Answer:
v= s/t
Explanation:
250 km/ h =69.44m/s
S1=2 times 69.44 ≈ 139m
Next 2.5 seconds:
S2 = 100m
Average speed:
v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.
If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.
For the wave you described . . .
Frequency = (speed) / (wavelength)
= (6 m/s) / (12 m)
= 0.5 / sec
= 0.5 Hz .
(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
Answer:
v = 5.15 m/s
Explanation:
At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N
As the cable tension is less than this value, the car must be accelerating downward.
7730 = 984(9.81 - a)
a = 1.95 m/s²
kinematic equations s = ut + ½at² and v = u + at
-5.00 = u(4.00) + ½(-1.95)4.00²
u = 2.65 m/s the car's initial velocity was upward at 2.65 m/s
v = 2.65 + (-1.95)(4.00)
v = -5.15 m/s
Density =mass/volume
density= (0.044kg/(11 x10^-6)
4000kg/m^3
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
