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Debora [2.8K]
3 years ago
12

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

te. Part A Calculate the average power delivered by the engine at this rotation rate.
Physics
1 answer:
jekas [21]3 years ago
5 0

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

P=\tau\omega

Where

\tau = Torque

\omega =Angular speed

Our values are given by

\tau = 630Nm

\omega = 3200rev/min

The angular velocity must be transformed into radians per second then

\omega = 3200rev/min (\frac{2\pi rad}{60s})

\omega = 335.103rad/s

Replacing,

P=(630)(335.103)

P = 211.11*10^3W

P = 211.1kW

The average power delivered by the engine at this rotation rate is 211.1kW

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A 30.0 g mass of iron at 24.5°C is heated to 45.0°C. The theoretical specific
diamong [38]

Answer:

276.135 J

Explanation:

Given that:

mass of Fe = 30.0 g

initial temperature = 24.5°C

final temperature = 45.0°C

specific heat of Fe = 0.449 J/g°C

We can determine the thermal energy added by using the formula;

Q = mcΔT

Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C

Q = 276.135 J

8 0
3 years ago
Gauss’ law: a. Relates the surface charge density to the electric field.b. Relates the electric field at points on a closed surf
Mamont248 [21]

Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

8 0
3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
What is the solution to the problem expressed to the correct number of significant figures?
oksian1 [2.3K]

Answer:

73.72

Explanation:

For this subtraction problem, the answer or solution is expressed to the least precise of the numbers we are trying to subtract.

The least precise number is the number with the lowest significant numbers:

105.4 - 31.681

105.4  has 4 significant numbers

31.681 has 5 significant numbers

  So;

             105.4  

        -      31.681

        ------------------

                 73.719

          ----------------

The solution is therefore  73.72

7 0
3 years ago
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