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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 fisherman moves from one end of a boat to another
Naily [24]

Answer:

That isnt a question so no one will know the answer to what you are talking about. I suggest adding a sceenshot or picture of the question.

5 0
2 years ago
if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?
madreJ [45]

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

7 0
3 years ago
A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.30 m from the slits. If the bright
kumpel [21]

Answer:

λ = 596 nm.

Explanation:

Fringe width = λ D / d

λ is wave length , D is screen distance and d is slit separation.

Putting the values

1.62 x 10⁻² =(  λ x 5.3 ) / .195 x 10⁻³

\lambda=\frac{1.62\times10^{-2}\times195\times10^{-6}}{5.3}

λ = 596 nm.

8 0
3 years ago
Angular velocity and linear velocity have the same dimensions. This statement is:
iragen [17]
I think it would be C: Sometimes true and sometimes false. It varies.

Hope this helped, have a great day! :D
5 0
3 years ago
A mass of 0.56 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic mot
NemiM [27]

Answer:

(a) 0.42 m

(b) 20.16 N/m

(c) - 0.42 m

(d) - 0.21 m

(e) 17.3 s

Solution:

As per the question:

Mass, m = 0.56 kg

x(t) = (0.42 m)cos[cos(6 rad/s)t]

Now,

The general eqn is:

x(t) = Acos\omega t

where

A = Amplitude

\omega = angular frequency

t = time

Now, on comparing the given eqn with the general eqn:

(a) The amplitude of oscillation:

A = 0.42 m

(b) Spring constant k is given by:

\omega = \sqrt{k}{m}

\omega^{2} = \frac{k}{m}

Thus

k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m

(c) Position after one half period:

x(t) = 0.42cos\pi = - 0.42\ m

(d) After one third of the period:

x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m

(e) Time taken to get at x = - 0.10 m:

-0.10 = 0.42cos6t

6t = co^{- 1} \frac{- 0.10}{0.42}

t = 17.3 s

7 0
3 years ago
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