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3 years ago

50 J of work was performed in 20 seconds. How much power was used to perform this task? A. 0.4 W B. 2.5 W C. 4 W D. 24.5 W

Physics

1 answer:

Ugo [173]3 years ago

5 0

Answer:

B. 2.5 W

Explanation:

Power: This can be defined as the rate at which energy used or it i the rate at which work is done. The S.I unit of power is Watt (W).

Mathematically,

Power = Energy or work/Time

P = W/t ........................Equation 1

Where P = power used to perform the task, W = work, t = time.

Given: W = 50 J, t = 20 s.

Substitute into equation 1

P = 50/20

P = 2.5 W.

Hence the power is 2.5 W.

The right option is B. 2.5 W

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A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet

puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

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4 years ago

What is the SI unit used to measure the temperature of a substance?

ss7ja [257]

Answer:

Kelvin

Explanation:

Kelvin is the universal and scientific unit for temperature as Celsius and Fahrenheit temperatures we use in everyday situations

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A system that can be affected by the outside environment, by an exchange of matter or energy is ______.

larisa86 [58]

$\sf{Answer:}$

A system that can be affected by the outside environment, by an exchange of matter or energy is an open physical system .

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Question 8

viktelen [127]

Answer: D(t) = $8.e^{-0.4t}.cos(\frac{\pi }{6}.t )$

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y = $a.sin(\omega.t)$ or y = $a.cos(\omega.t)$

where:

|a| is initil displacement

$\frac{2.\pi}{\omega}$ is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

$y=a.e^{-ct}.cos(\omega.t)$ or $y=a.e^{-ct}.sin(\omega.t)$

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

$y=a.e^{-ct}.cos(\omega.t)$

period = $\frac{2.\pi}{\omega}$

12 = $\frac{2.\pi}{\omega}$

ω = $\frac{\pi}{6}$

Replacing values:

$D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$

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