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Pavlova-9 [17]
3 years ago
7

If the mass of an object on Earth is 18 grams, what would the mass of the same object be on the Moon?

Physics
1 answer:
Tanya [424]3 years ago
4 0

Answer:18 grams

Explanation:

The mass of an object doesn't change,it is the same everywhere irrespective of it location,therefore if the mass of an object in earth is 18 grams, the mass of the object in the moon will also be 18 grams

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An artist wants to create a metal sculpture using a mold so that his artwork can be readily mass produced. He wants his sculptur
lukranit [14]

Answer:NO

Explanation:

No the mold should not be of the same size as that of sculpture because the material from which molds is made may shrink or expand depending upon its properties .

For example grey cast iron shrinks on cooling.

We need to make mold bigger in general so that if there is a need of finishing it can be done easily without altering the size of sculpture.

5 0
3 years ago
If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
A dentist uses a concave mirror (focal length 2.5 cm) to examine some teeth. If the distance from the object to the mirror is 1.
AysviL [449]

Answer:

2.28

Explanation:

From mirror formula,

1/f = 1/u+1/v .......... Equation 1

Where f = focal length of the mirror, v = image distance, u = object distance.

Note: The focal length mirror is positive.

make v the subject of the equation,

v = fu/(u-f)............ Equation 2

Given: f = 2.5 cm, u = 1.4 cm

Substitute into equation 2

v = 2.5(1.4)/(1.4-2.5)

v = 3.5/-1.1

v = -3.2 cm.

Note: v is negative because it is a virtual image.

But,

Magnification = image distance/object distance

M = v/u

Where M = magnification.

Given: v = 3.2 cm, u = 1.4 cm

M = 3.2/1.4

M = 2.28.

Thus the magnification of the tooth = 2.28.

3 0
3 years ago
A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.
Vesnalui [34]

Answer:

Normal, Gravity, Friction, and Air Resistance.

Explanation:

When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:

<u>Normal force:</u> It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.

<u>Gravity:</u> Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.

<u>Friction: </u>Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.

<u>Air resistance:</u> air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.

Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."

4 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}&#10; 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
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