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Serga [27]
1 year ago
12

Pls help me solve these questions.​

Physics
1 answer:
aliina [53]1 year ago
5 0

The value of constant in terms of Force, k, A, V and ρ is  x = ln (F /kAρV ).

<h3>What is opposing force?</h3>

The force which tends to oppose motion of fluid in forward direction like shear stress in liquids and friction force in solids

A body moving through the air at speed V, experiences the retarding force.

Force = kAρVˣ

where A is the surface area, ρ is the density of air and k is the numerical constant.

Taking natural logarithm both sides,

ln F = ln (kAρVˣ)

ln F = x ln (kAρV)

x = ln F / ln (kAρV)

x= ln (F /kAρV )

x= ln F -  ln (kAρV)

Thus, the value of constant is deduced in terms of Force , k, A, V and ρ.

Learn about opposing force.

brainly.com/question/13191643

#SPJ1

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Which situation describes the highest rate of power?
Darya [45]

Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.

Explanation :

Power : It is defined a the rate of doing work per unit time.

Formula used :

P=\frac{w}{t}

where,

P = power

w = work done

t = time

Now we have to determine the rate of power for the following options.

(A) A machine does 200 joules of work in 10 seconds.

P=\frac{200}{10}=20W

(B) A machine does 400 joules of work in 10 seconds.

 P=\frac{400}{10}=40W

(C) A machine does 200 joules of work in 5 seconds.

 P=\frac{200}{5}=40W

(D) A machine does 400 joules of work in 5 seconds.

P=\frac{400}{5}=80W

From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.

Hence, the correct option is, (D)

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3 years ago
Which describes the characteristics of a liquid?
Talja [164]
What happens when you pour water in a glass? It takes the shape of the glass. This means that water can't have a fixed shape or volume

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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

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V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

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