Answer:
a) dh/dt = -44.56*10⁻⁴ cm/s
b) dr/dt = -17.82*10⁻⁴ cm/s
Explanation:
Given:
Q = dV/dt = -35 cm³/s
R = 1.00 m
H = 2.50 m
if h = 125 cm
a) dh/dt = ?
b) dr/dt = ?
We know that
V = π*r²*h/3
and
tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5
⇒ h = (5/2)*r
⇒ r = (2/5)*h
If we apply
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π
a) if r = (2/5)*h
⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π
⇒ (4/25)(3*h²)(dh/dt) = -105/π
⇒ dh/dt = -875/(4π*h²)
b) if h = (5/2)*r
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π
⇒ (5/2)*(3*r²)(dr/dt) = -105/π
⇒ dr/dt = -14/(π*r²)
Now, using h = 125 cm
dh/dt = -875/(4π*h²) = -875/(4π*(125)²)
⇒ dh/dt = -44.56*10⁻⁴ cm/s
then
h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)
⇒ r = 50 cm
⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)
⇒ dr/dt = -17.82*10⁻⁴ cm/s
Answer:
D
Explanation:
calculate effective resistance of the resistors
= 80 + 120
= 200
overall resistance = 48 ohms
so overall current flowing through circuit
I = V / R
I = 12 / 200
I = 0.06 A
since circuit is in series, current is same at every point so...
potential difference across 80 = R x I
= 80 x 0.06
answer is potential difference = 4.8 V
hope this helps
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I just learned this chapter recently
First you need to draw the picture of the problem to better understand it. Like the one bellow.
In this task you have 2 sides of triangle and we can calculate angle between them. Angle between them is 120 - 35 = 85 degrees.
Once you have those 3 variables you can calculate third side of triangle using cosine law.
a - second displacement
b - first displacement
c- resultant displacement.

now we just need to calculate this.

a = 196
now, we use cosine law again to find the angle between second and first displacement.

degrees
The angle marked with "?" in the graph is our direction angle. We will call it


Second displacement has magnitude of 196 and a direction of -14.64 with positive x axis
The Sun emits electromagnetic radiation at many wavelengths across the EM spectrum.