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jarptica [38.1K]
2 years ago
7

What is the weight (in pounds) of a 7.0 kilogram bowling ball on earths surface

Physics
1 answer:
alekssr [168]2 years ago
3 0

A bowling ball of 7.0 kg will be weighed around 15.54 lbs (pounds) at the surface of the Earth.

<u>Explanation:</u>

The weight of the bowling ball as given in the question,

W=7.0 \mathrm{kg}

Since we know that the force of gravity acting on an object at the surface of the Earth is 2.22 lb i.e. the object will be weighed as 2.22 lb at the surface of the Earth. Similarly, here in this case;

The weight of the bowling ball,

W=7.0 \mathrm{kg} \times 2.22 \frac{\mathrm{lb}}{\mathrm{kg}}

W=15.54 l b

Hence, the bowling ball will be weighed as 15.54 lbs at the surface of the Earth.

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A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot
denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

P_r = 345 Watt

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

P_r = 345 Watt

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

\Delta V = E + i r

here we have

E = 12 V

i = 56 A

r = 0.11

now we have

\Delta V = 12 + (0.11)(56) = 18.16 V

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

Part c)

Rate of energy conversion into EMF is given as

P_{emf} = i E

P_{emf} = (56)(12)

P_{emf} = 672 Watt

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

V = E - i r

V = 12 - (56)(0.11)

V = 12 - 6.16 = 5.84 V

Part e)

now the rate of energy dissipation is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

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Explanation:

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Answer:

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