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Snezhnost [94]
4 years ago
7

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
Physics
1 answer:
yuradex [85]4 years ago
3 0
Since

Electric potential energy = qV

Where V = Ed

Hence
Electric potential energy = q(Ed) --- (1)

Since E = 1.0 * 10^3 N/C
d = 0.10 m
q = 4 * 10^-6 C

Plug in the values in (1)
(1) => Electric potential energy =  4 * 10^-6(1.0 * 10^3 * 0.10)
Electric potential energy = 400 μJ
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How many times bigger is the earth than the moon?
liraira [26]
The Earth is around 4 times as big as the moon.
7 0
3 years ago
A bowling ball with a mass of 9 kg is moving at a speed of 4 m/s. What is its
Ilia_Sergeevich [38]
I’m pretty sure it’s B
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A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa
aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

P=A\epsilon\sigma T^{4}

Here, A is the area, \sigma is the stefan's constant,\epsilon is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}

Here, A_{1} is the area of the sphere, A_{2} is the area of the cube,T_{1}  is the temperature of the sphere and T_{2}  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

A_{1}T_{1}^{4}=A_{2}T_{2}^{4}

The area of the sphere is A_{1}=4\pi \times r^{2}.

Here, r is the radius of the sphere.

The area of the cube is A_{2}=6\times a^{2}.

Here, a is the edge of the cube.

Put A_{1}=4\pi \times r^{2} and A_{2}=6\times a^{2}.

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

V_{1}=V_{2}

Here,V_{1},V_{1} are the volumes of the sphere and the cube.

\frac{4}{3}\pi r^{3}=a^{3}

\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}  

Put this value in the equation (1).

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

Put T_{1}=500°c.

T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}

T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

5 0
3 years ago
1. Fill in the blanks. (3 pts)
Usimov [2.4K]

Answer:

a) mass

b) Newtons

c) momentum formula where p stands for momentum, m stands for mass, and v stands for velocity

Hope this helps!

4 0
3 years ago
Read 2 more answers
Throw a rock horizontally at 22 m/s from a 315 m tall building. How far from the base of the building will it land?​
creativ13 [48]

Answer:

176.44 m

Explanation:

y = (vertical vi)t + (1/2)gt^2

315 = 0 + (1/2)(9.8)t^2

t = 8.02s

x = (horizontal vi)t + (1/2)at^2

x = (22)(8.02) + 0

x = 176.44 m

6 0
3 years ago
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