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3 years ago

The star named Canopus has a declination of approximately –52°. Which of these statements is correct about Canopus?

1 answer:

6 0

It’s A, it’s fifty two degrees above the celestial equator, you can also look up the same question and see if any of them match

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What the opposite of suspension physical science physical science

Svetlanka [38]

A colloid I think. Don’t hold it against me if I’m wrong my dude.

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3 years ago

You want to calculate the density of a marble what is one piece of data that you need A the weight B the volume C how fast the m

8090 [49]

C I think but I’m not really sure....

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3 years ago

Read 2 more answers

Someone please help with this

SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

$\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}$

Factoring out 2/3:

$\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}$

Substituting the original force:

$F'=\frac{2}{3}F$

The new force is 2/3 of the original force

7 0

3 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

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3 years ago

A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the v

zhenek [66]

Answer:

yyyggggggggggggggthhgggggyyyyy

Explanation:

yyygggggggggggg

4 0

3 years ago