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3 years ago

15

A 600 kg elevator starts from rest. It moves upward for 4.50 s with constant acceleration until it reaches its cruising speed, 1

.75 m/s. (a) What is the average power of the elevator motor during this period

1 answer:

viktelen [127]3 years ago

3 0

Answer:

408.33watts

Explanation:

Power is expressed according to the formula

Power = work done/time

Power = force × distance/time

Power = Force× velocity

Find the force

F = mv/t

F = 600×1.75/4.5

F = 1050/4.5

F=233.33N

Get the power

Power = 233.33×1.75

Power = 408.33watts

Hence the average power of the elevator motor during this period is 408.33m/s

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Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant

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Complete Question

The complete question(reference (chegg)) is shown on the first uploaded image

Answer:

The magnitude of the resultant force is $F = 199.64 \ N$

The direction of the resultant force is $\theta = 4.1075^o$ from the horizontal plane

Explanation:

Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be $Fcos(\theta )$ while if the force is moving away from the angle then the resolved force is $Fsin (\theta )$

Now from the diagram let resolve the forces to their horizontal component

So

$\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)$

$\sum F_x = 199.128 \ N$

Now resolving these force into their vertical component can be mathematically evaluated as

$\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)$

$\sum F_{y} = 14.30$

Now the resultant force is mathematically evaluated as

$F = \sqrt{F_x^2 + F_y^2}$

substituting values

$F = \sqrt{199.128^2 + 14.3^2}$

$F = 199.64 \ N$

The direction of the resultant force is evaluated as

$\theta = tan^{-1}[\frac{F_y}{F_x} ]$

substituting values

$\theta = tan^{-1}[\frac{ 14.3}{199.128} ]$

$\theta = 4.1075^o$ from the horizontal plane

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4 years ago

The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver

luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time = 2 hours

Speed = 200km

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Time = 4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0$

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100$

Lets find the speed at stage 1

Speed = $\frac{Distance }{Time}$

Speed = $\frac{80}{1}$

Speed 1= 80 km/hr

The speed at stage 2

Speed = $\frac{Distance }{Time}$

Speed = $\frac{200}{2}$

Speed 2 = 100 km/hr

The speed at stage 3

Speed = $\frac{Distance }{Time}$

Speed = $\frac{x}{4}$

Speed 3 = $\frac{x}{4}$

we kow that average is ,

$\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100$

$\frac{ 80 + 100+ \frac{x}{4} }{3} = 100$

$\frac{ 180 + \frac{x}{4} }{3} = 100$

$\frac{ \frac{720 +x}{4} }{3} = 100$

$\frac{720 +x}{4} \times \frac{1}{3} = 100$

$\frac{720 +x}{12} = 100$

$720 +x = 100 \times 12$

$720 +x = 1200$

$x = 1200- 720$

x = 480

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