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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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Answer:

p= kg m/s

Explanation:

Momentum is a vector quantity that represents the "amount of motion" of an object.

Mathematically, the momentum of an object is given by

p=mv

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m is the mass of the object

v is the velocity

Since momentum is a vector, it also has a direction, which is the same as the velocity.

Therefore, if we have two objects, the total momentum of the two objects will be obtained from the vector sum of the individual momenta of the two objects.

In this problem we have:

p_A=  kg m/s is the momentum of object A

p_B = kg m/s is the momentum of object B

Therefore, the total momentum of objects A and B can be obtained by adding each components of A to the corresponding component of B, so:

p_x = 20 +6 = 26 kg m/s\\p_y = -6 +6 = 0 kg m/s\\p_z = 0 + 0 = 0 kg m/s

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p= kg m/s

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3 years ago
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A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.
mars1129 [50]

Answer:

a) v = 31.67 cm / s , b)   v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

       w = √ K / m

       w = √ 23.2 / 0.37

       w = 7,918 rad / s

a) the expression against the movement is

        x = A cos (wt + Ф)

Speed ​​is

        v = dx / dt = - A w sin (wt + Ф)

 The maximum speed occurs for cos = ± 1

        v = A w

        v = 4.0 7,918

        v = 31.67 cm / s

b) as the object is released from rest

        0 = -A w sin (0+ Фi)

        sin Ф = 0

         Ф = 0

The equation is

        x = 4.0 cos (7,918 t)

        v = -4.0 7,918 sin (7,918 t)

        v = - 31.67 sin (7.918t)

     

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

          7,918 t = cos⁻¹ 1.5 / 4.0

          t = 1,186 / 7,918

          t = 0.1498 s

We look for speed

         v = -31.67 sin (7,918 0.1498)

         v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

         v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

         31.67 / 2 = 31.67 sin ( 7,918 t)

          7.918t = sin⁻¹ 0.5

         t = 0.5236 / 7.918

         t = 0.06613

With this time let's look for displacement

         x = 4.0 cos (7,918 0.06613)

        x = 3.46 cm

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3 years ago
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