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3 years ago

The field is strongest at which point?

Physics

1 answer:

Licemer1 [7]3 years ago

4 0

Answer:The poles

Explanation:

The field is strongest at the poles

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In a ____ wave, pasticles of the medium move perpendicular<br> to the direction of the wave.

GrogVix [38]

The answer is transverse

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3 years ago

14. Which of the following is not an example of work being done?

Natali5045456 [20]

Answer:

B. holding a coffee mug

Explanation:

Something must move a distance for work to be done.

7 0

3 years ago

What is the unit vector along ⃗ = hat − hat + hat ?

Elanso [62]

Answer:

1 hat

Explanation:

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3 years ago

A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s

vitfil [10]

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

$v^2-u^2=2as$

Applying values as:

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s$

<u>The speed with which the ball hit the ground = 38.87 m/s</u>

3 0

3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago