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3 years ago

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A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net for

ce on the probe is zero, what can be said about the relative masses of the moons? a. Moon A is twice as massive as moon B. b. Moon A has the same mass as moon B. c. Moon A is four times as massive as moon B. d. Moon A is half as massive as moon B.

1 answer:

Dennis_Churaev [7]3 years ago

3 0

Answer:

c. Moon A is four times as massive as moon B

Explanation:

Let's assume the:

mass of the object = $m\,kilogram$

mass of the moon A = $M_A\,kilogram$

mass of the moon B = $M_B\,kilogram$

distance between the center of masses of the object and moon B = $r\,meters$

According to the given condition the object is twice as far from moon A as it is from moon B

∴distance between the center of masses of the object and moon B = $2r\,meters$

<u>As we know, gravitational force of attraction is given by:</u>

$F=G\frac{m_1.m_2}{r^2}$

<em>According to the condition</em>

Force on m due to $M_B=$ Force on m due to $M_A$

$G\frac{m.M_A}{(2r)^2} =G\frac{m.M_B}{(r)^2}$

$\frac{M_A}{4r^2} =\frac{M_B}{r^2}$

$M_A=4M_B$

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A well hole having a diameter of 5 cm is to be cut into the earth to a depth of 75 m. Determine the total work (in joules) requi

guapka [62]

Answer:

total work is 99.138 kJ

Explanation:

given data

diameter = 5 cm

depth = 75 m

density = 1830 kg/m³

to find out

the total work

solution

we know mass of volume is

volume = $\frac{\pi}{4} d^2 dx$

volume = $\frac{\pi}{4} d^2 1830 dx$

so

work required to rise the mass to the height of x m

dw = $\frac{\pi}{4} d^2 1830$ gx dx

so total work is integrate it with 0 to 75

w = $\int\limits^{75}_{0} {\frac{\pi}{4} d^2 1830 gx dx}$

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w = 99138.53 J

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3 years ago

A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

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3 years ago

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bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

$f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}$

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An isolated system , it does not allow any matter or energy to be exchanged

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