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3 years ago

At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's

Physics

2 answers:

hram777 [196]3 years ago

5 0

Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

Ira Lisetskai [31]3 years ago

3 0

Answer:

D(t)=50 (0.8)^t

Explanation:

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Which of the following solid fuels has the highest heating value?

Makovka662 [10]

I wanna say its A . I could be wrong but im almost 100 percent sure that its A wood

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3 years ago

A dancer lifts his partner above his head with an acceleration of 2.5 m/s squared the dancer. Exerts a force of 200N what is the

elena-s [515]

F = ma
F = 200N, a = 2.5m/s^2

200N = (2.5)m
m = 80kg

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3 years ago

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The amount by which a machine multiplies input force is known as

fiasKO [112]

Mechanical advantage is the amount by which a machine multiplies input force. It is a measure of the force amplification using a tool or machine system.

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4 years ago

A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?

harina [27]

Answer:

impulse = 8820 kg· $\frac{m}{s}$ or 8820 N·s

Explanation:

Impulse J is equal to the average force $F_{av}$ multiplied by the elapsed time Δt or in equation form, J = $F_{av}$ Δt

As long as your force of 450 N is constant then that value is your average force $F_{av}$ and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg· $\frac{m}{s}$ or 8820 N·s.

6 0

3 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:

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

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

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

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

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

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:

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

$\Delta W = 7475.69 joules$

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:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

3 years ago