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2 years ago

If the electric field is zero everywhere inside a region of space

1 answer:

5 0

My Answer: <span>If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.</span> If this is a true/false question, the correct answer would be false.

Hope I helped! :D

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A doctor pushes the plunger of syringe down and then pulls it up to a draw liquid into a syringe? give reason please help me wit

ruslelena [56]

Answer:

When the doctor has the syringe, it is full of air. So, now after the doctor pushes the plunger, hence the air gets released into the medicine container. After this the doctor then takes the plunger and pull it back. Now , the air gets pulled up back into the syringe, but not only does the air come in but also the medicine because of the pressure build up.

If my answer helped, please mark me as the brainliest!!

Thank You!

4 0

2 years ago

An overtone that's a whole number multiple of the fundamental frequency of a string is called a

alina1380 [7]

The correct answer in this case is B. Harmonic.

The remaining answers refer to other parts of a musical composition, and some such as the pitch can even be found in the study of human voice.

6 0

2 years ago

A student jobs around a square park two times. Starting and ending at the gate to the park. The square jogging track is 40 meter

olga nikolaevna [1]

The answer would be the first one

3 0

2 years ago

Read 2 more answers

You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler

scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge, equate electrostatic force to weight

we have F = W

so $\frac{KQq}{r^2} =mg$

putting all value in equation,

$=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8$

solving for q

$q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}$

or q=1.4*10^{-9}C

5 0

2 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

2 years ago