The energy of the ski lift at the base is kinetic energy:
where m is the mass of the ski lift+the people carried, and
is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is
where
and h is the height at the top of the hill.
So, since the total energy must conserve, we have
and so
from which we find the height:
Sound at 70 dB is 70 dB louder than the human reference level. That's 10⁷ times as much as the reference sound power.
Sound at 73 dB is 73 dB louder than the human reference level. That's 10⁷.³ or 2 x 10⁷ times as much as the reference sound power.
Sound at 80 dB is 80 dB louder than the human reference level. That's 10⁸ or 10 x 10⁷ times as much as the reference sound power.
Now we can adumup:
Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)
Intensity = (13 x 10⁷) times the sound power reference intensity.
Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)
Intensity = 70 + 10 log(13)
Intensity = 70 + 10 (1.114)
Intensity = 70 + 11.14
Intensity = <em>81.14 dB</em>
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Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct. For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.
Sry i was knowing the answer i forgot ;(
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x 
m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x 
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x x 3 x )/2
=120m
The coefficient of friction between the road and the car's tire is determined as 0.78.
<h3>Acceleration of the car</h3>
The acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
- u is the initial velocity = 97 km/h = 26.94 m/s
a = (26.94)²/(2 x 47)
a = 7.72 m/s²
<h3>Coefficient of friction</h3>
μ = a/g
μ = (7.72)/9.8
μ = 0.78
Learn more about coefficient of friction here: brainly.com/question/14121363
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