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2 years ago

6

Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is t

he weight of the object on the moon?

1 answer:

uysha [10]2 years ago

5 0

<u>I have assumed a weight of 120 N on Earth.</u>

Answer:

<em>The object weighs 20 N on the moon</em>

Explanation:

Weight

The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.

The formula to calculate the weight is:

W = m.g

If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:

$g_m=1/6 g_e$

Dividing by ge:

$g_m/g_e = 1/6$

An object of weight We=120 N on planet Earth has a mass of:

$m = 120 / g_e$

Multiplying by gm:

$m.g_m=120 g_m/g_e$

Substituting the ratio of accelerations of gravity:

$m.g_m=120 * 1/6$

Since m.gm is the weight on the Moon Wm:

$W_m=20~N$

The object weighs 20 N on the moon

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When you and a friend move a couch to another room you exert a force of 75 N over 5m how much work will you do

sesenic [268]

The work done is 375 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this problem,

F = 75 N is the force applied to the couch

d = 5 m is the displacement

Assuming the force applied to the couch is parallel to the motion, $\theta=0$

And so, the work done is

$W=(75)(5)(cos 0)=375 J$

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3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

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2 years ago

Which of the following provides evidence that geologic activity occurred in the past on Earth's Moon?

Bingel [31]

Answer: d) the presence of solidified lava flows on the Moon

Explanation:

A geological activity means an occurrence of event such as volcanic eruption, earthquake, sedimentation, erosion etc. The revolution of the Moon around the Earth , the axial tilt of the Moon or the phases of the Moon are not surface features. hence, these events cannot provide the evidence of geological activity in the past of Moon.

The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.

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2 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

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3 years ago

The vertical normal stress increase caused by a point load of 10 kN acting on the ground surface at a point 1m vertically below

Tomtit [17]

Answer:

(b) 4.775 kN

Explanation:

see the attached file

8 0

2 years ago