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Westkost [7]
2 years ago
7

What is the force applied to a baseball that has a mass of 142kg and has a acceleration of 30m/s to the power of 2

Physics
1 answer:
aev [14]2 years ago
7 0
  • Mass=m=142kg
  • Acceleration=a=30m/s
  • Force=F

Using Newton's second law

\\ \sf\Rrightarrow F=ma

\\ \sf\Rrightarrow F=142(30)

\\ \sf\Rrightarrow F=4260N

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If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
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Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J

Therefore, the work done by the lawnmower is 236.14 J.

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You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work
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Answer:

1)   F = 71.6 N , 2)    F = 120.2 N , 3)  P_m=  68.600 Pa, 4)  V = 2.4210-5 m³

Explanation:

This is a problem of fluid mechanics, to find the force we must use its definition

         P = F / A

         F = P A

The area of ​​the circular pipe is

         A = π r² = π d²/4

The pressure is given by the expression

         P = P_atm + ρ g h

1) the force on the outer side is

       P = P_atm

we substitute in the expression of force

         F = P_atm π d² / 4

         

let's calculate

         F = 1,013 10⁵ π 0.03²/4

         F = 7.16 10¹ N

         F = 71.6 N

2) the force on the inner side

  the pressure

       P = P_atm + ρ g h

       P = 1,015 10⁵ + 1,000 9.8 7

        P = 1,701 10⁵ Pa

        F = 1,701 10⁵ π 0.03² / 4

        F = 1,202 10²

        F = 120.2 N

3) manometric pressure

       Pm = ρ g h

       P m = 1000 9.8  7

       P_m=  68.600 Pa

4) In this part they ask for the volume that comes out in time t= 3 h

   to calculate this volume we can use the flow ratio

         Q = A v

          V t = A v

          V = A v / t

sent is the velocity of the water that comes out, to calculate it we use the Bernoulli equation

we will use index 1 for the lake surface and ionice 2 apa the position of the plug

           P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

As the lake has much more capacity than the pipeline, the velocity of the surface of the lake is peeling, in this case we approximate it steel

           (P₁-P₂) + ρ G (y₁ -y₂) = ρ g v₂²

           1000 9.8 v₂² = ρ g h + 1000 9.8 (7-0)

             9800 v₂² = 1000 9.8 7 + 68600

              v₂ = √ (137200)

               v₂ = 370.4 m / s

             t = 3 h (3600s / h) = 10800 s

     

we substitute in the volume equation

             V = π d²/4   370.4 / 10800

             V = 2.4210-5 m³

4 0
2 years ago
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