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Gelneren [198K]

3 years ago

5

Volume of water shrinks between__

2 answers:

Lena [83]3 years ago

8 0

Answer:

when liquid water is cooled it contracts like one would expect until a temperature of approximately 4 degrees Celsius is reached. After that it expands slightly until it reaches the freezing point and then when it freeze it expands by Approximately 9percent

lorasvet [3.4K]3 years ago

5 0

Answer:

When liquid water is cooled, it contracts like one would expect until a temperature of approximately 4 degrees Celsius is reached. After that, it expands slightly until it reaches the freezing point, and then when it freezes it expands by approximately 9%.

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A man in a strength competition pulls an 18-wheel truck 3.10 m in 20.5 s. There is a cable that is attached to his body that exe

larisa [96]

Answer:

114.32195122 but Round your answer to three significant figures.) is 114

Explanation:

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1. How far did she travel in total?

Aloiza [94]

Why give the ecuación

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3 years ago

Sound travels at a rate of 340 m/s in all directs through the air. Matt rings a very loud bell at one location, and Steve hears

DENIUS [597]

Answer:

It will take about 1.32 seconds to travel to his location.

Explanation:

Considering the sound travels at 340 m/s, then if a person is at a distance of 450 m m from the bell, we can use the velocity formula to find the answer;

$velocity=\frac{distance}{time} \\340\,\frac{m}{s} =\frac{450\,\,m}{t} \\t=\frac{450}{340} \,s\\t\approx 1.32\,\,s$

4 0

3 years ago

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux

Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

$\Phi = EA cos \theta$

where

E is the electric field

A is the cross-sectional area

$\theta$ is the angle between the direction of the electric field and the normal to A

The flux is maximum when $\theta=0^{\circ}$ , so we are in this situation and therefore $cos \theta =1$ , so we can write

$\Phi = EA$

Here we have:

$\Phi = 7.50\cdot 10^5 N/m^2 C$ is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

$A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2$

And so, we can find the magnitude of the electric field:

$E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C$

3 0

3 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

<u>Now change in internal energy:</u>

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

B.

<u>Now heat added to the system:</u>

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

7 0

3 years ago