Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.
In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.
Answer:
= 2.33
Explanation:
.According to snell's law:
n1sin i = n2sin r ,
where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,
i is angle of incidence,
r is angle of refraction.
Given that,
n1 = 1,
i = 51 degrees,
r = 19.5 degrees. ,
n2= ?
So,
1*sin 51 = n2 sin 19.5
=> n2 = sin51 / sin19.5
= 2.33
All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>
Answer:
f ’= 97.0 Hz
Explanation:
This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer
in this case the source is the police cases that go to vs = 160 km / h
and the observer is vo = 120 km / h
the relationship of the doppler effect is
f ’= f₀ (v + v₀ / v- 
)
let's reduce the magnitude to the SI system
v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s
v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s
we substitute in the equation of the Doppler effect
f ‘= 100 (330+ 33.33 / 330-44.44)
f ’= 97.0 Hz
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