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2 years ago

14

Accorrding to Kepler's third law, a planet whose distance from the Sun is 2 A.U. would have an orbital period of how many Earth-

years

1 answer:

Lilit [14]2 years ago

4 0

Answer:

hi

Explanation:

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White light is an _____________ mixture of all the colors.<br> a) observable b) unequal c) equal

Musya8 [376]

Answer:

unequal

Explanation:

White light is a combination of all colors in the color spectrum.

6 0

3 years ago

Read 2 more answers

irakobra [83]

Answer:

The mass of the block, M =T/(3a +g) Kg

Explanation:

Given,

The upward acceleration of the block a = 3a

The constant force acting on the block, F₀ = Ma = 3Ma

The mass of the block, M = ?

In an Atwood's machine, the upward force of the block is given by the relation

Ma = T - Mg

M x 3a = T - Ma

3Ma + Mg = T

M = T/(3a +g) Kg

Where 'T' is the tension of the string.

Hence, the mass of the block in Atwood's machine is, M = T/(3a +g) Kg

3 0

3 years ago

Read 2 more answers

The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.M/s. What is the velocity

AysviL [449]

Ok so the equation for momentum is:
v=p/m

So you would do:
7800/1200=6.5

So the answer is:
6.5 m/s

Hope this helps :)

6 0

3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago

As the solar system formed, most of its mass became concentrated in one part of it. What is that that part?

sertanlavr [38]

That was sun as some smaller masses formed planets and other remaining formed sun

4 0

3 years ago