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2 years ago

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The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. the heat increased the _

____ of the air in the room, which means an increase in the air's molecular kinetic energy.

1 answer:

Levart [38]2 years ago

4 0

<span>The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. the heat increased the temperature of the air in the room, which means an increase in the air's molecular kinetic energy.

When heat leaves a system, that energy is transferred to its surroundings. Since the air is the surroundings, the heat is transferred to the air increasing the temperature. This causes an increase in the individual air molecules' energy.</span>

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a car travel the first 20km with a speed of 40km/h and the next 40km with a speed of 80km/h . find the average speed

REY [17]

Answer:

average speed is 60km/h

Explanation:

you sum up the speed attained in each distance covered and divide it by 2 to get your answer

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2 years ago

The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the

Maru [420]

Answer: $(48.41,\ 52.19)$

Explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : $n=9$

Sample mean : $\ovreline{x}=50.3\text{ decibels}$

Standard deviation : $\sigma=2.9\text{ decibels }$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=z_{0.025}=1.96$

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

$50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)$

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2 years ago

8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

3 0

2 years ago

Nataliya [291]

I think is between A&B
I think I would answer B

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:

<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:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

2 years ago