Answer:
12.56 A.
Explanation:
The magnetic field of a conductor carrying current is give as
H = I/2πr ............................... Equation 1
Where H = Magnetic Field, I = current, r = distance, and π = pie
Making I the subject of the equation,
I = 2πrH............... Equation 2
Given: H = 1 T, r = 2 m.
Constant: π = 3.14
Substitute into equation 2
I = 2×3.14×2×1
I = 12.56 A.
Hence, the magnetic field = 12.56 A.
Answer:
(i) The wavelength is 0.985 m
(ii) The frequency of the wave is 36.84 Hz
Explanation:
Given;
mass of the string, m = 0.0133 kg
tensional force on the string, T = 8.89 N
length of the string, L = 1.97 m
Velocity of the wave is:
(i) The wavelength:
Fourth harmonic of a string with two nodes, the wavelength is given as,
L = 2λ
λ = L/2
λ = 1.97 / 2
λ = 0.985 m
(ii) Frequency of the wave is:
v = fλ
f = v / λ
f = 36.29 / 0.985
f = 36.84 Hz
Answer: 1175 J
Explanation:
Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."
Given
Spring constant, k = 102 N/m
Extension of the hose, x = 4.8 m
from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m
Work done =
W = 1/2 k [x(i)² - x(f)²]
Since x(f) = 0, then
W = 1/2 k x(i)²
W = 1/2 * 102 * 4.8²
W = 1/2 * 102 * 23.04
W = 1/2 * 2350.08
W = 1175.04
W = 1175 J
Therefore, the hose does a work of exactly 1175 J on the balloon
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