The electric field 30cm from a van de Graaff generator is measured to be 28,300N/C. What is the charge of the van de Graaf?

pogonyaev

<h2>Law 1:</h2><h3>An object already in motion stays in motion, unless acted upon by a force.</h3><h3 /><h2>Law 2:</h2><h3> $f=ma$ </h3><h3>f = forces on an object</h3><h3>m = mass of that object</h3><h3>a = acceleration of that object</h3><h3 /><h2>Law 3:</h2><h3>Everything has an equal and opposite reaction.</h3><h3 /><h3>Hope this helps!</h3>

5 0

3 years ago

PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!

gladu [14]

Here given that x is inversely depends of y

so as we increase the value of y so due to inverse dependency it will decrease the value of x

So here we can also say that when x inversely depends on y

so the product of x and y will remain constant here

so here the graph should be like this that if we increase the quantity on x axis then it will decrease the other quantity on y axis

<u><em>So here best appropriate graph must be option A</em></u>

3 0

3 years ago

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

6 0

3 years ago

A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g

Sliva [168]

To solve for force, you need to get the product of mass and acceleration.
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

$a = \frac{vf-vi}{t}$

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.

So we can put it into our formula now:

$a = \frac{0m/s-50m/s}{20ms}$

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.

$20ms$ x $\frac{1s}{1000ms}$ = $\frac{20s}{1000} = 0.02s$

So your new time is 0.02s. Now we put this time into the formula:

$a = \frac{0m/s-50m/s}{0.02s}$
$a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2}$

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.

So now we have our acceleration. Now using this, we can get our force.

$F= ma$

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is $kg.m/ s^{2}$ but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.

$150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg$

Our new mass is 0.150kg.

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 $m/ s^{2}$

Now that all our units match, we can put that into our formula:

$F= ma$
$F= (0.150kg)(-2,500m/s^{2})$
$F = -375kg.m/ s^{2} or -375N$

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.

4 0

3 years ago

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

expeople1 [14]

Answer: $56.87m/s^{2}$