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Sergeeva-Olga [200]
3 years ago
9

professional baseball player nolan ryan could pitch a baseball at approximately 160.0 km/h. at that average velocity, how long d

id it take a ball thrown by ryan to reach home plate, which is 18.4 m from the pitcher’s mound? compare this with the average reaction time of a human to a visual stimulus, which is 0.25 s.
Physics
2 answers:
Margarita [4]3 years ago
6 0

Answer:

0.41 s

Explanation:

Parameters given:

Speed of ball, v = 160 km/h

Distance between Ryan and the home plate, d = 18.4 m = 0.0184 km

Speed is given as the time rate of change of distance. Mathematically:

Speed = \frac{distance}{time} \\\\\\v = \frac{d}{t}

Therefore, time will be:

t = \frac{d}{v}

t = \frac{0.0184}{160} \\\\\\t = 0.000115 hr = 0.41 s

The time it will take the ball to reach the home plate is 0.41 s.

Comparing this to the average reaction time of humans to visual stimulus (0.25 s), we see that the time it will take the ball is greater than the reaction time.

This implies that an average person should be able to track the motion of the ball as it moves towards the home plate.

prohojiy [21]3 years ago
4 0

Answer:

t = 0.4 s

Explanation:

Assuming that we can neglect the influence of gravity, the ball travels at a constant speed along a straight line, between the pitcher's mound and the home plate.

We can apply the definition of average velocity, as follows:

v = xf-xo /tfin-to

If we choose to = 0, and xo = 0 (coincident with the mound's  location), we can simply put:

t = xf / v (1)

As we have x and v in different units, we can convert v from km/h to m/s, as follows:

v= 160 Km/h * (1 h/ 3600 sec) * (1000 m / 1 Km) = 44.4 m/s

Replacing in (1):

t = 18.4 m / 44.4 m/s = 0.4 s

As the average reaction time of a human to a visual stimulus, is roughly 0.25 sec, the time needed for the ball to travel is just 1.25 of this time.

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Factors Affecting the Value of g

As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

As we go at large heights, value of g decreases.

To Calculate the Value of g

Value of universal gravitational constant, G = 6.7 × 10–11 N m2/ kg2,

Mass of the earth, M = 6 × 1024 kg, and

Radius of the earth, R = 6.4 × 106 m

Putting all these values in equation (iii), we get:

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Difference between Gravitation Constant (G) and Gravitational Acceleration (g)

S. No.

Gravitation Constant (G)

Gravitational acceleration (g)

1.

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Its value is 9.8 m/s2.

2.

It is a scalar quantity.

It is a vactor quantity.

3.

Its value remains constant always and everywhere.

Its value varies at various places.

4.

Its unit is Nm2/kg2.

Its unit is m/s2.

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Weight of an object is given as,

Hence, weight of the object on the moon = (1/6) × its weight on the earth.

Try the following questions:

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Q4. Differentiate between weight and mass.

Q5. Why is the weight of an object on the moon 1/6th its weight on the earth??

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