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2 years ago

11

Does mass change when volume does

1 answer:

alukav5142 [94]2 years ago

6 0

Answer:

Yes

Explanation:

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hai điện tích điểm bằng nhau có độ lớn 6.10^-8C đặt cách nhau một khoảng là r trong dầu hỏa có hằng số điện môi E=2,1. Lực tương

nadezda [96]

Answer:

hai điện tích điểm bằng nhau có độ lớn 6.10^-8C đặt cách nhau một khoảng là r trong dầu hỏa có hằng số điện môi E=2,1. Lực tương tác tĩnh điện giữa chúng laf3,85.10^-4. tìm r

5 0

2 years ago

In a ray diagram showing refraction, the incident ray and the refracted ray...

Amanda [17]

C because of the positioning of the initial ray of refraction

4 0

2 years ago

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)= $3.2 m/s^2$

Speed of car after 4.5 s

using equation of motion

v=u+at

$v=4.92+3.2\times 4.5=4.92+14.4$

v=19.32 m/s

Displacement of the car after 4.5 s

$v^2-u^2=2as$

$19.32^2-4.92^2=2\times 3.2\times s$

$349.05=2\times 3.2\times s$

s=54.54 m

4 0

2 years ago

According to Bernoulli's fluid formula a An increase in the speed will lower the internal pressure b An increase in the speed wi

lorasvet [3.4K]

Answer:

a An increase in the speed will lower the internal pressure

Explanation:

Bernoulli's fluid formula

$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_1+\frac{1}{2}\rho v_2^2+\rho gh_2$

where

P = Pressure

ρ = Density of fluid

g = Acceleration due to gravity

h = Height

v = Velocity of fluid

If there is no change in height then we get

$P_1+\frac{1}{2}\rho v_1^2=P_1+\frac{1}{2}\rho v_2^2\\\Rightarrow P+\frac{1}{2}\rho v^2=constant$

According to the Bernoulli's principle when the speed of the fluid is larger in a region of streamline flow the pressure is smaller in that region. From the above equation it can be seen that increase in speed should simultaneously reduce pressure in order for their sum to be constant.

5 0

2 years ago

imagine that 501 people are present in a movie theater of volume 8.00 x10^3 that is sealed shut so no air can escape. Each perso

Semenov [28]

Answer:

The temperature of air will increase by $\Delta T=41044.967\ K$

Explanation:

Given:

no. of person in a theater, $n=501$

volume of air in the theater, $V=8\times 10^3\ m^3$

rate of heat given off by each person, $P=110\ J.s^{-1}$

duration of movie, $t=2\ hr=7200\ s$

initial pressure in the theater, $p_i=1.01\times 10^5\ Pa$

initial temperature in the theater, $T_i=20+273=293\ K$

specific heat capacity of air at the given conditions, $c=1.0061\ J.kg^{-1}.K^{-1}$

<u>The total quantity of heat released by the total people in the theater during the movie:</u>

$Q=n.P.t$

$Q=501\times 110\times 7200$

$Q=396792000\ J$

<u>Form the relation of heat capacity:</u>

$Q=m.c.\Delta T$

∵ $p_i.V=m.R.T$

$Q=(\frac{p_i.V}{R.T}) \times c\times (T_f-T_i)$

$396792000=(\frac{1.01\times 10^5\times 8\times 10^3}{287\times 293}) \times 1.0061\times (T_f-293)$

$T_f=41337.967\ K$

Change in temperature of air:

$\Delta T=41044.967\ K$

8 0

2 years ago