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diamong [38]
3 years ago
14

A person travels by car from one city to another with different constant speeds between pair of cities. She drives for 36 min at

80.5 km/h, 10 min at 92.7 km/h, and 54.3 min at 35.6 km/h, and spends 20.9 min eating lunch and buying gas. Find the distance between the initial and final cities along this route. Answer in units of km
Physics
1 answer:
Softa [21]3 years ago
3 0

 Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:

let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance 

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During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de
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Answer: Smaller than ; larger than

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

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(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

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Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

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Put the value into the formula

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v_{2}=\sqrt{2\times55.915}

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The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
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3 years ago
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