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Aleksandr [31]
3 years ago
11

A construction worker pushes a cement block with a 750 N force for 10 minutes without the cement block moving. How much work did

the construction worker do during the 10 minutes
Physics
2 answers:
lutik1710 [3]3 years ago
8 0
The work done by a system on a different body is equal to the product of the force exerted and the distance that the body has move in parallel to the force exerted. In this item, we have to determine first the distance and multiply it with the given force equal to 750N. 
luda_lava [24]3 years ago
8 0

Answer:

Zero

Explanation:

Work done is defined as the product of force applied and the displacement traveled by the object.

Work = force applied in the direction of displacement x displacement

The SI unit of work is Joule. Work may be positive, negative or zero.

here, force applied = 750 N, displacement = 0 m

So, work done = 750 x 0 = 0 Joule

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alina1380 [7]

Answer:

what is the action and reaction ?

Explanation:

   

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A hockey puck slides on the ice and eventually stops. How would Newton interpret this behavior?
azamat

That would be a the first law of newton's laws of motion because it stops from an external force

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A book is at rest on a table. Identify the correct free-body diagram for this situation.
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Show a picture .. of what your doing
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3 years ago
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Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0
3 years ago
Read 2 more answers
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
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