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Tema [17]
2 years ago
9

Jake lifted a 10-kg bell by a distance of 0.50 meters in 0.50 seconds. How much power did Jake use?​

Physics
1 answer:
Reil [10]2 years ago
8 0

Answer:

10 watts

Explanation:

first calculate work.

Work =force×distance cos thita

10Kg×0.50M cos 0= 5joules

Therefore, Power=Work÷ Time

Therefore, 5joules÷0.50s=10 watts.

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How is the acceleration of an object in uniform circular motion constant?
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What is the new volume of the gas if the pressure on 350 L of oxygen
Iteru [2.4K]

Answer:

420 L

Explanation:

Applying Boyle's Law,

PV = P'V'.................... Equation 1

Where P = Initial pressure, P' = Final pressure, V = Initial volume, V' = Final volume.

make V' the subject of the equation

V' = PV/P'.................... Equation 2

From the question,

Given: P = 720 mmHg, V = 350 L, P' = 600 mmHg

Substitute these values into equation 2

V' = (720×350)/600

V' = 252000/600

V' = 420 L

7 0
3 years ago
A single-turn current loop, carrying a current of 3.50 a, is in the shape of a right triangle with sides 50.0, 120, and 130 cm.
dimulka [17.4K]
Thank your very much
3 0
3 years ago
A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is
Maru [420]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q  = -5.02×10^{-9}C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8×10^{-2}m) (21.8×10^{-2}m)  = 4.75×10^{-4}m²

A  = 4.75×10^{-4}m²

Surface charge density = Q/A

Surface charge density = (-5.02×10^{-9}C)/(4.75×10^{-4}m²)

Surface charge density = -1.057×10^{-5} Cm^{-2}

3 0
3 years ago
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