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Natali [406]
4 years ago
7

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. What is the time necessary for the projec

tile to reach the ground below?
Physics
1 answer:
baherus [9]4 years ago
5 0
The horizontal speed has no effect on the answer. 

It doesn't matter whether you flick a marble horizontally from the roof,
fire a high-power rifle horizontally from the roof, drive a school bus straight
off the roof, or drop a bowling ball from the roof with zero horizontal speed. 
Their vertical speed is completely determined by gravity, (and it happens to
be the same for all of them).

Handy dandy formula for the distance covered by anything that starts out
with zero speed and accelerates to the end:

            Distance = (1/2) (acceleration) x (time)²

If the beginning of the journey is on Earth, then the acceleration is
9.8 m/s² ... the acceleration of gravity on Earth.  We'll assume that
the 55-meter rooftop in the question is part of a building on Earth.

                       55 meters  =  (1/2) (9.8 m/s²) x (time)²           

Divide each side
by  4.9 m/s² :            55 m / 4.9 m/s²  =  (time)²

                                 (time)²  =  (55/4.9)  sec²

Square-root
each side:                time  =  √(55/4.9 sec²)

                                           =      3.35 sec  .
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Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?
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70kg

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3 0
3 years ago
-Para lograr que una pieza de 0,300 kg de cierto metal aumente su temperatura desde 40°C a 60 °C ha sido necesario suministrarle
marishachu [46]

Answer:

parte a) el calor específico es c = 0.383 J /(gr*K)

parte b) la temperatura inicial es T inicial= 52.72 °C

Explanation:

para el primer punto la formula para el calor Q es:

Q = m * c * ( T final - T inicial )

donde

m= masa de la pieza = 0.300 kg = 300 gr

Q = flujo de energía en forma de calor= 2299 J

c = calor específico

T final = temperatura final =40°C

T inicial = temperatura inicial = 60 °C

entonces

Q = m * c * ( T final - T inicial )

c = Q / [ m* ( T final - T inicial ) = 2299 J/[ 300 gr  * ( 60 °C - 40°C )]

= 0.383 J /(gr*K)

c = 0.383 J /(gr*K)

para el segundo punto usamos la misma formula

Q = m * c * ( T final - T inicial )

pero

m= 200 gr= 0.200 kg

c=459.8 J/(kg*K) , Q =20.900 J , T final = 280 °C

Q = m * c * ( T final - T inicial )

T inicial = T final - Q/(m*c)  =280 °C - 20.900 J/(459.8 J/(kg*K)* 0.200 kg) = 52.72 °C

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