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2 years ago

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A toaster oven uses 65,000 joules of energy in 40sec to toast a piece of bread. What is the power of the oven?

1 answer:

storchak [24]2 years ago

3 0

Answer:

Power = 1625 Watts

Explanation:

Given the following data;

Energy = 65,000 J

Time = 40secs

To find the power;

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

Substituting into the equation, we have

Power = 65,000/40

Power = 1625 Watts

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If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

salantis [7]

Answer:

$v = r\omega$

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

$\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r$

$v = r\omega$

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2 years ago

Which statements accurately describe mechanical waves? Check all that apply.

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Answer:

A

Explanation:

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1 year ago

A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and

Airida [17]

Answer:

Explanation:

Given

$N_1=1 rev/s$

angular velocity $\omega =2\pi N_1=6.284 rad/s$

Combined moment of inertia of stool,student and bricks $=6\ kg.m^2$

Now student pull off his hands so as to increase its speed to suppose $N_2$ rev/s

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After Pulling off hands so final moment of inertia is

$I_2=2\ kg-m^2$

Conserving angular momentum as no external torque is applied

$I_1\omega _1=I_2\omega _2$

$6\times 6.284=2\times \omega _2$

$\omega _2=18.85\ rad/s$

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3 years ago

Which of the following is an impact of trampling of beach grass by humans and vehicles?

grigory [225]

Increase in sea water pollution

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2 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

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2 years ago