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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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nataly862011 [7]

Answer:

0.031 W

Explanation:

The power used is equal to the rate of work done:

P=\frac{W}{t}

where

P is the power

W is the work done

t is the time taken to do the work W

In this problem, we have:

W = 900 J is the work done by the motor

t = 8 h is the time taken

We have to convert the time into SI units; keeping in mind that

1 hour = 3600 s

We have

t=8\cdot 3600 =28,800 s

And therefore, the power used is

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3 years ago
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Plz answer fast the question
alexgriva [62]

Answer:

Angle of incidence = 20°

Angle of reflection = 20°

Explanation:

Applying,

The first Law of Refraction: The incident ray, the reflected ray and the normal at the point of incidence all lies in the plane.

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Angle of incidence = 90-70

Angle of incidence = 20°

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A stone is thrown vertically upwards with a speed of 30.0 m/s.
matrenka [14]

a)

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a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

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Using the equation

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inserting the values

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b)

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v₀ = initial speed of stone at the time of launch = 30.0 m/s

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