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Juliette [100K]
3 years ago
8

The gravitational strength at the poles is greater than the gravitational strength at the equator. What will happen to an object

when it moves from the poles to the equator? Group of answer choices A. Its mass will increase. B. Its mass will decrease. C. Its weight will increase. D. Its weight will decrease.
Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

C

Explanation:

Because this same question was on my test last week and I got it correct

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8. List three temperature scales. Then write the boiling and freezing
Tomtit [17]

Explanation:

For most temperature scales, the boiling point of water and the freezing point is used to calibrate it.

Three known temperature scales;

  • Kelvin scale
  • Celcius scale
  • Fahrenheit scale

                               Kelvin scale              Celcius scale          Fahrenheit scale

Freezing point             273K                            0°C                        32°F

Melting point                373K                          100°C                     212°F

5 0
2 years ago
What are successfulness of the Competition policy in South Africa?​
Mars2501 [29]

Answer:

The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.

Explanation:

hope this helps you

6 0
3 years ago
The proportion of carbon-14 in an organism is useful in figuring out the age of that organism after it dies because...?
nata0808 [166]
I believe the correct answer from the choices listed above is option D. The proportion of carbon-14 in an organism is useful in figuring out the age of that organism after it dies because <span>the proportion of carbon-14 slowly decreases after the death of the organism. Hope this answers the question.</span>
4 0
3 years ago
Indicate whether the statement is true or false. an astronaut weighs the same on earth as in space.
tino4ka555 [31]
Weight is different (but mass is the same)
6 0
2 years ago
Read 2 more answers
A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
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