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mel-nik [20]
3 years ago
8

A particular circular track on level ground has a circumference of 400 m. a walker, traveling counterclockwise and starting at t

he southernmost point of the track, completes one lap in 800 s, ending his circular trip by arriving again at the southernmost point. find the average speed and the average velocity for the walker's trip around the track.
Physics
1 answer:
Sidana [21]3 years ago
3 0

Answer:

Average speed: 0.5 m/s. Average velocity: 0

Explanation:

Average speed is given by:

v=\frac{d}{t}

where

d is the total distance covered (the length, of one lap of the track, so d = 400 m)

t is the time taken to cover that distance (so, t = 800 s)

Substituting,

v=\frac{400}{800}=0.5 m/s

Instead, average velocity is defined as

v=\frac{d}{t}

where this time,

d is the displacement, which is the vector connecting the starting point to the final point of the motion

t is still the time taken (800 s)

However, in this case the walker starts and finishes his trip at the same point: therefore, the displacement is zero (d=0), and this means that the average velocity is zero as well.

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Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

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mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

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vf = 26.4 m/s

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imagine riding a single-speed bicycle. why do you have to push harders on the pedals to start the bicycle moving than to keep it
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One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
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Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

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v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

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Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

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fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

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A \rightarrow \frac{8A}{9}

As we know,

Energy will be:

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On putting the estimated values, we get

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⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

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          =0.246

3 0
3 years ago
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