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3 years ago

How does the solar wind affect earth

1 answer:

8 0

It affects it by the intense clouds of the high energy particles that it often contains which are produced by solar storms.
~ I hope this helps you!!~

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Which scientist invented a model of the atom that most closely resembles the modern electron cloud model

Alex Ar [27]

Bohr, he invented the Bohr model which is the basis for the beginning of quantum physics.

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3 years ago

A spaceship travels at approximately 0. 25c with respect to earth. this means 24 hours at the spaceship will be x hours to us on

Blababa [14]

The time passed on earth is mathematically given as

t' = 24.79 hrs

<h3>What is the time passed on earth?</h3>

Generally, the equation for is time mathematically given as

$t' = \gamma t$

Where

$\gamma = Lorentz\ factor \\\\\gamma = 1/ \sqrt {(1 - v^2/c^2)}$

$t' = t/ \sqrt {(1 - v^2/c^2)}$

Therefore

$t' = 24/ \sqrt {(1 - (0.25c/c)^2) }$

$t'= 24/ \sqrt {(1 - 0.25^2)$

t' = 24.79 hrs

In conclusion, the time passed on earth

t' = 24.79 hrs

Read more about the time

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2 years ago

The mass number is determined by which particles?

BartSMP [9]

I am not completely sure, but I believe that it depends on the total mass of the Protons and Neutrons

7 0

3 years ago

Read 2 more answers

A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of

damaskus [11]

The parallel component is given by
F=180cos(25)=163.14N

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3 years ago

What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass

Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

$F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz$

The frequency of the simple pendulum is given as;

$F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m$

Thus, the length of the simple pendulum is 0.25 m.

8 0

2 years ago