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3 years ago

If the moon disappeared what effect would this have on the earths tides

2 answers:

iogann1982 [59]3 years ago

4 0

The moon has a huge impact on the tides. With the Moon gone, the oceans would become much calmer. The Sun still has its effect on them known as the solar tides. Surfers wouldn't be completely devoid of waves.

Sliva [168]3 years ago

3 0

There would no longer be any tides.

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Identify the chemical change below

lesya [120]

Imma say go with C (:

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3 years ago

If the pressure exerted by the liquid of density 500kg/m3 is 20,000Pa. What is height of liquid column.

gregori [183]

The height of the liquid column is 4.08 metres.

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3 years ago

How is light described?

Lelu [443]

Light as a wave: Light can be described (modeled) as an electromagnetic wave. ... This changing magnetic field then creates a changing electric field and BOOM - you have light. Unlike many other waves (sound, water waves, waves in a football stadium), light does not need a medium to “wave” in.

Explanation:

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3 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago

An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.

Marianna [84]

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

<u>Given the following data;</u>

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

4 0

3 years ago