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2 years ago

A woman lifts a 300 newton child a distance of 1.5 meters in 0.75 seconds. What is her power output in lifting the child?

Physics

1 answer:

Nitella [24]2 years ago

7 0

Power = Work done / time

Work done = Force * Distance
= 300 N * 1.5 m = 450 J

Power = 450 / 0.75 = 600 Watts.

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In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.

solniwko [45]

The answer would be 48

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3 years ago

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A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t

const2013 [10]

Answer:

$\theta=12.19^{\circ}$

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

$sin\theta =\dfrac{u}{v}$

Now by putting the values in the above equation we get

$sin\theta =\dfrac{30}{142}$

$sin\theta=0.21$

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Therefore the angle will be 12.19° .

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3 years ago

Do the following calculation and express the final answer in

sammy [17]

Answer:

Explanation:

I got the same thing. So, i don't know but good luck

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1 year ago

A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated to

Soloha48 [4]

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

$f_{beat} = |f_2-f_1|$

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

$f_{beat} = |f_2-f_1|$

$5Hz = |f_2-440Hz|$

$f_2 = 440 \pm 5$

$f_2 = 445Hz or 435Hz$

B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

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3 years ago

Moment of inertia Wheel If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at

Vinil7 [7]

Answer:

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

I = Moment of inertia = 0.3 kgm²

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-0.72\times 2\pi}{12}\\\Rightarrow \alpha=-0.376991118431\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.3\times (-0.376991118431)\\\Rightarrow \tau=-0.113097335529\ Nm$

The magnitude of torque is 0.113097335529 Nm

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3 years ago