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2 years ago

Please Help!!! It's for a quiz!!

Physics

1 answer:

Sphinxa [80]2 years ago

3 0

Answer:

38.8 m/s

Explanation:

Force F(x) = 6 - 2x + 6x²

work

$W=\displaystyle\int_{0}^{13.9}F(x)dx=\displaystyle\int_{0}^{13.9}(6-2x+6x^2)dx$

$=6x-x^2+2x^3|_{0}^{13.9}\\=5261 J$

W = mv²/2=7v²/2 = 3.5v² = 5261

v = 38.8 m/s

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4 years ago

An Olympic diver is on a diving platform 5.40 m above the water. To start her dive, she runs off of the platform with a speed of

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Answer:

10.36m/s

Explanation:

From the question we are given;

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3 years ago

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3 years ago

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4 years ago

A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

$F_{c} = F_{G}$

$\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}$

$v = \sqrt{\frac{gr^{2}}{r+h}$

Where:

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r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km

$v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s$

b) The period of its revolution is:

$T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h$

c) The gravitational force acting on it is given by:

$F = \frac{GMm}{(r + h)^{2}}$

Where:

M is the Earth's mass = 5.97x10²⁴ kg

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

$F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N$

I hope it helps you!

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3 years ago