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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature $(T_i)=300 K$

$P_i=10 atm$

$P_f=2 atm$

Work done in iso-thermal process $=P_iV_iln\frac{P_i}{P_f}$

$P_i$ =initial pressure

$P_f$ =Final Pressure

$W=10\times 2.463\times ln\frac{10}{2}=39.64 J$

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

$V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L$

$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$

8 0

3 years ago

A 22kg Accelerates at a rate of 2.3 m/s. What is the magnitude of the net force acting on the bike?

Tcecarenko [31]

magnitude of the net force = mass x acceleraton

= 22 x 2.3

=50.6 N

7 0

3 years ago

A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3

Klio2033 [76]

Answer:

$\theta=30^{\circ}$

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

$F=ILB\ sin\theta$

$sin\theta=\dfrac{F}{ILB}$

$sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}$

$\theta=30^{\circ}$

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0

2 years ago

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

3 0

2 years ago

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

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