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icang [17]
3 years ago
12

The distance from home plate to the pitchers mound is 18.5 meters. For a pitcher capable of throwing at 3.85m/s(86mi/hr), how mu

ch time elapses between the release of the ball and the ball passing home plate ?
Physics
1 answer:
Taya2010 [7]3 years ago
7 0

Given that:

Distance , s = 18.5 m

Velocity , v = 3.85 m/s

Time , t =?

Since,

Velocity = distance/time

or

Time= distance/velocity

time= 18.5/ 3.85

time= 4.8 s

So the time elapse between the release of the ball and the ball passing home plate is 4.8 seconds.

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A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

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Answer:

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Explanation:

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red hot is hot, but other colours are even hotter. stars may be hotter than red hot.

there is also something calle the red shift.

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