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likoan [24]
3 years ago
14

A major-league pitcher can throw a baseball in excess of 44.7 m/s. If a ball is thrown horizontally at this speed, how much will

it drop by the time it reaches the catcher who is 16.1 m away from the point of release?
Physics
1 answer:
Sloan [31]3 years ago
6 0
<h2>The baseball drops by 0.64 meter.</h2>

Explanation:

Consider the horizontal motion of ball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 44.7 m/s

        Acceleration, a = 0 m/s²  

        Displacement, s = 16.1 m      

     Substituting

                      s = ut + 0.5 at²

                      16.1 = 44.7 x t + 0.5 x 0 x t²

                      t = 0.36 s

      Time taken to travel 16.1 m is 0.36  seconds

Now we need to find how much ball travel vertically during this 0.36 seconds.

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 0.36 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 0.36 + 0.5 x 9.81 x 0.36²

                      s = 0.64 m

     The baseball drops by 0.64 meter.

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GaryK [48]

Answer:

39,200 Joules

3920 watts

Explanation:

5 0
4 years ago
What is solid, geometric shape formed by a minerals?
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crystal hope this helps
6 0
4 years ago
Help with 2 Physics questions, WILL CHOOSE BRAINLIEST
Tju [1.3M]

1) D

2) D.) Greater than \theta_c

Explanation:

1)

The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.

This phenomenon occurs only if two conditions are satisfied:

  • The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
  • The angle of incidence is greater than a certain angle called critical angle

In picture 1, we have 4 different diagrams. In the diagrams:

  • The red arrow represents the incident ray
  • The green arrow represents the refracted ray
  • The blue arrow represents the reflected ray

Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)

2)

As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be

D.) Greater than \theta_c

3 0
3 years ago
We investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on solid ground with a
AlexFokin [52]

Answer:

Time  is 14.8 s and cannot landing

Explanation:

This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing  are the same

Calculate the time to stop, where it has zero speed

       Vf² = Vo² + a t

       t = - Vo² / a

       t = - 110²/(-7.42)

       t = 14.8 s

This is the time it takes to stop the jet

Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero

Vf² = Vo² + 2 to X

X = -Vo² / 2 a

X = -110² / 2 (-7.42)

X = 815.4 m

Since this distance is greater than the length of the runway, the jet cannot stop

Let's calculate the speed you should have to stop on a track of this size

Vo² = 2 a X

Vo = √ (2 7.42 800)

Vo = 109 m / s

It is conclusion the jet must lose some speed to land on this track

4 0
3 years ago
An astronaut whose mass is 80. kg and is initially at rest carries an empty oxygen tank with a mass of 10. kg. He throws the tan
sergeinik [125]

Answer:

0.25 m/s

Explanation:

From the law of conservation of momentum

Mu+ mu = Mv' + m v

M= mass of the astronaut = 80 kg

m= mass of the oxygen tank= 10 Kg

v= speedof the tank 2 m/s

u= initial velocity of the system= 0

If we substitute the values, we have

( 80× 0 )+(10×0)= [(80 x v )+ (10 x 2)]

0= 80v + 20

-20=80v

v= -0.25 m/s ( we have a negative value because the astronaut and the motion of the cylinder are in opposite direction)

Hence the velocity the astronaut start to move off into space is 0.25 m/s

6 0
3 years ago
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