Answer:
B. The truck and mosquito exert the same size force on each other.
Explanation:
Newton's third law (law of action-reaction) states that
"When an object A exerts a force (action) on an object B, then object B exerts an equal and opposite force (reaction) on object A"
In this case, we can call
object A = the truck
object B = the mosquito
Thereforce according to Newton's third law, the force exerted by the truck on the mosquito is equal in magnitude to the force exerted by the mosquito on the truck (and in opposite direction).
The reason for which the mosquito will experience much more damage is the fact that the mosquito's mass is much smaller than the truck's mass, and since the acceleration is inversely proportional to the mass:
the mosquito will experience a much larger deceleration than the truck, therefore much more damage.
Ideal Gas Law PV = nRT
THE GASEOUS STATE
Pressure atm
Volume liters
n moles
R L atm mol^-1 K^-1
Temperature Kelvin
pv = rt
divide both sides by v
pv/v = rt/v
p = rt/v
answer: p = rt/v
Ideal Gas Law: Density
PV = NRT
PV = mass/(mw)RT
mass/V = P (MW)/RT = density
Molar Mass:
Ideal Gas Law PV = NRT
PV = mass/(MW) RT
MW = mass * RT/PV
Measures of Gases:
Daltons Law of Partial Pressures; is the total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Total = P_ A + P_ B
P_ A V = n_ A RT
P_ B V = n_ B R T
Partial Pressures in Gas Mixtures:
P_ total = P_ A + P_ B
P_ A = n_ A RT/V P_ B = n_ B RTV
P_ total = P_ A + P_ B = n_ total RT/V
For Ideal Gasses:
P_ A = n_ A RT/V P_ total = n_ toatal RT/V
P_ A/P_ total = n_ A RTV/n_ total RTV
= n_ A/n_ total = X_ A
Therefore, P_ A = X_ A P_ total.
PV = nRT
P pressure
V volume
n Number of moles
R Gas Constant
T temperture (Kelvin.).
Hope that helps!!!!!! Have a great day : )
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
Answer:
The number is 
Explanation:
From the question we are told that
The wavelength is 
The length of the glass plates is 
The distance between the plates (radius of wire ) = 
Generally the condition for constructive interference in a film is mathematically represented as
![2 * t = [m + \frac{1}{2} ]\lambda](https://tex.z-dn.net/?f=2%20%2A%20%20t%20%20%3D%20%5Bm%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5D%5Clambda)
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
![2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}](https://tex.z-dn.net/?f=2%20%2A%20%202%20%2A%20d%20%20%20%3D%20%5Bm%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5D%20520%20%2A10%5E%7B-9%7D)
=> ![2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}](https://tex.z-dn.net/?f=2%20%2A%20%202%20%2A%20%20%20%282.8%20%2A10%5E%7B-5%7D%29%20%3D%20%5Bm%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5D%20520%20%2A10%5E%7B-9%7D)
=>
given that we start counting m from zero
it means that the number of bright fringes that would appear is
=> 
=>
