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icang [17]
3 years ago
9

Gravitational force on the moon is only 1/6 that of the gravitational force on earth. what would be the weight of a 10-kg object

on the moon and on earth? what would be its mass on the moon and on earth?
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
6 0
According to Newton's Second Law of Motion, the force is equal to the mass of an object multiplied by its acceleration. When you talk about gravitational force, the acceleration referred to here is the acceleration due to gravity. This is very familiar to us in physics. The acceleration due to gravity on Earth is equal to 9.81 m/s². It actually depends on the location. According to the Universal Law of Gravitation:

F = Gm₁m₂/d²

The force is a factor of the product of two masses and their distance from each other. The G is a constant called the universal gravitational constants. So, gravitational force is actually a relative force exerted by one body to another. 

Going back the Second Law of Motion, we can modify the equation to:

F = mg

Since it is mentioned that the gravity on the moon is only 1/6 of the Earth, then the gravity for moon is:
g,moon = 1/6(9.81) = 1.635 m/s²
So, let's compare the weight of the object with a mass of 10 kg. The weight is actually the force due to gravity pulling you towards the center of the body.

Weight on Earth = (10 kg)(9.81 m/s²) = 98.1 N
Weight on Moon = (10 kg)(1.635 m/s²) = 16.35 N

The mass, on the other hand, is not affected by gravity. It is always constant. Therefore, the mass of the object on the moon is the same with its mass on the Earth.
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If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu
zubka84 [21]

Answer:

minimum speed is 15.35 m/s

frictional coefficient  is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and   minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² =  236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 =  0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245  = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient  is 0.26

4 0
3 years ago
Read 2 more answers
DNA’s diameter is 0.000000002 meters. What is this measurement in scientific notation? a. 2.0 x 109 m
NARA [144]

Answer:

0.000000002 m=2.0*10⁻⁹ m

Explanation:

Scientific notation allows us to write very large or very small numbers in abbreviated form. This notation simply consists of multiplying by a power of base 10 with a positive or negative exponent.

A number written in scientific notation has the form:

a*10ⁿ

where:

  • the coefficient a has a value such that 1 ≤ a <10
  • n is an integer.  Represents the number of times the decimal point is shifted. It is always a whole number, positive if it is shifted to the left, negative if it is shifted to the right.

So to write the number 0.000000002 in scientific notation, the following steps are performed:

  • The decimal point is moved to the right as many spaces until it reaches the right of the first digit.
  • This number is then written, which will be the coefficient a in the expression of the previous product. So a=2.0
  • The base 10 is written with the exponent equal to the number of spaces that the comma moves. So n=9. But this is a negative number because the comma shifts to the right.

So, you get: <u><em>0.000000002 m=2.0*10⁻⁹ m</em></u>

3 0
3 years ago
Why do stars appear so steady when viewed from the surface of moon or<br> by an astronaut in space?
Nutka1998 [239]

Answer:

Explanation:

The light from these little disks is also refracted by Earth's atmosphere, as it travels toward our eyes. That's because, in the direction of any horizon, you're looking through more atmosphere than when you look overhead. If you could see stars and planets from outer space, both would shine steadily.

Answer From GauthMath If you like it then please heart it and comment thanks.

4 0
2 years ago
Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such
Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity  = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0
3 years ago
The acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. Th
Lorico [155]
1/3 the weight than it is on earth, duh
3 0
3 years ago
Read 2 more answers
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