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ikadub [295]
3 years ago
8

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the h

orizontal velocity component of the ball?

Physics
1 answer:
dybincka [34]3 years ago
6 0

Given,

A player kicks a soccer hits at an angle of 30° at a speed of 26 m/s

We can resolute the trajectory of soccer into horizontal and vertical components.(Please see the attached file)

We can have,

Horizontal velocity component of ball= 26cos(30°)  = 26×(√3÷2) = 22.51 m/s

And vertical velocity component of ball = 26sin(26°) = 26×(1÷2) = 13 m/s


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Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

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This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

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3 years ago
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Force of gravity

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3 years ago
Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i
shepuryov [24]
<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}    (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

m_{1} and m_{2} will be now 2m_{1} and 2m_{2}

r will be now 2r

Let's rewrite the equation (1) with this new values:

F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}    (2)

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F=4G\frac{m_{1}2m_{2}}{4r^2}    

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As we can see, equation (3) is the same as equation (1).

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3 years ago
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Verrrrry interesting !
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the mass of the Moon ... that point is inside the Earth, and it looks a lot
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When the new body arrives to replace the lightweight Moon, that point
will be a lot closer to the new companion ... maybe even inside it. 
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Earth is orbiting it.
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Answer:

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