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3 years ago

A sled starts from rest,

Physics

1 answer:

Marrrta [24]3 years ago

5 0

Answer:

25.6 m/s

Explanation:

Draw a free body diagram of the sled. There are two forces acting on the sled:

Normal force pushing perpendicular to the hill

Weight force pulling straight down

Take sum of the forces parallel to the hill:

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 38.0°)

a = 6.03 m/s²

Given:

v₀ = 0 m/s

a = 6.03 m/s²

t = 4.24 s

Find: v

v = at + v₀

v = (6.03 m/s²) (4.24 s) + (0 m/s)

v = 25.6 m/s

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gogolik [260]

Answer:

A. Zero

Explanation:

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Since the normal of the area is parallel to the magnetic field, θ = 0

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3 years ago

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Alecsey [184]

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3 years ago

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svetlana [45]

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4 years ago

Read 2 more answers

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: 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.

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

A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in

Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

4 0

3 years ago