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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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Kepler’s
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We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
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M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)


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Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²          

where k = 31 N/m  (spring constant)

s = 0.15 m  (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²    

   

where Ek = Ep = 3.4875J

m = 0.08 kg  (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g          

where F = unknown (frictional force)

∪ = 0.4   (coefficient of friction)

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g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a                  

where F = 0.31392   (frictional force)

m = 0.08 kg   (mass of block)

a = unknown  (acceleration)

a = -3.924 m/s²      -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as            

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u = 2.9526 m/s (initial velocity

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s = unknown (distance traveled)

s = 1.11 meters

3 0
3 years ago
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