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3 years ago

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A fan powered by a 0.5 hp motor and delivers air at a rate of 85 m3 / min. Calculate the speed of the air moving by the fan if t

he air has a density of 1.18 kg / m3. Help me please

1 answer:

Elenna [48]3 years ago

4 0

s the following graph a bipartite graph or not, and why?

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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

Balanced equations account for the conservation of mass.

uranmaximum [27]

Answer:

fff

Explanation:

f

6 0

3 years ago

Read 2 more answers

What is the force if the mass is 75kg and the acceleration is 24.5m/s^2

Snowcat [4.5K]

<u>Given;</u>

mass m = 75 kg

acceleration a = 24.5 ms²

<em>F = ma </em>

F = 75 kg * 24.5 ms²

= 1837.5 kg ms².

4 0

3 years ago

When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106J/kg and the boiling point is 120∘C. At th

Vikki [24]

Answer:

2033219.05 J

Explanation:

V = Volume

P = Pressure = 2 atm

m = Mass of water = 1 kg

$L_v$ = Heat of vaporization = $2.2\times 10^6\ J/kg$

Work done in an isobaric system is given by

$W=-P\Delta V\\\Rightarrow W=-2\times 101325\times (1\times 10^{-3}-0.824)\\\Rightarrow W=166780.95\ J$

Work done is 166780.95 J

Change in internal energy is given by

$\Delta U=Q-W$

Heat is given by

$Q=mL_v\\\Rightarrow Q=1\times 2.2\times 10^6\\\Rightarrow Q=2.2\times 10^6\ J$

$\Delta U=2.2\times 10^6-166780.95\\\Rightarrow \Delta U=2033219.05\ J$

The increase in internal energy of the water is 2033219.05 J

6 0

3 years ago

The tire above has a recommended tire pressure of 35 PSI, however, its current pressure is only 26 PSI. Which of the following c

Dima020 [189]

Answer:

option B

Explanation:

It is given that in the tire the recommended pressure is 35 PSI however the current pressure is only 26 PSI which means that pressure in the tire is less than the recommended so the chances of blowout of the tire gets eliminated hence option A is not correct.

Having pressure less in the tire can lead to the Unstable handling of the vehicle.

so correct answer is option B

7 0

3 years ago