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matrenka [14]
3 years ago
12

What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?

Physics
1 answer:
Artist 52 [7]3 years ago
3 0

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

U=k\frac{q_1q_2}{r}    (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J

Hence, the energy between charges is 0.413 J

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What is a law in physics?​
Rina8888 [55]

Answer:

By nature, laws of Physics are stated facts which have been deduced and derived based on empirical observations. Simply put, the world around us works in a certain way, and physical laws are a way of classifying that “working.”

6 0
3 years ago
Buoyancy is a force that always acts in an
Margarita [4]

Buoyancy is a force that always acts in an upward direction exerted by a fluid on a body placed in the fluid

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7 0
3 years ago
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a ball is thrown horizontally from a 20 m high building with a speed of 5.0 m/s. How far from the base of the building does the
kipiarov [429]
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
7 0
3 years ago
the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s
12345 [234]

The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

If you need to learn more about displacement click here:

brainly.com/question/28370322

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

4 0
2 years ago
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take
vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)}  )\\T=1.38*10^{6} seconds

3 0
3 years ago
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