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Debora [2.8K]
3 years ago
14

Based on the data, which of the following explains the relationship between the distance of the planet from the Sun and its plan

etary motion around the sun?
A. The planets with the least distance from the sun experience a longer time of revolution and shorter days of rotation.

B. The planets with the least distance from the sun experience a shorter time of revolution and shortest days of rotation.

C. The planets with the greatest distance from the sun experience a longer time of revolution and longer days of rotation.

D. The planets with the greatest distance from the sun experience a longer time of revolution and shorter days of rotation.

Physics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

.

Explanation:

...

max2010maxim [7]3 years ago
3 0
Answer is B! Just think about mercury or Venus! Mercury takes 88 earth days to rotate around the sun and Venus takes 225. The farther you go out into space, the longer it takes to orbit the sun, because there’s a longer distance. It takes Pluto 248 years, and Neptune takes 165 years:)
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Hi can you help me with this please?
bagirrra123 [75]

Answer:

4s

Explanation:

My assumption would be 4s since 25 going into 100 would be 4? hope that helped..

5 0
3 years ago
when two resistors are wired in series with a 12 v battery, the current through the battery is 0.31 a. when they are wired in pa
liberstina [14]

Let  R₁  and R₂ be the  two Resistance

R₁ - Resistance of 1st resistor

R₂ - Resistance of 2nd resistor

V = Voltage = 12V

I = O.31A ( in series)

I  = 1.6 A ( in parallel)

when two resistance connected in series

Rs = R₁ + R₂

V = IRs = 12 /0.31 V

R₁+R₂ = 38.700Ω   (equation1.)

When two resistance connected in parallel

    R_{p} =\frac { R_{1} R_{2}}{R_{1} + R_{2}  }

V = I Rp

V=  I (R₂ R₁ /R₁ +R₂)

R₁ R₂ = 290.25 Ω  

As we know

(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2

Using above values we get,

 (R₁ - R₂ ) ²  = (38.70) ²  - 4x 290.25

 (R₁ - R₂ ) ²   = 336.69  Ω  

R₁ - R₂ = 18.34     (equation 2 )

Using equation 1 and 2 we get

R₁+ R₂ = 38.70 52

R₁ - R₂ = 18.34

R₁ = 28.535 Ω  

Using the value of  R₁ in equation 1 we get

R₂ = 38·700 -  28.535

RR₂ = 10.125 Ω  

R₁ = 28. 535 v  and   R₂  10.125 Ω  

To know more about resistance in series and parallel:

brainly.com/question/27882579

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7 0
1 year ago
A block of mass
Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

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6 0
2 years ago
Energy transformation in wound spring of a toy car?<br>give your own answrr​
BaLLatris [955]

Answer:

The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy (i.e potential energy due to change in shape). ... During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

Explanation:

7 0
3 years ago
Read 2 more answers
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
3 years ago
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