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3 years ago

7

Of orbital radius, orbital velocity, orbital circumference, or radial area swept through, which of these stays constant accordin

g to one of Kepler’s Laws?
a. orbital distance
b. radial area swept through
c. none of these
d. orbital velocity

1 answer:

kow [346]3 years ago

5 0

Answer:

dyow

Explanation:

sorrrey

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found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it

evablogger [386]

(h + .16) m g = 1/2 k x^2 total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2 PE released = PE of spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

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3 years ago

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

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3 years ago

How are electric and magnetic forces the same? How are they different?

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They both have a certain force. They are different because that force is different

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2 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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2 years ago

A box is pulled with a horizontal force of 500N and moves 5m what is work done

dalvyx [7]

The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

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3 years ago