Question

A hot plate of area 0.125 m2
is pulled at 0.25 m/s with respect to
stationary parallel plate 1 mm distant from it. Space between the plates
containing water of viscosity 0.001 Ns/m2
, find the force necessary to
maintain this velocity and also the power required.

Answer 1

Answer:

i don't know because not able to understand the question

try to explain it well

Answer 2

Answer:

The force necessary to maintain this velocity and also the power required is 3.125×10⁻²N and 7.8125×10⁻³ J/s respectively.

Explanation:

The force is calculated as,

$F=\eta A\frac{v}{t}$      (1)

Where,

F=force necessary to maintain velocity

η=coefficient of viscosity

A=area of the plate

v=velocity with which the plate is pulled

t=thickness of the plate

From the question we have,

The coefficient of viscosity=0.001 Ns/m²=10⁻³ Ns/m²

The area of the plate=0.125 m²

The velocity with which the plate is pulled=0.25m/s

The thickness of the plate=1mm=1×10⁻³m

When all the necessary values are entered into equation (1), we obtain;

$F=10^-^3\times 0.125\times \frac{0.25}{1\times 10^-^3}$

$F=0.03125$

$F=3.125\times 10^-^2\ N$   (2)

The power required is calculated as,

$P=F.v$    (3)

When we enter the values into equation (3), we obtain;

$P=3.125\times 10^-^2\times 0.25$

$P=7.8125\times 10^-^3\ J/s$    (4)

Hence, the force necessary to maintain this velocity and also the power required is 3.125×10⁻²N and 7.8125×10⁻³ J/s respectively.

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