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3 years ago

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Which of the following is occurring while a satellite is in orbit around Earth? O It is continuously pulling away from Earth It

is continuously falling toward the surface of Earth. It stays in a constant speed orbit where it was oriented from the start. It stays in the same orbit orientation traveling at variable speeds.

1 answer:

hoa [83]3 years ago

4 0

AnswerIt is continuously falling towards the surface of the earth

Explanation:

since gravity from earth is the thing that keeps it constantly in orbit

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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field directed

Hatshy [7]

Answer:

0.114m

Explanation:

From the general expression for the radius of the proton's resulting orbit, we have

$r=\frac{mv}{qB}$

where q is is the charge of the proton $1.6*10^{-19}C$

m is the mass of the proton $1.67*10^{-27}kg$

B is the magnetic field $0.040T$

and v i the speed.

to determine the speed, we use the expression

Kinetic Energy= $qV$

$1/2mv^{2}=qV$

where <em>V </em>is the voltage value i.e 1.0kv

and v is the speed

Hence, from simple rearrangement we have the speed v to be

$v=\sqrt{\frac{2Vq}{m}} \\$

if we substitute value, we have

$v=\sqrt{\frac{2*1000*1.6*10^{-19} }{1.67*10^{-27}}} \\$

carrying out careful arithmetic we arrive at

$v=4.38*10^{5} m/s$ .

using the value for the speed in the expression for the radius of the orbit as stated earlier, we have

$r=\frac{1.67*10^{-27}*4.38*10^{5}}{1.6*10^{-19}*0.04} \\$

$r=0.114m$

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3 years ago

Would you buy a pair of shoes from the same company that sold Jack his shoes? Why or why not?

LenKa [72]

Answer:

Maybe

Explanation:

It depends. If he says the shoes sucked than no because they suck-

But if he really liked the shoes and said it was really good than yes

I would buy the shoes

8 0

3 years ago

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What is the difference between subjective and objective mineral properties?

schepotkina [342]

Subjective refers to something that takes place within the mind of the person. It is modified based on individual bias.

Objective refers to defining things or expressing them without any distortion. They are expressed only as perceived.

Finally, mineral is anything composed of matter apart from animals and plants.

7 0

4 years ago

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A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

<u>Energy:</u>

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

<u>MRUV:</u>

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

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4 years ago

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