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3 years ago

A lawnmower engine running for 20 minutes does 4560000 j of work. What is the power output of the engine

Physics

1 answer:

Alenkasestr [34]3 years ago

3 0

<em>Answer: </em>

tim e (t) = 20 min.

= 20 × 60 = 1200 s ,

Work ( W) = 4560000 J

= 4560 KJ ,

Determine:

Power output (P) = Work ÷ time

= 4560 ÷ 1200

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A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During th

fgiga [73]

Answer:

the final energy of the system is 35.5 kJ.

Explanation:

Given;

initial energy of the system, E₁ = 10 kJ

heat transferred to the system, q₁ 30 kJ

Heat lost to the surrounding, q₂ = 5kJ

heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

E₂ = 10 kJ + 25.5 kJ

E₂ = 35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.

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2 years ago

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3 years ago

What is the measure of an average kinetic energy??

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3 years ago

1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?

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Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

Mass m = 1.75 kg

Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

$K.E=\frac{1}{2}mv^2$

where

m is the mass (kg)

v is the velocity (m/s)

K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2$

$K.E=1458\times 1.75$

$K.E=2551.5$ J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

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2 years ago

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

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Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

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

we know that

force = pressure * area

therefore force is

$F =(\frac{T_{1}}{T_{2}}*P_{1})A$

$F =(\frac{378}{278}*101325)(17*10^{-2})^{2}$

F = 3.98 kN

5 0

3 years ago