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Bingel [31]
3 years ago
7

A force that pushes or pulls is known as A. A reaction force. B. An expected force. C. A positive force. D. An applied force.

Physics
2 answers:
otez555 [7]3 years ago
7 0

Answer:

the answer is an applied force

Explanation:

Julli [10]3 years ago
6 0

I believe it’s D, an applied force.

A reaction force seems to refer to Newton’s third law, but is relatively vague to be the answer to this question.

An expected force isn’t a concept nor the name of any subject of forces that is taught within the physics textbooks.

A positive force refers to direction, a negative force can have less, equal, or even more magnitude than the positive force, thus it contradicts itself. As that’s still a force that can push or pull.

So I believe it to be D.

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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Gnesinka [82]

Answer:

a).a_p=-2.39x10^{-12} rad/s^2

b).t=1016298.8 years

c).T_i=80.58x10^{-3}s

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

a_p=\frac{dw}{dt}

w=\frac{2\pi}{t}

a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}

dT=0.0808s

dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s

Replacing

a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2

b).

If the pulsar will continue to decelerate at this rate, it will  stop rotating at time:

t=\frac{w}{a_p}

w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s

t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s

t=1016298.8 years

c).

582 years ago to 2019

1437

T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s

5 0
4 years ago
A team of scientists wants to conduct a study on an endangered animal inthe wild. Which rule should the study follow in order to
cricket20 [7]

Answer:

imma have to say C I'm not sure tho

3 0
3 years ago
Write the word equation, balanced symbol equations and ionic equation got magnesium carbonate
Finger [1]

Explanation:

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Here you divide the four by the two and transfer the two toMg. In reactions like this oxidation numbers are exchanged.

7 0
3 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

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3 years ago
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Alliptical galaxies lack anything resembling the disk of a spiral galaxy.



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4 years ago
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