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Karolina [17]
3 years ago
9

A garbage truck has a mass of 2100 kg. It starts from rest and travels 75 meters in 15 seconds. The truck uniformly accelerates

the entire time. What net force is applied to the truck?
Physics
1 answer:
Ivanshal [37]3 years ago
3 0

Answer: F=(2100)(0.33)=693N

Explanation:

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Answer:

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Both cars start from the same point. Which describes the motion shown?
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After a collision, the y-component of momentum of a car is 4.50 meters/second, and the x-component of its momentum is 9.80 meter
Natalka [10]
The components are at a right angle so the resultant is R^2 = 4.50^2 + 9.80^2 R = 10.78 or 10.8

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~Lena~
5 0
3 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

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3 years ago
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