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3 years ago

15

A man weighing 490N on earth weighs 81.7N on the moon.His mass on the moon is kg

2 answers:

Ymorist [56]3 years ago

6 0

His mass is 50 kg . . . on the Moon, on the Earth, in the capsule rocketing between them, and on Halley's comet if he ever goes there.

Veseljchak [2.6K]3 years ago

6 0

Answer:

50kg

Explanation:

The mass of a man on the Earth and the moon is the same, so we only need to use the 490N weight on Earth.

The formula for mass is m=w/g. Plugging in the values, we can determine that m=490N/9.8m/s^2. (g is gravitational constant, which is approximately 9.8m/s^2 on Earth)

The reason that the unit in the answer is in kg is because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the m/s^2 in the gravity constant, so it leaves kg.

m=50kg

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An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the

andrew-mc [135]

Answer:

50 W

Explanation:

Case 1

Power = V * I

100 = 220 * I

I = $\frac{100}{220}$ A

Case 2

P = V * I

P = 110 * $\frac{100}{220}$

P = 50 W

I think the answer is 50 W

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A piece of metal has a mass of 9.00 kg. If it displaces water that fills a container 10.0 cm x 10.0 cm x 10.0 cm, what is the ma

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3 years ago

180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperat

Zolol [24]

Answer : The final temperature of the mixture is $91.9^oC$

Explanation :

First we have to calculate the mass of water.

Mass = Density × Volume

Density of water = 1.00 g/mL

Mass = 1.00 g/mL × 180 cm³ = 180 g

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of hot water (liquid) = $4.18J/g^oC$

$c_2$ = specific heat of ice (solid)= $2.10J/g^oC$

$m_1$ = mass of hot water = 180 g

$m_2$ = mass of ice = 20 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of hot water = $97^oC$

$T_2$ = initial temperature of ice = $0^oC$

Now put all the given values in the above formula, we get

$(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC$

$T_f=91.9^oC$

Therefore, the final temperature of the mixture is $91.9^oC$

8 0

3 years ago