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insens350 [35]
4 years ago
5

find the coefficient of kinetic friction for a 10 kg box being dragged steadily across the surface with a force of 2.0 Newtons​

Physics
1 answer:
Mazyrski [523]4 years ago
5 0

Answer:

0.02

Explanation:

Force, F=\mu_k N where N is normal reaction and coefficient of kinetic friction is \mu_k. Also, N is equivalent to mg ie N=mg where m is mass of an object and g is acceleration due to gravity. Making \muthe subject of the formula then

\mu_k=\frac {F}{mg}

Substituting F with 2 N, m with 10 kg and g with 9.81 then

\mu_k=\frac {2}{10\times 9.81}=0.0203873598369\approx 0.02

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3.4\cdot 10^{-19} J

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In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

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A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

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Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

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