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Mila [183]
2 years ago
14

A skateboader is accelerating forward how would his acceleration change if he had more mass

Physics
1 answer:
Masteriza [31]2 years ago
6 0
Answer: If the force stays the same, the acceleration would decrease
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Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps. 
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3 years ago
Explain why a diverging lens is used to correct nearsightedness (difficulty seeing objects far away).
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Answer:

so that it can diverge the light to make sure that it focused on the ratina and the image is formed.

Explanation:

nearsightedness is when the light is focused in front of the ratina and for an image to be formed in the eye, the light must be focused on the ratina so to correct that we use the diverging lenses so that it will diverge the light and allow the cornea and the lens to converge it so it is focused on the ratina.

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2 years ago
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A car of mass 150 kg is riding down at constant velocity. What is the normal force?
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Answer:

1500N

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constant velocity means acceleration = 0

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normal force = mg = 150 × 10 = 1500N

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3 years ago
The electric potential at points in an xy plane is given by V = (2.0 V/m2)x2 -(3.5 V/m2)y2. What are (a) the magnitude of the el
kenny6666 [7]
<h2>Well Formatted Question:</h2>

The electric potential at points in an xy plane is given by V = (2.0 V/m²)x² - (3.5 V/m²)y². What are:

(a) the magnitude of the electric field at the point (4.1 m, 2.8 m) and;

(b) the angle that the field there makes with the positive x direction.

<h2>Answer:</h2>

(a) 25.56 V/m

(b) 129.92°

<h2>Explanation:</h2><h2></h2>

Write the electric potential in vector form as follows;

V = (2.0)x² i  - (3.5)y² j              ------------------(i)

Where;

i and j are unit vectors in the x and y direction respectively

The electric field (E) is given as the negative derivative of electric potential (V) with respect to x and y as follows;

E = - derivative [V]

E = - ([2 * 2.0x] i - [2 * 3.5 y] j )

E = - (4.0x i - 7.0y j )

E = - 4.0x i + 7.0y j        -----------------------------(ii)

Now to get the electric field at point (4.1m, 2.8m), substitute the values of x = 4.1m and y = 2.8m into equation (ii) as follows;

E = -4.0(4.1) i + 7.0(2.8) j

E = - 16.4i + 19.6j        ---------------------(iii)

The magnitude of E ( | E |) is calculated as follows;

| E | = \sqrt{(-16.4)^2 + (19.6)^2}

| E | = \sqrt{(268.96 + 384.16)}

| E | = \sqrt{(653.12)}

| E | = 25.56V/m

Therefore, the magnitude of the electric field is 25.56V/m

(b) The direction, θ, of the electric field is given as

tan θ = (-19.6 / 16.4)

tan θ = - 1.195

θ = tan⁻¹ (-1.195)

θ = -50.08°

The negative sign shows that it is measured with respect to the -x axis.

To get the equivalent as measured with respect to the +ve axis, note that from equation (iii), it is evident that the electric field is at -i and +j counterclockwise from the +x-axis.

Therefore, add 180° to -50.08° to measure the direction counterclockwise from the positive x direction as follows;

true angle = 180°  - 50.08° = 129.92°

Therefore the angle that the field makes with the positive x axis is 129.92°

5 0
2 years ago
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