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2 years ago

A bowling ball has a weight of 70 Newton’s on earth. What is it’s mass on earth? What’s it mass on the moon?

1 answer:

3 0

Weight = (mass) x (gravity)

70 N = (mass) x (9.8 m/s²)

Divide each side by (9.8 m/s²) , and you have

mass = 70 N / 9.8 m/s² = 7.14 kg.
___________________________

Mass on the moon:

Mass doesn't change. It's a number that belongs to the bowling ball,
no matter where the ball goes. If the mass of the bowling ball is 7.14 kg
anywhere, then it's 7.14 kg everywhere ... on Earth, on the moon, on Mars, rolling around in the trunk of my car, or floating in intergalactic space.

However, WEIGHT depends on the gravity wherever the ball happens to be
at the moment.

The acceleration of gravity on the moon is 1.622 m/s².
So the WEIGHT of the ball on the moon is

(7.14 kg) x (1.622 m/s²) = 11.58 Newtons

That's only about 16% of its weight on Earth.

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2 years ago

The difference between the speed of sound n air at 0°C and the speed of sound in air at 20°c is that... A. Cooler air molecules

sergejj [24]

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2 years ago

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Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic

Anni [7]

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass × g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction

Fa = 74.97 × 0.10

Fa = 7.49 N

and

force for moving constant velocity is

Fa = weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

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3 years ago

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Describe how impacts between objects in the solar system have helped create notable features in the solar system such as craters

konstantin123 [22]

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2 years ago

A certain string just breaks when it is under400N of tension.

OLEGan [10]

Answer:

The string will break with a speed of 20 m/s.

Explanation:

It is given that,

Tension at which the string just breaks, T = 400 N

Mass of the stone, m = 10 kg

Radius of the circle, r = 10 m

We need to find the speed at which the string will break. The boy continuously increases the speed of the stone. The tension acting on the stone is equal to the centripetal force. It is a force that acts towards the center of circle. It is given by :

$T=F=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{Tr}{m}}$

$v=\sqrt{\dfrac{400\ N\times 10\ m}{10\ kg}}$

v = 20 m/s

So, the string will break with a speed of 20 m/s. Hence, this is the required solution.

4 0

3 years ago