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
The acceleration is given as:
a = g sin(30°) where g is the gravitational acceleration
For g = 10 m/s^2, we get
a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2
3 kilometers, it is just 5/60 or 1/12 multiplied by 36.
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Answer:
Force constant, k = 653.3 N/m
Explanation:
It is given that,
Weight of the bag of oranges on a scale, W = 22.3 N
Let m is the mass of the bag of oranges,
m = 2.27 kg
Frequency of the oscillation of the scale, f = 2.7 Hz
We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :
k = 653.3 N/m
So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.
