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OLga [1]
3 years ago
8

What do we call the tide that occurs when constructive interference between lunar and solar bulges produces a large tidal range?

Physics
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

Spring Tide

Explanation:

Normal tidal sizes become slightly larger during complete or new moons— which arise when the Moon, the sun, and the moon are almost in a straight line. It occurs two times a month. The moon looks new (dark) when it's exactly between the sun and the Earth. Once the Moon would be between the sun and earth the moon appears complete. On both situations, the sun's gravitational pull is "added up" to the Earth's gravitational pull n the moon, allowing the oceans to boom a little more than normal. This equals higher tides are marginally higher and low tides are somewhat lower than usual.

This is called as spring tide.

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please help me and give an explanation
Zanzabum

Explanation:

The moon orbits the Earth once every 27.322 days. It also takes approximately 27 days for the moon to rotate once on its axis. As a result, the moon does not seem to be spinning but appears to observers from Earth to be keeping almost perfectly still. Scientists call this synchronous rotation

3 0
3 years ago
What is the difference between washing of clothes at home and washing by dry cleaning at laundry
Arlecino [84]
The dry cleaning process uses chemicals to clean the clothes. It’s “dry” because it doesn't use water, as in normal wet laundering. Most laundries use chemicals to remove grease and stains from clothing.
8 0
3 years ago
What would happen if a group of force of 45 N compete against a group of 50 N
larisa [96]
The 50N group of force would be greater
So whichever object is being pulled will be pulled towards the 50N force
7 0
3 years ago
A charge Q accumulates on the hollow metallic dome, of radius R, of a Van de Graaff generator. A probe measures the electric fie
andre [41]

The electric field changes by a factor 4/9

Explanation:

The electric field at a point outside a hollow charged sphere is identical to that produced by a single point charge Q, therefore its strength is given by

E=k\frac{Q}{r^2}

where

k is the Coulomb constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

In this problem, we have a charge Q accumulated on the dome of the generator of radius R. The electric field strength at a distance of 2R from the sphere centre is

E=\frac{kQ}{(2R)^2}=\frac{1}{2}\frac{kQ}{R^2}

Then, the probe is moved to a new distance of

r = 3R

From the centre of the sphere: therefore, the new electric field will be

E'=\frac{kQ}{(3R)^2}=\frac{1}{9}\frac{kQ}{R^2}

So, the electric field has changed by a factor

\frac{E'}{E}=\frac{1/9}{1/4}=\frac{4}{9}

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
An object of mass m is dropped from a height h above the surface of a planet of mass M and radius R. Find the speed of the objec
Shtirlitz [24]

Answer:

v=\sqrt{\frac{2GMh}{R^{2}}}

Explanation:

mass of object = m

Mass of planet = M

Radius of planet = R

Height = h

Let the speed of the object as it hits the earth's surface is v.

the value of acceleration due to gravity

g = G M / R^2

where, g is the universal gravitational constant.

Use third equation of motion

v^{2}=u^{2}+2gh

where, u is the initial velocity which is equal to zero.

So, v^{2}=0 + 2 \times \frac{GM}{R^{2}}\times h

v=\sqrt{\frac{2GMh}{R^{2}}}

8 0
4 years ago
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