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Jet001 [13]
2 years ago
10

What is the electric potential energy of a 4. 0 uc charge placed at that point where the electric potential at that point in spa

ce is 8 v?
Physics
2 answers:
dexar [7]2 years ago
6 0

The electric potential at that point in space is 8 v is 32uj

what is electric potential energy?

Potential energy is the energy within an object relative to its position and proximity to other objects within a field.

The electric potential of a point charge is V=kQ/rV=kQ/r , or V = k Q / rV=kQ/r . Electric potential is a scalar while the electric field is a vector. While the total electric field can be computed by adding individual fields as vectors, the voltage resulting from a collection of point charges can be calculated by adding voltages as integers.

It is decided to set the potential at infinity to zero.

but V for a point charge reduces with distance while decreasing with distance squared. Remember that the electric field E is a vector, but the electric potential V is a scalar with no direction.

The electric energy is (8×4)=32 uj

learn more about electric potential from here: https//brainly.com/question/28167853

#SPJ4

nexus9112 [7]2 years ago
3 0

The electric energy is 32 uj

The electric potential of a point charge is V=kQ/r, or V = k Q / r. Electric potential is a scalar while the electric field is a vector. While the total electric field can be computed by adding individual fields as vectors, the voltage resulting from a collection of point charges can be calculated by adding voltages as integers.

It is decided to set the potential at infinity to zero. E=Fq=kQr^2, but V for a point charge reduces with distance while decreasing with distance squared. Remember that the electric field E is a vector, but the electric potential V is a scalar with no direction.

The electric energy is (8×4)=32 uj

Learn more about electric energy  herebrainly.com/question/24403138

#4219.

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we would be at (4,0)
7 0
3 years ago
The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of
Aleks04 [339]

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}

(1)

y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s

(2)

y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s

(3)

y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

5 0
3 years ago
One method of determining the location of the center of gravity of a person is to weigh the person as he/she lies on a board of
andrey2020 [161]

Answer:

x = 1.018 m

Explanation:

given,

height of man = 190 cm

                       = 1.9 m

scale reading on left = 450 N

scale reading on the right = 390 N

Let center of gravity of man be x distance from feet, feet is on right side.

For system to be in equilibrium moment about center should be equal to zero.

∑M = 0

now,

450(1.9 - x ) - 390 × x = 0

450(1.9 - x ) = 390 × x

855 - 450 x = 390 x

840 x = 855

 x = \dfrac{855}{840}

 x = 1.018 m

hence, point of center of gravity from feet is equal to x = 1.018 m

7 0
4 years ago
a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)
rodikova [14]

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)

Now, using the given information, we have:

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Calculating, we have:

N_{x}=mg*Sin(\alpha)

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Hence, we have that the x-component of the normal force is equal to  <u>1706.45 N.</u>

Have a nice day!

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3 years ago
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