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olga_2 [115]
3 years ago
14

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air? The softball expe

riences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through. It also experiences a downward pull because of .
Physics
2 answers:
sasho [114]3 years ago
7 0
<h2>The different forces acting on the ball while its in air</h2>

Amy throws a softball through the air. Applied, drag and gravitational forces are acting on the ball while it’s in the air. The softball experiences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through.

It also experiences a downward pull because earth has the property to attract everything which is on the earth towards it. The ball is moving in the air but earth applies force on the ball to get back on the ground. Hence, in this way, gravitational force applies.

There is also a drag force which results due to friction that is present in the air. It resist to move ball in the air and there will also be applied force which is given by a person who throws by applying force.    

ivolga24 [154]3 years ago
7 0

Answer:

The downward pull is the gravitational pull of the earth.

Explanation:

The ball experiences three types of forces.

1. Applied force by Amy.

2. Dragging force due to the air friction which resists the motion of the ball.

3. The gravitational pull which acts in downward direction applied by the earth.

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3 years ago
ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav
Montano1993 [528]

Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

           

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0
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