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Amiraneli [1.4K]
3 years ago
13

A baseball pitcher throws a fastball horizontally at a speed of 43.0 m/s. Ignoring air resistance, how far does the ball drop be

tween the pitcher’s mound and home plate, 60 ft 6 in away?. Express your answer using SI units.
Physics
2 answers:
Nady [450]3 years ago
7 0
Based on your problem where as ask for the distance of the ball drop between the pitchers mound and the home plate and with a given of the speed of ball is 43m/s and the homeplates is 60.6ft away. Based on my step by step procedure and also considering the value of gravity by 9.8m/s^2 i came up with the distance of 144m away
ehidna [41]3 years ago
4 0

Answer:

The distance covered by the ball between the pitcher’s mound and home plate is 0.864 meters.

Explanation:

Given that,

Horizontal velocity of the football, v = 43 m/s

In this case, we need o find the distance covered by the ball between the pitcher’s mound and home plate, 60 ft 6 in away. It implies,

d=60\ ft\ 6\ in\\\\d=60.5\ ft\\\\d=8.44\ m

Let t is the time taken by the football to cover that much distance. So,

t=\dfrac{d}{v}\\\\t=\dfrac{18.44}{43}\\\\t=0.42\ s

Initial velocity of the ball in vertical direction is equal to 0. So, the distance covered by the ball between the pitcher’s mound and home plate is given by :

y=\dfrac{1}{2}gt^2\\\\y=\dfrac{1}{2}\times 9.8\times (0.42)^2\\\\y=0.864\ m

So, the distance covered by the ball between the pitcher’s mound and home plate is 0.864 meters. Hence, this is the required solution.

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Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

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v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

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2 years ago
An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what hap
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An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.  

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The distance - time graph of the humming bird is missing, so i have attached it.

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