# Answer to Question #30812 in Mechanics | Relativity for disha

Question #30812

assuming that the largest mass that can be moved by a flowing river depends on the velocity of flow,density of river water,and acceleration due to gravity,show that the mass varies as the sixth power of velocity of flow.

Expert's answer

We posit that

M = C v^a * p^b * g^c

with C a dimensionless constant, and a , b and c (dimensionless) exponents.

Let us determine what values of a, b and c are allowed to make the formula dimensionally correct.

v is s speed, so length/time ( L / T) .

p is a density, so mass/volume (m / L^3)

g is an acceleration, so length / time^2 ( L / T^2)

This means that in terms of dimensions, the equation is

m = (L / T)^a * (m / L^3)^b * ( L / T^2)^c

So, combining powers in each quantity (L, m, T):

m = L^(a - 3b + c) * m^b * T^(-a -2c)

Whence we conclude that

b = 1

a - 3b + c = 0

-a -2c = 0

I.e.

b = 1

a + c = 3

a + 2 c = 0

The latter two equations are solved by

c = -3

a = 6

So

M = C * v^6 * p / g^3

M = C v^a * p^b * g^c

with C a dimensionless constant, and a , b and c (dimensionless) exponents.

Let us determine what values of a, b and c are allowed to make the formula dimensionally correct.

v is s speed, so length/time ( L / T) .

p is a density, so mass/volume (m / L^3)

g is an acceleration, so length / time^2 ( L / T^2)

This means that in terms of dimensions, the equation is

m = (L / T)^a * (m / L^3)^b * ( L / T^2)^c

So, combining powers in each quantity (L, m, T):

m = L^(a - 3b + c) * m^b * T^(-a -2c)

Whence we conclude that

b = 1

a - 3b + c = 0

-a -2c = 0

I.e.

b = 1

a + c = 3

a + 2 c = 0

The latter two equations are solved by

c = -3

a = 6

So

M = C * v^6 * p / g^3

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