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3 years ago

If an object weighs 300 N on earth, what is it’s mass on the moon?

Physics

1 answer:

inna [77]3 years ago

4 0

Answer:

The mass of the object on the Moon (and anywhere else) is about 30.61kg. Please see more detail below.

Explanation:

Weight is the gravitational force exerted on the object and is a function of mass and gravitational acceleration:

(weight) = (mass) x (gravitational acceleration)

We are to find the mass, knowing the weight on Earth to be 300N:

(mass) = (weight on Earth) / (gravitational acceleration on Earth) = 300N / 9.8 m/s^2 = 30.61 kg

The mass of the object is 30.61kg.

The mass of the object is independent of gravity. Therefore the answer to the question "What is its mass on the Moon" is 30.61kg.

If the question were what is its weight on the Moon, the answer would be

(weight on Moon) = (mass) x (grav.accel. on Moon) = 30.61kg x 1.62 m/s^2 = 49.59N

which is about 1/6 of the object's weight on the Earth.

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G (where g=9.8 m/s2). If an object’s mass is m=10. kg, what is its weight?

Semenov [28]

$F_{w} =m*g \\ F_{w} =10*9.8=98N$

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4 years ago

In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr

NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

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

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is $t$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

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

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is $h$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

7 0

3 years ago

What is the average velocity of the car from t=5 seconds to t=12 seconds?

Debora [2.8K]

The question is not complete

4 0

3 years ago

Please help on this one someone

Nonamiya [84]

Valence electrons are the electrons in the outermost energy level of an atom — in the energy level that is farthest away from the nucleus.

I think it's A.

6 0

4 years ago

two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma

nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force = 3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos $\theta$

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)

35 = √49 x²

x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0

3 years ago