x (5 kg) x (speed)²

Divide each side by 2.5 kg:

                   17.64 joules / 2.5 kg  =  speed²

Write out the units of joules:

                   17.64 kg-m²/s²  /  2.5 kg  =  speed²

                   (17.64 / 2.5) (m²/s²)  =  speed²

                       7.056 m²/s²  =  speed²

Take the square root
of each side:              Speed = √(7.056 m²/s²)

                                           =    2.656 m/s .

Looking through the choices, we're overjoyed to see
that one if them is  ' 2.7 m/s '.  Surely that's IT !
_______________________________

Note:
The question asked for the pendulum's 'velocity', but our (my) calculation
only yielded the speed.

In order to describe a velocity, the direction of the motion must be known,
and the question doesn't give any information on exactly how the pendulum
is hanging, and how it's swinging. 

We know that at the bottom of its swing, the motion is completely horizontal,
but we have no clue as to what direction.  So all we can discuss is its speed.
6 0
3 years ago
how fast would a(n) 73 kgkg man need to run in order to have the same kinetic energy as an 8.0 gg bullet fired at 430 m/sm/s ?
Mars2501 [29]

Answer:

an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,

v=4.0m/s

Explanation:

Typically, the mathematical formula for kinetic energy is as follows:

KE=\frac{1}{2} *m*v^{2}

K.E= kinetic energy

M=mass

V= speed

so,

KE=1/2*0.008*(420)*420

K.E=672.4J

 Therefore,

v*v=672.4/85

v*v=15.821

v=4.0m/s

Therefore

v=4.0m/s

What is Kinetic Energy?

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

To know more about Kinetic Energy? Visit to: -

brainly.com/question/22174271

#SPJ4

8 0
1 year ago
You go rock climbing with a pack that ways 79 N and you reach a height of 30 m. How much work did you do to lift your pack? If y
Vanyuwa [196]
When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).
In order to climb, you need to work against work done by gravity on the pack.
Hence work done by you = work done by gravity on pack 
                                        = Force x displacement = 70 x 30 = 2100 J.
 
4 0
3 years ago
PLEASE HELP <br> GREATLY APPRECIATED IF YOU DO
Yuliya22 [10]

Answer:

a. 38400j

Explanation:

p= mgh

2400×10×1.6

3 0
3 years ago
A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime
Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0
3 years ago
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