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vlada-n [284]
3 years ago
7

A positive charge of 6.0 x 10-4 C is in an electric field that exerts a force of 4.5 x 10 -4 on it. What is the strength of the

electric field?
Physics
1 answer:
Yuki888 [10]3 years ago
6 0

Electric field is defined as force per unit charge.

So it is given by

F = q E

now we can find electric field by

E = \frac{F}{q}

E = \frac{4.5*10^{-4}}{6 * 10^{-4}}

E = 0.75 N/C

So field strength is 0.75 N/C.

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A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
If a cart goes around a turn at 20 km/h ,what remains constant
Vitek1552 [10]

Answer: 4.speed

Explanation:

In this case, we know that the cart remains at a constant 20km/h.

Now, one could say that "the velocity remains constant, because it always is 20km/h"

But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.

But the module of the velocity, the speed, remains constant at 20km/h.

Then the correct option is 4, speed.

7 0
2 years ago
How high is an object lifted if the total work done on it is 125 J and the force required to lift the object is 25 N? Use the eq
mixas84 [53]
The answer would be C. 5m
This is because to find d, you would need to divide W (125 J) by F (25 N).
Hope this helps!
8 0
3 years ago
E-mail is usually a poor choice for the distribution of sensitive electronic files because
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D. email is saved on the server that transfers is
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2 years ago
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A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma
Oksanka [162]

Answer:

Explanation:

The momentum of the first piece = m v =- m x 31 i kg m/s in - x direction  direction

The momentum of the second piece = -m x 31 j kg m /s in Y - direction

Total momentum = - 31 m( i + j )

To conserve momentum , the third piece must have momentum equal to this

and opposite to it

So momentum of the third piece  = 3m x V = 31 m ( i +j )

V = 31/3 ( i + j ) =

Magnitude of velocity V = √2 x 31/ 3  = 14.6 m / s

Its direction will be towards the vector i + j  ie 45° from x - axis in positive direction

4 0
3 years ago
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