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sergey [27]
4 years ago
5

A foul ball is hit straight up into the air with a speed of about 25 m/s. ? how high does it go? (b) how long is it in the air?

Physics
1 answer:
nikdorinn [45]4 years ago
5 0

To solve this problem, we make use of the equations of motion:

vf^2 = v0^2 – 2 g h

vf = v0 – g t

h = v0 t – 0.5 g t^2

 

A. Finding for max height:

Max height occurs when vf = 0, therefore:

0^2 = 25^2 – 2 (9.8) h

h = 31.89 m

 

B. Finding for t:

t = t(up) + t(down)

 

vf = v0 – g t(up)

0 = 25 – 9.8 t(up)

t(up) = 2.55 s

 

h = v0 t(down) – 0.5 g t(down)^2

31.89 = 0.5 (9.8) t(down)^2

t(down)^2 = 6.51

t(down) = 2.55 s

 

so,

<span>t = 5.1 s</span>

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shutvik [7]

Answer:

Since this is a linear equation

y = m x + b     or

U = m F + b     is a linear equation

when ΔF = (212 - 32) = 180

and ΔU = (60 - (-15)) = 75

m = 75 / 180 = 2.4 if converting F to U and a = .417

U = .417 F + b

If F = 32 then U = -15 and

-15 = .417 * 32 + b

b = -15 - 13.3 = -28.3 and our equation becomes

U = .417 F - 28.3

Check: let F = 212

U = .417 * 212 - 28.3 = 60          as it should

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2 years ago
A 90kg cannon fires 100kg shell with muzzle velocity of 75m/s. Calculate the recoil velocity of the cannon relative to the groun
Gala2k [10]

Answer: 83.8 m/s

Explanation:

momentum of shell = momentum of cannon

M(s) × V(s) = M(c) × V(c)

M(s) = mass of shell, V(s) = velocity of shell

M(c) = mass of cannon, V(c) = velocity of cannon

100kg × 75m/s = 90kg × V(c)

7500 = 90 × V(c)

7500 ÷ 90 = V(c)

83.3 = V(c)

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3 years ago
Choose the 200 kg refrigerator. set the applied force to 400 n (to the right). be sure friction is turned off. what is the net f
Firlakuza [10]

The net force acting on the refrigerator is 400 N to the right.

<h3></h3><h3>FURTHER EXPLANATION</h3>

The net force or resultant force is the sum of all the forces acting on a body or an object in x and y axes.

  • Forces along the y-axis The forces that usually act on an object vertically (in the y-axis) are: gravitational force which is a downward force and the normal force which is an upward (perpendicular) force exerted by a surface on an object resting above it that keeps the object from falling.
  • Forces along the x-axis These include the force or forces applied to cause a left or right motion of an object along the horizontal plane (called the Applied Force) and the force that opposes the motion or friction.

In this problem the forces acting on the x and y - axes can be determined:

Along the x-axis:

  • gravitational force = -1960 N
  • normal force = +1960 N
  • Net force = -1960 N + 1960 N = 0

The gravitational force is the weight of the object obtained by multiplying the mass of the object (in kg) with the acceleration due to gravity, 9.8 m/s^2. It is given a negative (-) sign to indicate that it is a downward force.

Since the object is not falling through the surface, it can be assumed that the gravitational force and normal force are balanced. Hence, the size of the normal force is the same as the gravitational force but with the opposite direction indicated by the + sign for an upward force.

The forces along the x-axis are balanced  (i.e. net force is zero) so the object neither moves upward or downward.

Along the y-axis

  • applied force = +400 N
  • friction = 0
  • Net force = +400 N + 0 = +400 N

The applied force is +400 N. It is given a + sign to indicate that its direction is to the right.

The friction, as mentioned in the problem, is set to zero or "turned off".

The net force along the y-axis is +400. The forces are unbalanced so the object will move to the right as force is applied to it.

<h3>LEARN MORE</h3>
  • Balanced Forces brainly.com/question/760473
  • Unbalanced Forces brainly.com/question/4289010
  • Free body Diagrams brainly.com/question/12345810

Keywords: net force, resultant force

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Answer:

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Monica [59]

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hope this helps:)

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