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d1i1m1o1n [39]
3 years ago
5

A race car, traveling at constant speed, makes one lap around a circular track of radius 200 m. When the car has traveled halfwa

y around the track, what is the magnitude of the displacement from the starting point?
Physics
1 answer:
mario62 [17]3 years ago
8 0

The magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

Answer:

Explanation:

Since the race track is a circular track, the distance for one lap will be equal to the circumference of the circular track. And the circumference will be equal to the circumference of the circle.

Since the radius of the track is given as 200 m, then the circumference of the circular track will be

Circumference = 2πr = 2 × 3.14 × 200

So the circumference of the circular track = 1256 m.

So the starting point or position of the track is considered as zero and if the car has traveled half way means, the car has covered half of the circumference of the track.

As the circumference = 1256 m, then half of the circumference of the circle = 1256/2 = 256 m.

So the displacement is the measure of difference between the final position and initial position. As here the initial position is zero and the final position is the halfway around the track which is equal to 256 m.

Then Displacement = Final-Initial = 256-0= 256 m.

So the magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

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Answer:879.29 N-m

Explanation:

Given

mass of first child m_1=44 kg

distance of first child from tree is r_1=1 m

tree is inclined at an angle of \theta =27^{\circ}

mass of second child m_1=27 kg

distance of second child from tree is r_2=2.1 m

Weight of first child=m_1g=431.2 kg

Weight of second child=m_2g=264.6 kg

Torque of first child weight=m_1g\cos \theta \cdot r_1

T_1=44\times 9.8\times \cos 27\times 1=384.202 N-m

Torque of second child weight=m_2g\cos \theta \cdot r_2

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Net torque T_{net}=T_1+T_2=384.202+495.096=879.29 N-m

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Metal A will transfer heat to the water since it's temperature is higher than that of water.

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Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

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           P = I A

the area is of a sphere

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we substitute the ratio of intensities

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we calculate

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