In 41 s, a pump delivers 0.628 dm3 of oil into barrels on a platform 35 m above the pump intake pipe. Calculate the work done by the pump. The density of the oil is 0.776 g/cm3 . and the acceleration due to gravity is 9.8 m/s 2 . Answer in units of kJ.

0.167 kJ

Explanation:

Given:

Volume of oil delivered by the pump (V) = 0.628 dm³ = 628 cm³     [1 dm³ = 1000 cm³]

Density of the oil (d) = 0.776 g/cm³

Acceleration due to gravity (g) = 9.8 m/s²

Time for which pump runs (t) = 41 s

Height to which oil is raised (h) = 35 m

Now, we know that, mass of a body is equal to the product of its volume and density. Therefore,

mass of oil (m) = volume × density

Now, work done by the pump is equal to the increase in the potential energy of the oil being raised to the given height.

Increase in potential energy is given as:

Therefore, the work done by the pump is given as:

Work done = ΔU

Now, 1 J = 0.001 kJ

Therefore, 167.04 J = 167.04 × 0.001 = 0.167 kJ

Hence, work done by the pump in kilojoules is 0.167 kJ

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm