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3 years ago

600 J of work is done on a system in a process that decreases the thermal energy of the system by 300 J.

Physics

1 answer:

Daniel [21]3 years ago

6 0

Answer:

900 J

Explanation:

$\Delta U$ = Change in entropy = -300 J (Decrease)

$W$ = Work done = 600 J

$Q$ = Heat transferred

Change in internal energy is given by

$\Delta U=W+Q\\\Rightarrow Q=\Delta U-W\\\Rightarrow Q=-300-600\\\Rightarrow Q=-900\ J$

The heat transferred to or from the system as heat is -900 J.

900 J is transferred. As heat is transferred from the system the sign is negative.

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I need help on this

ZanzabumX [31]

What is the question asking?

6 0

3 years ago

A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that

kherson [118]

Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

v = 4.3772331227 × 10^7m/s

Approximately = 4.4 × 10^7 m/s

Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

4 0

3 years ago

If I get a spare in one frame and then roll a 1 and 5 in the next frame, what would my score for the spare be?

bonufazy [111]

The answer is letter choice B...

8 0

3 years ago

A parallel circuit has a 125 Volt battery connected with 3 resistors. R1= 20 Ω, R2= 100 Ω, and R3= 50 Ω. Find the total current

Musya8 [376]

Answer:

10A

Explanation:

to calculate r

1/R=1/20+1/100+1/50=2/25

r=25/2=12.5

I=125/12.5=10A

8 0

3 years ago

Read 2 more answers

Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di

jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at $20^0\ C$ is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

$d=\dfrac{\lambda}{2}$ ........(1)

Velocity of sound wave is given by :

$v=f\times \lambda$

$d=\dfrac{v}{2f}$

$d=\dfrac{343\ m/s}{2\times 686\ Hz}$

$d=0.25\ m$

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

$d=n\lambda$ ( n = integers )

Let n = 1

So, $d=\dfrac{v}{f}$

$d=\dfrac{343\ m/s}{686\ Hz}$

$d=0.5\ m$

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0

3 years ago

Read 2 more answers