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olganol [36]
3 years ago
14

A 0.358 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius

of 1.66 m. The particle is given an initial speed of 8.19 m/s. After one revolution, its speed has dropped to 6.29 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.
Physics
1 answer:
oee [108]3 years ago
4 0

Answer:

The energy loss due to friction in one revolution is 4.92 J.

Explanation:

Given that,

Mass of particle = 0.358 kg

Radius = 1.66 m

Initial speed =8.19 m/s

Final speed = 6.29 m/s

We need to calculate the kinetic energy

Using formula of kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{f}^2-v_{i}^2)

Put the value into the formula

\Delta K.E=\dfrac{1}{2}\times0.358\times(8.19^2-6.29^2)

\Delta K.E=4.92\ J

Hence, The energy loss due to friction in one revolution is 4.92 J.

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ahrayia [7]
<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:  

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D={C}_{d}\frac{\rho V^{2} }{2}A  (1)

Where:  

C_ {d} is the drag coefficient  

\rho is the density  of the fluid (air for example)

V is the velocity  

A is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its <u>weight </u>due to the gravity force W:  

W=m.g

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g is the acceleration due gravity  

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g  (4)

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As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0
3 years ago
A helium-filled balloon occupies a volume of 15 cubic meters at sea level. the balloon is released and raises to a point in the
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According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.

Therefore,

P_{1} V_{1} =P_{2} V_{2}

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V_{2} = \frac{P_{1}\times V_{1}  }{ P_{2} } = \frac{1\times 15}{0.75} =20 m^3

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A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a
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Answer with Explanation:

We are given that

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1 cm^2=10^{-4} m^2

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B=4t-2t^2

We know that magnetic flux

\phi=BA

Emf ,E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid

Current, I=\frac{E}{R}

Current, I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid

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Substitute t=1 s

Then, I=1.6\mid (1-1)\mid=0

Substitute

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Current, I=1.6\mid(1-2)\mid=1.6 A

8 0
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Answer:

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Explanation:

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