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3 years ago

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A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction

between the refrigerator and the floor is 230 N. How much work has been performed by the mover on the refrigerator?

1 answer:

masha68 [24]3 years ago

8 0

Answer:

1863 J

Explanation:

F = Force moving the object

f = Frictional force = 230 N

The forces in the horizontal direction have the force balance

$F-f=0\\\Rightarrow F=f\\\Rightarrow F=230\ \text{N}$

So, the force by which the object is being pushed is $230\ \text{N}$

s = The displacement of the object = 8.1 m

Work done is given by

$W=Fs\\\Rightarrow W=230\times 8.1\\\Rightarrow W=1863\ \text{J}$

The work done by the mover is 1863 J.

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A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the

stepan [7]

Answer:

m = 1.99 kg = 2 kg

Explanation:

The moment of inertia of a bicycle rim about it's center is given by the following formula:

$I = mr^{2}\\$

where,

I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²

r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2

r = 0.65 m/2 = 0.325 m

m = Mass of the Bicycle Rim = ?

Therefore,

$0.21\ kg.m^{2} = m(0.325\ m)^{2}\\m = \frac{0.21\ kg.m^{2}}{(0.325\ m)^{2}}\\$

<u>m = 1.99 kg = 2 kg</u>

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2 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

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3 years ago

Feliz [49]

40 seconds I’m pretty sure sorry if I’m wrong

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2 years ago

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17. (16 pts) You set up a two-slit experiment with a laser that produces light with a

Natali [406]

when the two waves interfere with eachother to make a dark spot the periodic difference of the two waves is π . the wave length for 2π is 600nm

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3 years ago

an open tank has the shape of a right circular cone (see figure). the tank is 8 feet across the top and 6 feet high. how much wo

Liono4ka [1.6K]

The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.

Given that, the tank is 8 feet across the top and 6 feet high

By the property of similar triangles, 4/6 = r/y

6r = 4y

r = 4/6*y = 2/3*y

Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²

The weight of each disc is m = ρw* A

m = 62.4* 4π/9*y² = 87.08*y²

The distance pumped is 6-y.

The work done in pumping the tank by pumping the water over the top edge is

W = 87.08 ∫(6-y)y² dy

W = 87.08 ∫(6y³ - y²) dy

W = 87.08 [6y⁴/4 - y³/3]

W = 87.08 [3y⁴/2- y³/3]

The limits are from 0 to 6.

W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs

The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.

To know more about work done:

brainly.com/question/16650139

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1 year ago