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Sedbober [7]
3 years ago
15

When would a scientist accept the alternative

Physics
1 answer:
vredina [299]3 years ago
3 0

Answer:

B.) if statistical results are significant

Explanation:

If a scientist accepts the null hypothesis, then there is enough evidence to reject the null hypothesis.

This means that the p-value is less than the significance level.

When the scientist accepts the alternative hypothesis, it means the statistical results are significant.

The correct answer is B.

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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
4 years ago
Read 2 more answers
What is the name of the force that is a combination of all forces that act on an object?
jarptica [38.1K]

Answer:

If the combination of all the forces acting on an object sums upto be zero, then the net force is known as <u>balanced force.</u>

If the combination of all the forces acting on an object is non-zero, then the net force is said to be <u>unbalanced force. </u>The unbalanced force causes the object change the state of rest or motion.

6 0
3 years ago
Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we
Scilla [17]

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>
  • Mass 1 = m₁
  • Mass 2 = m₂
  • Gravitational constant = G
  • Distance apart = r
  • Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>
  • Mass 1 = 2m₁
  • Mass 2 = 3m₂
  • Gravitational constant = G
  • Distance apart (r) = r
  • New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

6 0
2 years ago
A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

4 0
3 years ago
What is the Force acting on a ball if the mass of the ball is 2 kg and the acceleration of the ball is 5 m/s2?
hjlf

Answer:

please find attached file

Explanation:

5 0
3 years ago
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