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Scrat [10]
3 years ago
15

A projectile has a speed of √ gm/3r directed away from a planet, when it is a distance of 4R from the center of the planet. the

planet has mass M and radius R. will this projectile be able to escape from the gravitational attraction of the planet?
Physics
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer:

hat the speed of the rocket is lower than the minimum escape velocity whereby the ship cannot escape the planet

Explanation:

To find out if the projectile can escape the planet, let's find the minimum escape velocity using the concepts of energy.

Starting point. On the planet's surface

            Em∠₀ = k + U = ½ m v² - G mM / R²

End point far away

          Emf = U = - g m M / r²

          Em₀ = Emf

          ½ m v² - G m M / R² = -G m M / r²

          v² = G M (1 / R² -1 / r²)

Let's find the velocity for the height of the rocket r = 4R

          v =√GM (1 / R² - 1/16 R²) = √a GM / R²      0.968

This is the speed to escape planet

 

Let's compare this minimum escape velocity with the given value

             

            v = √GM /R²    1 /√ 3

            v = √GM / R 0.577

We can see that the speed of the rocket is lower than the minimum escape velocity whereby the ship cannot escape the planet

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A fat person weighing 80 Kg falls on a concrete floor from 2m. If whole of the mechanical energy is converted into heat energy,
sasho [114]

Answer:

The heat produced is 1568 J

Explanation:

The given parameters are;

The mass of the person, m = 80 kg

The height from which the person falls, h = 2 m

Mechanical Energy, ME = Potential Energy, PE + Kinetic Energy, KE

At the height, from where the person falls, the initial velocity of the person = 0 m/s

Therefore;

The initial kinetic energy, K.E. = 1/2·m·v² = 1/2 × 80 kg × (0 m/s)² = 0 J

From which we have;

The Mechanical Energy, M.E. = The initial Potential Energy, P.E. + 0 J

∴ The Mechanical Energy, M.E. = The initial Potential Energy, P.E.

The initial Potential Energy, P.E. = m·g·h

Where;

m = The mass of the person

g = The acceleration due to gravity ≈ 9.8 m/s²

∴ The initial Potential Energy, P.E. = 80 kg × 9.8 m/s² × 2 m = 1568 J

The Mechanical Energy, M.E. = The initial Potential Energy, P.E. = 1568 J

The whole mechanical energy is converted into heat energy, therefore, we have;

The Mechanical Energy, M.E. = The heat energy = 1568 J

The heat produced = The heat energy = 1568 J.

4 0
3 years ago
Pls help me idk what to do
Alika [10]
The dot would be the earthquake epicenter. That is where the earthquake originated.
6 0
4 years ago
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

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3 years ago
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You could jump high!
3 0
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Two small children decide it would be fun to toss a couple of large cats at each other. Cat A (7kg) is thrown at 7m/s and cat B
Alex777 [14]

Answer:

V=4.7m/s

Explanations:

Let Ma mass of cat A=7kg

Va velocity of cat A=7m/s

Mb mass of cat b=6.1kg

VB velocity of cat b=2m/s

From conservation of linear momentum

MaVa+MbVb=(Ma+Mb)V

7*7+6.1*2=(7+6.1)V

61.2=13.1V

V=4.7m/s

3 0
4 years ago
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