Ask question

Ask question

All categories

3 years ago

11

A system initially has an internal energy U of 504 J. It undergoes a process during which it releases 111 J of heat energy to th

e surroundings, and does work of 222 J. What is the final energy of the system, in J?

1 answer:

Likurg_2 [28]3 years ago

6 0

Answer:

615 J

Explanation:

internal energy (U) = 504 J

heat lost (q) = 111 J = - 111 J (negative sign is because heat is lost)

work done = 222 J

what is the final energy in the system

total energy = final energy - initial energy

final energy = total energy + initial energy

where

initial energy = 504 J

total energy = 222 - 111 = 111 J

final energy = 504 + 111 = 615 J

You might be interested in

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

If your mass is 20kg, and you stand on the scale (witch is your force), and the acceleration due to gravity is10m/s^2, what does

Phantasy [73]

Weight = (mass) x (gravity)

= 200 Newtons.

(About 44 pounds. You're very skinny.)

5 0

3 years ago

Read 2 more answers

Hi gir_ls join nkd-mbja-nuj

GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

3 0

3 years ago

Which symbol and unit of measurement are used for electric current?

Burka [1]

Answer: Symbol is I and unit A

Explanation: A represents Amperes

HOPE THIS HELPS!!!!!!!!

4 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

Dmitrij [34]

Answer:

$\lambda_2 = 573.3 nm$

Explanation:

As we know that the position of maximum intensity on the screen is given as

$y = \frac{N\lambda L}{d}$

here we know that

$\lambda$ = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

$\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}$

so here we have

$8 (645 nm) = 9 \lambda_2$

$\lambda_2 = 573.3 nm$

4 0

3 years ago