Answer to Question #283576 in Physics for Harrisonfaith

Question

Write down the dimensions of the following quantities: work, impulse, density, coefficient of viscosity, power, modulus of elasticity

Answer 1

(a) By the definition of work, we have:

$$W=Fd,$$

here, $F$ is the force, $d$ is the displacement.

Then, we can write the dimensions of work as follows:

$$W=[\dfrac{ML}{T^2}]\times [L]=[ML^2T^{-2}].$$

(b) By the definition of the impulse, we have:

$$J=Ft,$$

here, $F$ is the force, $t$ is the time.

Then, we can write the dimensions of impulse as follows:

$$J=[\dfrac{ML}{T^2}]\times[T]=[MLT^{-1}].$$

(c) By the definition of the density, we have:

$$\rho=\dfrac{m}{V},$$

here, $m$ is the mass, $V$ is the volume.

Then, we can write the dimensions of density as follows:

$$\rho=[\dfrac{M}{L^3}]=[ML^{-3}].$$

(d) By the definition of the coefficient of viscosity, we have:

$$\eta=\dfrac{F}{A\dfrac{dv}{dx}},$$

here, $F$ is the force, $A$ is the area, $\dfrac{dv}{dx}$ is the velocity gradient.

Then, we can write the dimensions of coefficient of viscosity as follows:

$$\eta=[\dfrac{ML}{T^{2}}]\times[L^{-2}]\times[(\dfrac{LT^{-1}}{L})^{-1}]=[ML^{-1}T^{-1}].$$

(e) By the definition of the power, we have:

$$P=\dfrac{W}{t}=\dfrac{Fd}{t},$$

here, $W$ is work, $F$ is the force, $d$ is the displacement, $t$ is the time.

Then, we can write the dimensions of power as follows:

$$P=[\dfrac{ML}{T^2}]\times[L]\times[T^{-1}]=[ML^2T^{-3}].$$

(f) By the definition of modulus of elasticity, we have:

$$Y=\dfrac{stress}{strain}=\dfrac{FL}{A\Delta L},$$

here, $F$ is the force, $L$ is the original length of the object, $A$ is the cross-sectional area, $\Delta L$ is the change in length of the object.

Then, we can write the dimensions of modulus of elasticity as follows:

$$Y=\dfrac{[MLT^{-2}]\times[L]}{[L^2]\times[L]}=[ML^{-1}T^{-2}].$$