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blondinia [14]
3 years ago
7

A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will

it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
Physics
1 answer:
SSSSS [86.1K]3 years ago
5 0

Answer:

Ball will drop by 0.82 meter.

Explanation:

Horizontal speed = 41 m/s

Horizontal displacement = 17 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    17 = 41 t + 0.5 x 0 x t²

     t = 0.41 s

Now we need to find how much vertical distance ball travels in 0.41 s.

Initial vertical speed = 0 m/s

Time = 0.41s

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 0 x 0.41 + 0.5 x 9.81 x 0.41²

    s = 0.82 m

So ball will drop by 0.82 meter.

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Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.  

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A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

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• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2  is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

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The law of conservation of energy states that the energy stored is constant:

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Q   =  Q1

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       = 21/2.2

      = 9.5

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