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kolbaska11 [484]
3 years ago
13

A student walks 4 blocks east, 1 blocks north, 2 block west and then 1 blocks south in an hour. Her average velocity is?​

Physics
1 answer:
arsen [322]3 years ago
4 0

Answer:

2 m/s

Explanation:

Given that a student walks 4 blocks east, 1 blocks north, 2 block west and then 1 blocks south in an hour.

The total time = 1 hour

The vertical displacement = 1 - 1

Vertical displacement = 0

Horizontal displacement = 4 - 2

Horizontal displacement = 2

Total displacement = sqrt (2^2 - 0^2)

Total Displacement = 2

Average velocity = displacement/time

average velocity = 2 / 1

Average velocity = 2 m/s

Therefore, her average velocity is 2 metres per second.

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A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed
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Answer:v=41.23 m/s

Explanation:

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T_{net}=m_1gr_1-m_2gr_2

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thus

T_{net}\times \theta =\frac{mv^2}{2}

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v= velocity\ at\ launch

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Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis
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Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

v^{2}=u^2+2as

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equalsT=2\times 30.51\approx61seconds

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 39.66m/s

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4 years ago
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