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3 years ago

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2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to ma

itain this Speed. Determine the viscosity of the fluid between the plates?

1 answer:

Marizza181 [45]3 years ago

5 0

Answer:

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So, μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s

μ = 6.54 × 10⁻⁵ Ns/m²

μ = 6.54 × 10⁻⁵ Pa-s

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A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A,

daser333 [38]

Answer:

<h2>16,931 turns</h2>

Explanation:

The magnetic field produced is expressed using the formula

$B = \frac{\mu_0NI}{L}$

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$\mu_0$ is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A

L is the length of the solenoid = 32 cm = 0.32 m

N is the number of turns in the solenoid.

Making N the subject of the formula from the equation above;

$B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\$

$N = \frac{BL}{\mu_0I}$

Substituting the give values to get N;

$N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21$

The number of turns the solenoid must have is approximately 16,931 turns

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3 years ago

A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

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Answer:

Explanation:

Given

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mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

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$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

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$n=\frac{4243.6686}{64.8}=65.48\approx 66$

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taurus [48]

Answer:

Work is done by the heart on the blood during this time is 0.04 J

Explanation:

Given :

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Initial kinetic energy of blood is determine by the relation:

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Applying work-energy theorem,

Work done = Change in kinetic energy

W = E₂ - E₁

$W=\frac{1}{2} m (v^{2}-u^{2})$

Substitute the suitable values in the above equation.

$W=\frac{1}{2}\times0.08\times (1^{2}-0^{2})$

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3 years ago

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Answer:

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Explanation:

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