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3 years ago

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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?

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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Answer:

$u=34cm$

Explanation:

From the question we are told that:

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Near point is $u=17 cm$

Therefore

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Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

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Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

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and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

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Explanation:

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3 years ago

Read 2 more answers

Please help! Will give brainliest. 10 points. Show work!

Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

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2 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

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3 years ago