01.01.2023

One of the fastest pitches ever thrown in Major League Baseball was by Aroldis Chapman and had a velocity of 105.1 miles/hour. How many seconds did it take this pitch to travel the 60 feet and 6 inches from the pitcher's mound to home plate

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24.06.2023, solved by verified expert
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t = 0.39 s

Explanation:

Assuming that the ball is launched horizontally, once in the air, if we neglect the resistance of the air, the ball moves at a constant speed, equal to the initial velocity, in this case, 105.1 mi/hr.In order to find time in seconds, it is advisable to convert the speed in mi/hr to m/s, as follows:

       One of the fastest pitches ever thrown in Major, №17886361, 01.01.2023 04:56

In the same way, it's advisable to convert 60' 6'' (60.5') to m, as follows:

       One of the fastest pitches ever thrown in Major, №17886361, 01.01.2023 04:56

Applying the definition of average velocity, we can find the time traveled by the ball from pitcher's mound to home plate, as follows:

       One of the fastest pitches ever thrown in Major, №17886361, 01.01.2023 04:56

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Physics
Step-by-step answer
P Answered by PhD

t = 0.39 s

Explanation:

Assuming that the ball is launched horizontally, once in the air, if we neglect the resistance of the air, the ball moves at a constant speed, equal to the initial velocity, in this case, 105.1 mi/hr.In order to find time in seconds, it is advisable to convert the speed in mi/hr to m/s, as follows:

       105.1 mi/hr * (\frac{1hr}{3600s})*\frac{1609m}{1mi} = 47.0 m/s (1)

In the same way, it's advisable to convert 60' 6'' (60.5') to m, as follows:

       60.5 ft * \frac{0.3058m}{1ft} = 18.5 m (2)

Applying the definition of average velocity, we can find the time traveled by the ball from pitcher's mound to home plate, as follows:

       t = \frac{18.5m}{47m/s} = 0.39 s (3)

Physics
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P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

Physics
Step-by-step answer
P Answered by PhD

The question specifies the diameter of the screw, therefore the IMA of this screw is 0.812? / 0.318 = 8.02

Physics
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P Answered by PhD
Answer:
7.25 secs.

Explanation:
First find the distance it takes to stop
s = [v^2-u^2]/2a = 0^2 - 8.7^2/2[-2.4] = 8.7^2/4.8
Next find the time it takes to go that distance , s = ut +[1/2] at^2
8.7^2/4.8 = 8.7t +[1/2] [ -2.4]t^2 , rearrange and
t^2 -[8.7/1.2]+ 8.7^2/[(1.2)(4.8)]=0 complete the square
[t - (8.7/2.4)]^2=0
t = 8.7/2.4 = 3.625 secs
At this stage the deceleration will push the object back in the direction it came from for another 3.625 secs when it will be 8.7 m/s again
Total time , T =2t = 7.25 secs.

Note:
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x.
Physics
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P Answered by PhD
First sum applied the Newton's second law motion: F = ma
Force = mass* acceleration
This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time,
force=(mass*velocity)/time
such that, (mass*velocity)/time=momentum/time
Therefore we get mass*velocity=momentum
Momentum=mass*velocity
Elephant mass=6300 kg; velocity=0.11 m/s
Momentum=6300*0.11
P=693 kg (m/s)
Dolphin mass=50 kg; velocity=10.4 m/s
Momentum=50*10.4
P=520 kg (m/s)
The elephant has more momentum(P) because it is large.
Physics
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P Answered by PhD
The change in temperature is 9.52°CExplanation:Since, the heat supplied by the electric kettle is totally used to increase the temperature of the water.Thus, from the law of conservation of energy can be stated as:Heat Supplied by Electric Kettle = Heat Absorbed by WaterHeat Supplied by Electric Kettle = m C ΔTwhere,Heat Supplied by Electric Kettle = 20,000 JMass of water = m = 0.5 kgSpecific Heat Capacity of Water = C = 4200 J/kg.°CChange in Temperature of Water = ΔTTherefore,20,000 J = (0.5 kg)(4200 J/kg.°C) ΔTΔT = 20,000 J/(2100 J/°C)ΔT = 9.52°C
Physics
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P Answered by PhD
Weight of barbell (m) = 100 kg
Uplifted to height (h) = 2m
Time taken= 1.5 s
Work done by Jordan = potential energy stored in barbell = mgh
= 100×2×9.8
= 1960J
Power = energy/time
= 1960/1.5
1306.67watts
Physics
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P Answered by PhD
Weight of jasmine (m) = 400 N
Height climbed on wall (h) = 5m
Total time taken in climbing = 5 sec
Work done in climbing the wall = rise in potential energy = mgh
= 400×9.8×51
= 19600J
Power generated by Jasmine = potential energy / time
= 19600/5
= 3920Watts

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