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Vladimir79 [104]

3 years ago

11

If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of

the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?

1 answer:

erik [133]3 years ago

6 0

Answer:

The sun will be located near the Gemini constellation at sunset

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A man runs 1200m on a straight line in 4 min . find his velocity.

luda_lava [24]

Answer:

5m/sec^2

Explanation:

Distance=1200m

Time=4 min

1=60sec

4=4 x 60

=240sec

Velocity=Distance/Time

Velocity=1200/240

Velocity=5m/sec^2

Mark me as brainliest

7 0

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4. A woman releases one egg every month for 37 years. Calculate how many

BigorU [14]

so, 444 eggs would have been released in 37yrs

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2 years ago

for a certain chemical reaction, the reactants contain 52 kj of potential energy and the products contain 32 kj how much energy

Luden [163]

B, is the answer. (20 kj is released)

7 0

2 years ago

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worty [1.4K]

Explanation:

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2 years ago

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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.

5 0

3 years ago