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konstantin123 [22]

3 years ago

7

A spherical tank with radius 4 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top.

Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)

1 answer:

kompoz [17]3 years ago

7 0

The answer & explanation for this question is given in the attachment below.

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Which of the following is NOT a characteristic of an inner planet?. . A.. rocky. . B.. solid surface. . C.. near the sun. . D..

sveticcg [70]

Being made mostly of gas is NOT a characteristic of an inner planet. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

3 0

3 years ago

Read 2 more answers

During an isothermal process, 10 j of heat is removed from an ideal gas. What is the work done by the gas in the process?

Black_prince [1.1K]

The work done in the isothermal process is 10 joule.

We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.

ΔU = 0

Hence,

ΔU = Q - W

0 = Q - W

Q = W

It means that the heat transferred is the same as the work done.

From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.

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8 0

1 year ago

Un objeto, que se encuentra a nivel del suelo, es lanzado verticalmente hacia arriba con una velocidad de 160 km/hr. ¿Qué altura

LiRa [457]

I don’t speak Spanish?

6 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe

RideAnS [48]

Sarah's acceleration is $-0.49 m/s^2$

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

$F_f = \mu mg$

where

$\mu$ is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

$F=ma$

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

$F=-\mu mg = ma$

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

$a=-\mu g$

And substituting the following values:

$\mu = 0.05$ (coefficient of friction)

$g=9.81 m/s^2$ (acceleration of gravity)

we find:

$a=-(0.05)(9.81)=-0.49 m/s^2$

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4 0

3 years ago