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dsp73
3 years ago
15

un jugador de beisbol batea un foul recto en el aire.la pelota deja el bate con una rapidez de 120 km/h. en ausencia de resisten

cia del aire. ¿cual sera la rapidez de la pelota cuando la atrape el catcher
Physics
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

120 km/h

Explanation:

This is a case of a parabolic motion. For this kind of motion the horizontal component of the velocity does not change in the complete trajectory. That is:

v_x=constant

Then, the velocity of the ball at the moment in which the catcher caught the ball is:

v=120\frac{km}{h}

the velocity of the ball is 120 km/h.

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A racecar driver has to hold on tightly when going around a banked curve. Approximately what is the centripetal force on a 2220.
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3 years ago
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?
Ilya [14]

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

<u>And now we have to write the temperature in kelvins.</u>

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

<u>Let us find it now.</u>

64°C + 273 = 337k

Therefore,

64°C ⇒ <u>337k</u>

8 0
1 year ago
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