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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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Two balls of the same shape and size have charges of +8 and-2. The balls are brought together, allowed to touch, and then separa

mr Goodwill [35]

Explanation:

Initially, two balls of the same shape and size have charges of +8 and-2. The balls are brought together, allowed to touch, and then separated.

When they are touched, the total charge is evenly distributed among the whole surface and they behaves as if it is a single piece. Net charge at this point is +8-2=6 C. When they are separated, each ball will have a charge of 6/2 = 3 C.

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3 years ago

Which type of energy is the energy of a moving tennis ball

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Answer:

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3 years ago

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Some hydrogen gas is enclosed within a chamber being held at 200^\ { C} with a volume of 0.025 \rm m^3. The chamber is fitted wi

vlada-n [284]

Answer:

The final volume is $0.039 m^3$

Explanation:

Initial temperature: $T1=200C$

Final temperature: $T2=200C$

Initial pressure: $P1=1.50 \times10^6 Pa$

Final pressure: $P2=0.950 \times10^6 Pa$

Initial volume: $V1=0.025m^{3}$

Final volume: $V2=?$

Assuming hydrogen gas as a perfect gas it satisfies the perfect gas equation:

$\frac{PV}{T}=nR$ (1)

With P the pressure, V the volume, T the temperature, R the perfect gas constant and n the number of moles. If no gas escapes the number of moles of the gas remain constant so the right side of equation (1) is a constant, that allows to equate:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

Subscript 2 referring to final state and 1 to initial state.

solving for V2:

$V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=\frac{(1.50 \times10^6)(0.025)(200)}{(200)(0.950 \times10^6)}$

$V_{2}=0.039 m^3$

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3 years ago

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

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Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is $\sqrt{a^2 + b^2}$

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

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3 years ago

Two balls, one twice as massive as the other, are dropped from the roof of a building (freefall). Just before hitting the ground

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2 years ago