How much time is needed for a 15,000 W engine to do 1,800,000 J of work? (Power: P = W/t)

A helicopter travels west at 80 mph. It is moving above a car traveling on a highway at 80 mph. Given this information, you can

gavmur [86]

Answer:

d. at the same velocity

Explanation:

I will assume the car is also travelling westward because it was stated that the helicopter was moving above the car. In that case, it depends where the observer is. If the observer is in the car, the helicopter would look like it is standing still ( because both objects have the same velocity). If the observer is on the side of the road, both objects would be travelling at the same velocity. Also recall that, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes but speed is the rate at which object covers distance and it is not direction wise. Hence velocity is the best option.

5 0

3 years ago

A centrifugal pump discharges 300 gpm against a head of 55 ft when the speed is 1500 rpm. The diameter of the impeller is 15.5 i

Ksivusya [100]

Answer: Question 1: Efficiency is 0.6944

Question 2: speed of similar pump is 2067rpm

Explanation:

Question 1:

Flow rate of pump 1 (Q1) = 300gpm

Flow rate of pump 2 (Q2) = 400gpm

Head of pump (H)= 55ft

Speed of pump1 (v1)= 1500rpm

Speed of pump2(v2) = ?

Diameter of impeller in pump 1= 15.5in = 0.3937m

Diameter of impeller in pump 2= 15in = 0.381

B.H.P= 6.0

Assuming cold water, S.G = 1.0

eff= (H x Q x S.G)/ 3960 x B.H.P

= (55x 300x 1)/3960x 6

= 0.6944

Question 2:

Q = A x V. (1)

A1 x v1 = A2 x V2. (2)

Since A1 = A2 = A ( since they are geometrically similar

A = Q1/V1 = Q2/V2. (3)

V1(m/s) = r x 2π x N(rpm)/60

= (0.3937x 2 x π x 1500)/2x 60

= 30.925m/s

Using equation (3)

V2 = (400 x 30.925)/300

= 41.2335m/s

To rpm:

N(rpm) = (60 x V(m/s))/2 x π x r

= (60 x 41.2335)/ 2× π × 0.1905

= 2067rpm.

6 0

3 years ago

A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at

fenix001 [56]

Answer:

a) $I_{RMS} = 4.79 A$

b) $PF = 0.908$

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, $cos \theta_{1} = 0.92$

$\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074$

$Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W$

For load 2

Active power, P₂ = 250 W

Power factor, $cos \theta_{2} = 0.8$

$\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87$

$Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W$

For load 3

Active power, P₃ = 250 W

Power factor, $cos \theta_{3} = 1$

$\theta_{3} = cos^{-1} 1\\\theta_{3} =0$

$Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W$

Calculate the total reactive power, $Q_{net} = 42.6 + 187.5 + 0$

$Q_{net} = 230.1 W$

Calculate the total active power, $P_{net} = 100 + 250 + 150 = 500 W$

$S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1$

$P_{net} = IVcos \theta_{net}$

$\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712$

V = 115 $V_{rms}$

$500 = I_{RMS} * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A$

b) Power factor of the composite load is $cos\theta_{net}$

$\theta_{net} = 24.712\\PF = cos 24.712\\PF = 0.908$

4 0

3 years ago

How is using a digital signal like turning on and off a regular light<br> switch?

Orlov [11]

Answer:

they both use 1 and 0

Explanation:

Flipping a switch shows I for light on and 0 for light off

8 0

3 years ago

Define electromagnetic radiation

Nat2105 [25]

Answer:

a kind of radiation including visible light, radio waves, gamma rays, and X-rays, in which electric and magnetic fields vary simultaneously

8 0

3 years ago