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3 years ago

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what is the electric potential energy of a charge that experiences a force of 4.6 x 10-5N when it is 5.8 x 10-5 m from the sourc

e of the electrical field

1 answer:

kvv77 [185]3 years ago

7 0

Answer:

what is this i am in class 7 i dont know and not read yet

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What is half-life? a. Half-life is a term that describes the time it takes for half of a particle to disintegrate c. Half-life i

Alisiya [41]

Well, half-life is the radioactivity of an identified isotope that decreases by half of the actual value.
So, your answer would be C.

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4 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

3 years ago

Why is more energy required to vaporise 1kg of water than to melt 1kg of ice?

KiRa [710]

Answer:

cause it takes a lotof energy to melt

Explanation:

7 0

3 years ago

Read 2 more answers

While following the directions on a treasure map a pirate walks 37.0 m north and then turns and walks 8.5 m east what is the mag

kogti [31]

Answer: $38\ m$

Explanation:

For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.

Then:

$d^2=\triangle x^2+\triangle y^2$

You can observe that the square of the displacement is equal to the sum of the square of the horizontal displacement and the square of the vertical displacement.

Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:

$\triangle x=37.0\ m\\\triangle y=8.5\ m$

Substituting values and solving for "d", you get:

$d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m$

8 0

3 years ago

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

jekas [21]

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$ Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$

7 0

3 years ago