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yawa3891 [41]
3 years ago
11

If we could easily create machines or systems with no friction, what could we use them for?

Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

We could make a frictionless generator.

Explanation:

This would change life by making the cost for energy next to zero

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The diagram shows a swinging pendulum.
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Answer:

A. The mechanical energy transforms to thermal energy as the pendulum slows and eventually stops moving.

Explanation:

took the quiz on edge

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What results from splitting an atomic nucleus
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A uranium-235 atom<span> absorbs a neutron and fissions into two new </span>atoms<span> (fission fragments), releasing three new neutrons and some binding energy. ... Several heavy elements, such as uranium, thorium, and plutonium, undergo both spontaneous fission, a form of radioactive decay and induced fission, a form of </span>nuclear<span> reaction.</span>
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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
Which object will be considered to be in free fall
Pani-rosa [81]
It depends on what they are 
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Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!
DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0
3 years ago
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