A bucket of water experiencing a gravitational force of 525 N is pulled from a water well. Net force in the Y direction is 45 N
1 answer:
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Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
(1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
(2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
i = 3.4 A
(c)
Current density is given as
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
E = 9.01 x 10⁻³ V/m