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JulsSmile [24]
3 years ago
13

An automatic dryer spins wet clothes at an angular speed of 5.2 rad/s. Starting from rest, the dryer reaches its operating speed

with an average angular acceleration of 3.2 rad/s². How long does it take the dryer to come up to speed?
Physics
1 answer:
Dvinal [7]3 years ago
6 0
<h2>Time taken by dryer to come up to speed is 1.625 seconds.</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u =  0 rad/s

     Final velocity, v = 5.2 rad/s    

     Time, t = ?

     Acceleration, a = 3.2 rad/s²

     Substituting

                      v = u + at  

                      5.2 = 0 + 3.2 x t

                      t = 1.625 s

Time taken by dryer to come up to speed is 1.625 seconds.

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Explain in your own words how volume is measured.
pav-90 [236]

Answer:

volume measured by pid^3 over 6 i think

Explanation:

7 0
3 years ago
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The valve in the exit pipe is closed . The density of water is 1000kg / m and the gravitational force on free fall of water is 1
Stels [109]

Answer:

300000

Explanation:

p=30x10x1000=30000pascal

7 0
3 years ago
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When hydrogen is fused into helium, energy is released from Choose one: A. the increase in pressure. B. the decrease in the grav
Tatiana [17]

The nuclear fusion of hydrogen atoms releases a huge amount of energy. So the correct choice is C. Conversion of mass to energy.

What is nuclear fusion?

When two small nuclei join to form a new nucleus, then this process is termed nuclear fusion. A huge amount of energy is released when there occurs nuclear fusion between the two nuclei. And a new element is formed.

It has been observed that the amount of energy released in nuclear fusion is equal to the mass difference between the mass of the formed nucleus and the total mass of old nuclei. Hence in the nuclear fusion of hydrogen nuclei to form a helium nucleus, the energy is released due to the conversion of mass into energy.

The pressure is increased to make the hydrogen atoms fuse but this change in pressure does not contribute to the energy released in the fusion of hydrogen.

The magnitude of the gravitational field is too low and it does not contribute to the energy released in the fusion of hydrogen.

The gravitational collapse does not occur between the two hydrogen atoms. This phenomenon occurs in celestial bodies so this also does not contribute to the energy released in the fusion of hydrogen.

Learn more about nuclear fusion here:

brainly.com/question/10165218

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8 0
2 years ago
Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has
a_sh-v [17]

Answer:

  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

         Em₀ = U = m g h

final point. Lower, just before the crash

         Em_f = K = ½ m v_{b}^2

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = \sqrt{2gh}

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ \sqrt{2gh}

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

          Em₀ = K = ½ 3m v²

Final point. Higher

          Em_f = U = (3m) g h'

          Em₀ = Em_f

          ½ 3m v² = 3m g h’

           

we substitute

         h’=  \frac{v^2}{2g}

         h’ =  \frac{1}{3^2} \  \frac{ 2gh}{2g}

         h’ = 1/9 h

6 0
3 years ago
Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en
allsm [11]

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

Δx = 0.5mv²/F

Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

Δx = 0.5(4.0)(2.0)²/F = 8/F

Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

4 0
3 years ago
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