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2 years ago

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An expanding gas does 153 J of work on its surroundings at a constant pressure of 1.01 atm. If the gas initially occupied 68.0 m

L, what is the final volume of the gas?

1 answer:

Natalija [7]2 years ago

3 0

Answer:

69.5mL

Explanation:

We are given that

Work done by gas on surroundings=w=153 J= $\frac{153}{101.3}Latm$

1Latm=101.3J

Pressure =P=1.01atm

Initial volume of gas=68 mL

We have to find the final volume of the gas.

We know that

Work done on constant pressure

$w=P\Delta V=P(V_2-V_1)$

Using the formula

$\frac{153}{101.3}=1.01(V_2-68)$

$V_2-68=\frac{153}{101.3\times 1.01}=1.5$

$V_2=68+1.5=69.5mL$

Hence, the final volume of gas=69.5mL

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Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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3 years ago

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

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3 years ago

A string under a tension of 68 N is used to whirl a rock in a horizontal circle of radius 3.7 m at a speed of 16.53 m/s. The str

alekssr [168]

Answer:

$F = 5253.7 N$

Explanation:

As we know that tension force in the string will be equal to the centripetal force on the string

so we will have

$T = \frac{mv^2}{L}$

now we have

$68 = \frac{m(16.53^2)}{3.7}$

now we have

$68 = 73.8 m$

$m = 0.92 kg$

now when string length is 0.896 m and its speed is 71.5 m/s then we will have

$F = \frac{mv^2}{r}$

$F = \frac{0.92(71.5^2)}{0.896}$

$F = 5253.7 N$

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