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kompoz [17]
2 years ago
14

A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction

between the box and the surface.
Physics
1 answer:
Anna11 [10]2 years ago
8 0
The weight of the box is 50x9.8 = 490 N. The force of friction is 100N. F= μΝ so coefficient = 100/490 = 0.20
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If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?
vodomira [7]

<u>Answer:</u>

 Option A is the correct answer.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

 First walking 1.2 km north,  displacement = 1.2 j km

 Secondly 1.6 km east, displacement = 1.6 i km

 Total displacement = (1.6 i + 1.2 j) km

 Magnitude = \sqrt{1.2^2+1.6^2} = 2 km

 Angle of resultant with positive X - axis = tan^{-1}(1.2/1.6)=36.87^0 = 36.87⁰ east of north.

 

5 0
2 years ago
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FrozenT [24]

the answer is (A) the movement of the magnet relative to the coil

4 0
3 years ago
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Pavel [41]
The answer to this is Animal habitation 
4 0
3 years ago
A particle moving on a circle has a velocity of 5 m/s and a normal acceleration of 10 m/s^2. What is the radius of the circle?
dybincka [34]

Answer:

Radius of the circle will be 2.5 m

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r=\frac{25}{10}=2.5m

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3 0
3 years ago
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2 years ago
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