Answer:
a) d = 7.62 10⁻⁶ m, b) l = 3.25 10⁴ m
Explanation:
Resistance is expressed by the formula
R = ρ l / A (1)
density is defined by
density = m / V
the volume of a wire is the cross section by the length
V = A l
we substitute
density = m / A l
A = m / density l
we substitute in 1
R = ρ l density l / m
R =ρ density l² / m
l = √ (R m /ρ density)
let's calculate the cable length
l = √(11.7 13.5 10⁻³ / (1.68 10⁻⁸ 8.9 10³))
l = √(10.56 10⁸)
l = 3.25 10⁴ m
now we can find the cable diameter with the density equation
A = m / density l
A = 13.5 10⁻³ / (8.9 10³ 3.25 10⁴)
A = 4,557 10⁻¹¹ m²
the area of the circle is
A = π r² = π d² / 4
d = √ (4A /π)
d = √ (4 4,557 10⁻¹¹/π)
d = 7.62 10⁻⁶ m
Because if we don't you'll be hungry, if you don't eat for about a month you'll die
Answer:
909.1 m
Explanation:
Rate of temperature increase with 100 m elevation = 1°C
h = Maximum Height
Adiabatic lapse rate = -0.65°C/100 m
We have the relation
The maximum height is 909.1 m
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.
Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms
Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A
Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.
Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A
Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
Answer:
Option (A) is correct.
Explanation:
A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is
mass of string, m = 0.00145 kg
Frequency, f = 120 Hz
wavelength = 0.6 m
Speed = frequency x wavelength
speed = 120 x 0.6 = 72 m/s
Let the tension is T.
Use the formula
Option (A) is correct.
