Three moons can fit inside the volume of the sun.
<h3>What is the moon?</h3>
The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.
Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;
8,000 miles/ 2,000 miles = 3
Hence, three moons can fit inside the volume of the sun.
Learn more about the moon:brainly.com/question/13538936
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Answer:
P = 450 J
Explanation:
Given that,
Mass of a child, m = 18 kg
The vertical distance from the top to the bottom of the slide is 2.5 metres.
The Gravitational field strength = 10 N/kg
We need to find the decrease in gravitational potential energy of the child sliding from the top to the bottom of the slide.
The formula for the gravitational potential energy is given by :
P = mgh
Substituting all the values,
P = 18 kg × 10 m/s² × 2.5 m
P = 450 J
Hence, the decrease in gravitational potential energy is 450 J.
this is an equation that you need to solve for motional emf. motional emf=vBL, where v is velocity in meters/second, B is magnetic field in Teslas and L is length or distance the rails are apart from each other. when we plug everything into the formula given above, we get: motional emf=5m/s*0.80T*0.20m. solving all this we get 0.8 volts. pretty sure that since they are giving you the direction of the field, they want to know which way the current will flow . since the conductor is moving from left to right the area of the field is increasing which means magnetic flux is increasing as Ф(magnetic flux)=B(magnetic field)*A(area)*cosФ(little phi is the angle to the normal. in this case little fee is 0 degrees so the cosФ doesn't matter). so ↑Ф=B↑A. if magnetic flux is increasing, the induced magnetic field is in the opposite direction as the original magnetic field meaning the induced magnetic field will be out of the page. using the right hand rule which says that if the field is in to the page, the current should go clockwise and if the field is out of the page, the current is counterclockwise so that means that the current should be going counter clockwise since the induced field is going out of the screen. the top of the conducting wire will have its current go to the left and the bottom of the conducting wire will have the current go to the right.
