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ANTONII [103]
3 years ago
9

What force is needed to move a barrel 45-m if 3600 J of work are accomplished?​

Physics
1 answer:
lesantik [10]3 years ago
8 0

Answer:

<h3>The answer is 80 N</h3>

Explanation:

The force acting on the object can be found by using the formula

f =  \frac{w}{d}  \\

where

d is the distance

w is the work done

We have

f =  \frac{3600}{45}  \\

We have the final answer as

<h3>80 N</h3>

Hope this helps you

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Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

Hope you understood

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3 years ago
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Which of the following could be used to create an open circuit?
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A switch
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Binding energy is the energy needed to
densk [106]

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Answer C

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2 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that th
iragen [17]

Answer:

 μ = 0.725

Explanation:

This problem refers to Newton's second law.

       F = ma

Let's write the equations on each axis

Y Axis

      N-W = 0

     N = W

    N = mg

X axis

   F-fr = ma

With the body not started moving its acceleration is zero

  F-fr = 0

  F = fr

The friction force equation is

  fr = μ N

  fr = μ m g

Let's replace and calculate

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   μ = 321 /45.2 9.8

   μ = 0.725

8 0
3 years ago
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