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never [62]
3 years ago
11

A boy and his dog are running. The dog has a mass of 20kg and the boy has a mass of 45kg.

Physics
1 answer:
Ad libitum [116K]3 years ago
4 0

Answer:

Explanation:

mass of boy =m1=45 kg

mass of dog =m2=20kg

the speed of boy=3 m/s

the speed of dog =3 m/s

K.E of boy  

K.E=1/2mv²

K.E=1/2(45)(3)²

K.E=202.5 J

K.E of dog

K.E=1/2mv²

K.E=1/2(20)(3)²

K.E=90 J

now the new speed of dog is 6 m/s

therefore new K.E of dog is

K.E=1/2(20)(6)²

K.E=360 j

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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.

5 0
3 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
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Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

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4 0
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Which planetery body has the greast gravitational pull?
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Answer should be the earth
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Answer:

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Given;

initial charge on the plastic rod, q₁ = 15nC

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let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

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The work done by the three students is 3,000 J.

The energy transferred in the process is 3,000 J.

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W = Fd

The work done by the three students is calculated as follows;

W = 300 x 10

W = 3,000 J

<h3>What is energy transfer?</h3>
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The energy transferred in the process is determined by work energy theorem.

E = W

E = 3,000 J

Learn more about work-energy theorem here: brainly.com/question/22236101

3 0
2 years ago
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