Answer:
c) The pitch of the two sirens sounds the same to you
Explanation:
The pitch does not depend on the distance of the object from the observer.
As per the given data
pitch = frequency
Frequency = 
= 
Hence, the pitch of the two sirens remains the same for the observer.
I believe the answer is x
You skipped over a number in the question, and you didn't tell me what my average speed is. Lucky for you, my average speed has NO EFFECT on the answer to the question.
When you calculate velocity, you only use the straight-line distance between the start-point and the end-point. It doesn't matter what route the thing took to get there, or how much ground it actually covered.
If I travel in a circle and stop at the same point I started from, then the size of the circle doesn't matter, and neither does my speed. The distance between my start-point and my end-point is zero, and my average velocity is zero.
Answer:
a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm
Explanation:
a) For this part we use the concept of density
ρ = m / V
As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats
m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg
V = m / ρ
V = 0.56 / 3.52
V = 0.159 cm3
We use the relation of the bulk module
B = P / (Δv/V)
ΔV = V P / B
ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹
ΔV = 2,118 10⁻⁸ m³
b) indicates that we approximate the diamond to a sphere
V = 4/3 π R³
For this part let's look for the initial radius
R₀ = ∛ ¾ V /π
R₀ = ∛ (¾ 0.159 /π)
R₀ = 0.3361 cm
Now we look for the final volume and with this the final radius
= V + ΔV
= 0.159 + 2.118 10⁻²
= 0.18018 cm3
= ∛ (¾ 0.18018 /π)
= 0.3504 cm
The radius increment is
ΔR = - R₀
ΔR = 0.3504 - 0.3361
ΔR= 0.0143 cm
Answer:
Correct answer is A.
The higher the enzyme, the higher the Vmax
Explanation:
Although, in the absence of enzyme, the rate of a reaction(Vmax) increase linearly with substrate concentration. The reaction rate is given as dp/dt.
The rate of a reaction involving enzyme also increases.
At low enzyme concentrations or high substrate concentrations, all of the available enzyme active sites could be occupied with substrates. Therefore, increasing the substrate concentration further will not change the rate of diffusion. In other words, there is some maximum reaction rate (Vmax) when all enzyme active sites are occupied. The reaction rate will increase with increasing substrate concentration, but must asymptotically approach the saturation rate, Vmax. Vmax is directly proportional to the total enzyme concentration, E
