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3 years ago

5

A flywheel flows from 250rpm to 150rpm in 4.2 seconds. How many revolutions occur during this time

1 answer:

Assoli18 [71]3 years ago

6 0

Answer:

7revolutions

Explanation:

Given parameters:

Initial revolution = 250rpm

Final revolution = 150rpm

Time = 4.2s

Unknown:

Number of revolutions that occur at this time = ?

Solution:

To solve this problem;

let us find the change in revolution = 250rpm - 150rpm = 100rpm

Convert the time to seconds;

60s makes 1 minute

4.2s will make $\frac{4.2}{60}$ = 0.07min

So;

The number of revolutions at this time = 100rpm x 0.07min

= 7revolutions

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Vikentia [17]

When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).
In order to climb, you need to work against work done by gravity on the pack.
Hence work done by you = work done by gravity on pack
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4 years ago

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

A

$x = 0.198456 \ m$

B

$h = 1.3061 \ m$

C

$v = 5.06 \ m/s$

D

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

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A 65-kg person walks from the ground to the roof of a 100 m tall building. How much gravitational potential energy does she have

ivolga24 [154]

I,think potential energy is mgh so 65*100*9,81

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3 years ago

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Arte-miy333 [17]

Answer:

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Explanation:

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3 years ago

A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which

Leviafan [203]

Answer:

The value is $a = -3.7 \ m/s^2$

Explanation:

From the question we are told that

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Generally the average acceleration is mathematically represented as

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=> $a = \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }$

=> $a = -3.7 \ m/s^2$

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