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jeka94
3 years ago
11

How long will your trip take if you travel 4000 m at an average speed of 8m/s

Physics
2 answers:
SpyIntel [72]3 years ago
8 0
  • Answer:

<em>500 sec</em>

<em>8 min 20 sec</em>

  • Explanation:

<em>Hi there !</em>

<em />

<em>8 m ................ 1 s </em>

<em>4000 m ........ x s</em>

<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>

<em />

<em>Good luck ! </em>

Andrews [41]3 years ago
8 0

Answer:8.3mins

Explanation:

S=Vt

where V=average speed

S=4000m V=8m/s

4000=8×t

t=4000/8

t=500s

t=500/60

t=8.3min

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
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Yes. Towards the center. 8210 N.

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Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

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Answer:

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- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

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What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

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                                  0 = m_1*V_1 - m_2*V_2

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- Apply Conservation of Energy on both skaters as follows:

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-Simplify:                      0.5*V_2^2 = u_k*g*s_2

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- simplify:                     (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

                                  m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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