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3 years ago

12in 4in 5in a. 40in2 b. 216in c. 256in d. 176in? What is the answer

Physics

1 answer:

daser333 [38]3 years ago

6 0

Answer:

Explanation:

because 20 + 20 = 40 and 40 = 8 * 5

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You have a box of filled with gas and by some process you double the number of atoms in the box, but now each one only has (on a

Fed [463]

Answer:

<u>Option-(D):</u>Thermal energy unchanged, temperature decreased.

Explanation:

<u>Thermal energy and Temperature,T:</u>

Thermal energy of the system is remained unchanged while the temperature,T of the objects present inside the system is more associated with the objects fundamental properties which each matter shows or exhibits in nature.

While the thermal energy is the flow or exchange of thermal or heat energy between the different objects in interaction to each other.

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3 years ago

WILL GIVE BRAINLY TO CORRECT ANSWER

11111nata11111 [884]

C is the answer to your question!

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3 years ago

Read 2 more answers

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

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4 years ago

Normal conversation has a sound level of about 60 db. How many times more intense must a 100-hz sound be compared to a 1000-hz s

vampirchik [111]

Answer:

5.65 times

Explanation:

60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.

But when the frequency of sound is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.

A decrease of 10 phon on phon- scale makes sound 2 times less loud

Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz

must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at 1000 Hz frequency.

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4 years ago

Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy

uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

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4 years ago