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3 years ago

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An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc

reases by only 500 J. During this process, how much heat flows into or out of the ideal gas? Enter a positive number to indicate a heat flow into the gas or a negative number to indicate a heat flow out of the gas.

1 answer:

Phantasy [73]3 years ago

3 0

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

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In a material for which o=5.0 S/m and e, =1 the electric field intensity is E- 250 sin 10" (V/m). Find the conduction and displa

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Answer:

Explanation:

Given that:

The conductivity of the material $\sigma$ = 5.0 s/m

$The \ relative \ permittivity \ of \ the \ material$ ${\varepsilon_r} = 1$ ; &

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(a) The conduction current density $( J_c) = \sigma E$

$= 5.0 \times 250 \ sin ( 10^{10} \ t )$

$\mathbf{ = 1250 \ sin (10 ^{10} t ) \ A/m^2 }$

(b) Displacement current density $(J_d) = \varepsilon _d * \dfrac{\delta E}{\delta t}$

Recall that:

$\varepsilon _o = 8.854 \times 10^{-12}$

∴

$(J_d) = 8.854 \times 10^{-12} \times \dfrac{d}{dt} \times (250 \ sin \ (10^{10} \ t))$

$(J_d) = 8.854 \times 10^{-12} \times 250 \times 10^{10} \times \ cos \ (10^{10} \ t)$

$(J_d) = 22.135 \ cos \ 10^{10} \ t \ A/m^2$

(c) The frequency at which $J_c \ and \ J_d$ will have the same magnitude is:

$f = \dfrac{\sigma}{2 \pi \varepsilon_o \varepsilon_r}$

By substitution

$f = 18 \times 10^9 \times \dfrac{\sigma }{\varepsilon_r}$

$f = 18\times 10^9 \times \dfrac{5}{1 }$

f = 90 GHz

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A cylindrical pulley with a mass of 4.4 kg, radius of 0.816 m and moment of inertia 1 2 M r2 is used to lower a bucket with a ma

satela [25.4K]

Answer:

5.85 m/s²

Explanation:

Mass of pulley (M) = 6.8 kg, mass of bucket (m) = 1.5 kg, radius of pulley (r) = 0.816 m, acceleration due to gravity (g) = 10 m/s².

Let the tension exerted on the pulley be T and the angular acceleration be α,

The angular acceleration (α) = a / r; where a is the linear acceleration.

mg - T = ma

T = m(g - a)

Also, for the disk:

Tr = Iα

I = moment of inertia = 0.5Mr²

m(g - a)r = 0.5Mr²(a/r)

m(g - a)r = 0.5Mra

2m(g - a) = Ma

2mg - 2ma = Ma

Ma + 2ma = 2mg

a(M + 2m) = 2mg

a = 2mg / (M + 2m)

Substituting:

a = 2(3.1 kg)(10 m/s²) / (4.4 + 2(3.1))

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What is the number of paths through which electricity can flow in a series circuit?

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A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

sattari [20]

Answer:

Magnetic force, $F=9.6\times 10^{-17}\ N$

Explanation:

It is given that,

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The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

$F=qvB\ sin\theta$

The magnetic field is oriented perpendicular to the proton’s velocity, $\theta=90^{\circ}$

$F=1.6\times 10^{-19}\times 300\times 2$

$F=9.6\times 10^{-17}\ N$

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is $9.6\times 10^{-17}\ N$ . Hence, this is the required solution.

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A lamp uses a 230 V mains supply and transfers 96 J of energy every second. Work out the current through the lamp. Give your ans

sertanlavr [38]

Answer:

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Explanation:

From the question,

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I = 96/230

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4 years ago