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Elis [28]
2 years ago
6

Help me please this is for physics

Physics
1 answer:
Yuri [45]2 years ago
3 0
<h2>Hello there! :)</h2>

It's a pleasure to be helping you today with your<u> physics question!</u>

Answer:

23.1m/s

Explanation:

We want to find the initial speed of the ball.

To do this, we have to apply the formula for the time of flight of a projectile:

T=\frac{2_{v0~sin 0} }{g}

where θ = angle of flight

g = acceleration due to gravity

v0 = initial speed

Therefore, substituting the given values into the formula, we have that:

\boxed{4.2=\frac{2~x~_{v0~sin63} }{9.8}}

⇒ 2 ×_{v0} ×0.8910= 9.8 × 4.2

⇒\boxed{{v0}=\frac{9.8~times~4.2}{2~times~0.8910}}

\boxed{{v0} =23.1m/s}

That is the initial speed of the ball.

<em />

<em>I hope this helps you!</em>

<em>Good Luck with your Assignment!</em>

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3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
We now have two tappers that are tapping the surface of the water together. They are tapping in sync (both are hitting the surfa
ollegr [7]

Answer:

The cork moves up and down.

Explanation:

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