Answer:
48.26 m
Explanation:
time to goes up (till stop for a while in the air - maximum height)
vt = vo + a t
0 = 15 + g . t
0 = 15 + (-9.8) . t
9.8t = 15
t = 1.531 s
so the time left to goes down is
4.0 - 1.531 = 2.469 s
height from the top of building can find it by using
vo =√(2gh)
15 = √(2)(9.8).h
15² = 19.6h
h = 225/19.6 = 11.48 m
so the distance of maximum height to the ground is
t = √(2H/g)
2.469 = √(2H/9.8)
2.469² = 2H/9.8
6.096 = 2H/9.8
2H = 6.096 x 9.8 = 59.74 m
so the vertical distance of the building (or the building height's is)
H - h = 59.74 - 11.48 = 48.26 m
Answer:
There are no appropriate descriptions on the list of choices you've provided.
Explanation:
Reverberation, in psychoacoustics and acoustics, is a persistence of sound after the sound is produced.[1] A reverberation, or reverb, is created when a sound or signal is reflected causing a large number of reflections to build up and then decay as the sound is absorbed by the surfaces of objects in the space – which could include furniture, people, and air.[2] This is most noticeable when the sound source stops but the reflections continue, decreasing in amplitude, until they reach zero amplitude.
U = 0, initial vertical velocity
v = 60 mph = 88 ft/s
Ignore air resistance and take g = 32 ft/s².
It t = time to attain 60 mph, then
(88 ft/s) = (32 ft/s²)*(t s)
t = 88/32 = 2.75 s
Answer: 2.75 s
Answer:
a) 
b) 
Explanation:
From the exercise we know that
From dynamics we know that the formula for average velocity is:
a) For the three intervals:
b) The average velocity for the entire motion can be calculate by the following formula:
Answer:
5.33 cm
Explanation:
The lens equation states that:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem,
p = 8 cm
q = 16 cm ( the sign is positive since the image is real, which means it is formed on the other side of the lens)
Substituting into the equation,
