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ludmilkaskok [199]
3 years ago
9

Move the Earth so it is one box from the Sun. (Note: one box side equals about 46,000,000 miles.) Do not change the length of th

e velocity vector. Predict what will happen to the Earth and Sun when you hit Play?
Physics
1 answer:
goblinko [34]3 years ago
3 0

Answer:

When we change the distance, the universal attraction force increases, so that the system is free to reach a new equilibrium, the linear speed of the earth must rise to the calculated value.

v = √ (G M / r)

Explanation:

For this exercise it is asked that if you maintain the linear speed of the Earth and bring it closer to the sun that would pass.

We pass the distance from Ro = 1.49 10¹¹ m to r = 0.736 10¹¹ m shortens, we write Newton's second law

             F = m a

where the force is the universal force of attraction

            F = G mM / r²

acceleration is central

            a = v² / r

           G m M / r² = m v² / r

            v = √ (G M / r)

When we change the distance, the universal attraction force increases, so that the system is free to reach a new equilibrium, the linear speed of the earth must rise to the calculated value.

We can compare this value with that of the normal orbit

           v₀ = √ (GM / R₀)

            v / v₀ =√ (Ro / r)

           v² r = v₀² R₀

 either of these two expressions gives the relations gives the change in velocity with the radius of the orbit

 

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A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe
Karolina [17]

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

F_{r}=kv^2

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}

Put the value into the formula

3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}

Put the value of k

3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}

1600-v^2=m\dfrac{d^2v}{dt^2}

At terminal velocity \dfrac{d^2v}{dt^2}=0

So, 1600-v^2=0

v=\sqrt{1600}

v=40\ ft/sec

Hence, The velocity is 40 ft/sec.

4 0
3 years ago
A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the fina
balu736 [363]

Answer:

\displaystyle \vec{d}=

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position  

\displaystyle P_1(x_1,y_1)

and then moves at another position at

\displaystyle P_2(x_2,y_2)

The displacement vector is directed from the first to the second position and can be found as

\displaystyle \vec{d}=

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

\displaystyle x=z\ cos\alpha

\displaystyle y=z\ sin\alpha

The question describes the situation where the initial point is the base of the mountain, where both components are zero

\displaystyle P_1(0,0)

The final point is given as a 520 m distance and a 32-degree angle, so  

\displaystyle x_2=520\ cos32^o= 440.99\ m

\displaystyle y_2=520\ sin32^o=275.6\ m

The displacement is

\displaystyle \vec{d}=

5 0
3 years ago
A driver is traveling eastward on a dirt road when she spots a pothole ahead. She slows her car from 14.0 m/s to 5.5 m/s in 6.0
Ivahew [28]

Answer:

- 1.42m/s²

Explanation:

Acceleration is defined as the change in velocity of a body with respect to time.

Acceleration = change in velocity/time

Change in velocity = final velocity - initial velocity

Acceleration = final velocity - initial velocity/time

Since she slows her car from 14.0 to 5.5m/s in 6seconds,

Initial velocity = 14m/s

Final velocity = 5.5m/s

Time = 6seconds

Substituting in the given formula, we will have

Acceleration = 5.5 - 14/6

Acceleration = - 8.5/6

Acceleration = - 1.42m/s²

The negative acceleration shows that the car decelerates.

8 0
3 years ago
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Bezzdna [24]

Explanation:

A. answer is the correct answer

6 0
3 years ago
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