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4 years ago

How do you measure potential and kinetic energy?

Physics

1 answer:

Ray Of Light [21]4 years ago

3 0

Answer:

potential energy is a stored energy or energy of position (gravitational).

Kinetic energy is a energy of motion.

Explanation:

in the formula K is for the kinetic and the P stand for the potential.

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Please select the word from the list that best fits the definition a puffy white cloud with a flat bottom

ankoles [38]

Cumulus? They are fluffy clouds that have flat bottoms. Is this what you need?

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4 years ago

Everglades National Park only protects the southern 1/5 of the Everglades but its still

Sergeeva-Olga [200]

25 should be the answer

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3 years ago

g HW 5.6.The Moon and the Earth both orbit the center of mass of the Earth-Moon system.Earth is 81 times more massive than the M

Law Incorporation [45]

Answer:

$R_{cm}$ = 4.688 106 m

Explanation:

The expression for the center of mass is

R cm = 1 / M ∑ $r_{i}$ $m_{i}$

Where M is the total mass of the system, ri and m1 are the distance and mass of each particle from a defined origin

Let's look for the total mass, the mass of the moon is m and that of the earth Me

M = $M_{e}$ + m

M = 81m + m

M = 82 m

The distance from the earth to the moon is R = 384.4 106 m

$R_{cm}$ = 1 / M (M R1 + m R)

$R_{cm}$ = 1 / 82m (0 + m R)

$R_{cm}$ = R / 82

$R_{cm}$ = 384.4 106/82

$R_{cm}$ = 4.688 106 m

The radius of the earth is

$R_{e}$ = 6371 103 m

$R_{e}$ = 6.371 106 m

We can see that the center of mass of the system is within the beating radius

6 0

3 years ago

What is the smallest resistance you can make by combining them?

Luda [366]

Answer:

166.67 ohm

Explanation:

Complete statement of the question is :

You have a collection of six 1.0 kO resistors. What is the smallest resistance you can make by combining them?

When we connect resistors in parallel, the combination resistance is always smaller than the smallest resistance value we combine.

Parallel Combination of resistances is given as

$\frac{1}{R_{p}} = \frac{1}{R_{1}} +\frac{1}{R_{2}} +\frac{1}{R_{3}} + ....$

where

$R_{p}$ = Equivalent resistance

and $R_{1} , R_{2}, R_{3}$ are resistances in parallel.

So for the above question, we have

$R_{1} = R_{2} = R_{3} = R_{4} = R_{5} = R_{6} = 1 kohm = 1000 ohm$

So parallel combination is given as

$\frac{1}{R_{p}} = \frac{1}{R_{1}} +\frac{1}{R_{2}} +\frac{1}{R_{3}} +\frac{1}{R_{4}} +\frac{1}{R_{5}} +\frac{1}{R_{6}}\\\frac{1}{R_{p}} = \frac{1}{1000} +\frac{1}{1000} +\frac{1}{1000} +\frac{1}{1000} +\frac{1}{1000} +\frac{1}{1000}\\R_{p} = \frac{1000}{6} \\R_{p} = 166.67 ohm$

So the smallest resistance is 166.67 ohm

8 0

4 years ago

To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the co

ANEK [815]

Answer:

1) Option D is correct.

The electric field inside a conductor is always zero.

2) Option A is correct.

The charge density inside the conductor is 0.

3) Charge density on the surface of the conductor at that point = η = -E ε₀

Explanation:

1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.

Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.

2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.

3) Surface Charge density = η = (q/A)

But electric field is given as

E = (-q/2πε₀r²)

q = -E (2πε₀r²)

η = (q/A) = -E (2πε₀r²)/A

For an elemental point on the surface,

A = 2πrl = 2πr²

So,

η = -E ε₀

Hope this Helps!!!

4 0

4 years ago