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Natali [406]
4 years ago
7

Select three standard units of the metric system.

Physics
1 answer:
dolphi86 [110]4 years ago
8 0
Centimeter, meter, liter
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What unit describes thr magnitude, or size of a force?
Bess [88]
N - adica Newtonul.
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5 0
3 years ago
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As a football player moves in a straight line [displacement (3.00 mm)i^i^ - (6.50 mm)j^j^], an opponent exerts a constant force
Brilliant_brown [7]

Answer:

Work done, W=(0.378i-1.092j)\ J

Explanation:

Displacement,

d=(3i-6.5j)\ mm\\\\d=(0.003i-0.0065j)\ m

Force, F=(126i+168j)\ N

Work done by the opponent do on the football player is given by :

W=F{\cdot} d\\\\W=(126i+168j){\cdot} (0.003i-0.0065j)\ m\\\\W=(0.378i-1.092j)\ J

So, the work done by the opponent do on the football player is  (0.378i-1.092j)\ J.

4 0
3 years ago
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Which statements about acceleration are true? <br> You can chose 2 or more answer
dangina [55]
Answers A, C, and D.
3 0
4 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

5 0
3 years ago
At one instant, an object moving downward in free fall on the Earth has a speed of 50 m/s. Its speed 1 second later is
maksim [4K]

Answer:

velocity 1 second later = 59.8 m/s

Explanation:

Velocity final = Velocity initial + acceleration • time

Velocity final = 50m/s + 9.8m/s/s • 1s

Velocity final = 59.8 m/s

4 0
2 years ago
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