Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
Vapor pressure<span> or equilibrium </span>vapor pressure<span> is defined as the </span>pressure<span> exerted by a </span>vapor<span> in thermodynamic equilibrium with its condensed phases at a certain temperature. It is independent with atmospheric pressure since it does not change by changing the atmospheric pressure only. </span>
Answer:
The correct option is a
Explanation:
From the question we are told that
The mass of the block is 
The height of the vertical drop is 
Generally from the law of energy conservation , the potential energy at the top of the slide is equal to the kinetic energy at the point after sliding this can be mathematically represented as
i.e 
=> 
=> 
=> 
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
