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Aleksandr [31]
3 years ago
15

A sled of mass 15 kg slides across the ground with a friction force of 80.85 N. What is k between the sled and the ground?

Physics
2 answers:
Dvinal [7]3 years ago
8 0
The expression for the frictional force between the sled and the ground is:
F=k m g
where k is the coefficient of friction, m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration.

The friction force in our problem is F=80.85 N. The mass of the object is m=15 kg. Re-arranging the formula, we can find the value of k:
k= \frac{F}{mg}= \frac{80.85 N}{(15 kg)(9.81 m/s^2)}=0.55
ki77a [65]3 years ago
3 0

0.55 is correct - APEX

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Explanation:

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This is given by:

fo = V +-Vo/ V +-Vo × source

Where fo= observed frequency

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Vo= vo it of the observer

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Given:

Observed frequency of the approaching train fo1= 452Hz

The observed frequency of train= fo2= 442Hz

Velocity of sound= 334m/s

Velocity of source=?

Train approaching the observer is given by:

fo1= V/(V - Vs)× source ...eq1

Train passes the student is given by:

fo2= V/(V - Vs)×source ...eq2

Divide eq1 by eq2

452/442 = (343+Vs)/(343 - Vs)

1.02 =(343+Vs)/(343 -Vs)

Cross multiply

1.02(343- Vs) = 343 + Vs

350.76 - 1.02Vs = 343 + Vs

Collecting like terms

350.76 -343= 1.02Vs+ Vs

7.76 = 2.02Vs

Vs= 7.76/2.02

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4 0
3 years ago
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ω = 0.571 rad/s

Explanation:

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<span>soo, t = 1 sec</span>
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ipn [44]

Answer:

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Download pdf
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