Answer: 3.84dB
Explanation:
Since person A is talking 1.2dB louder than B, we will have
A = 1.2B... (1)
Similarly, person C is talking 3.2 dB louder than person A, we have
C = 3.2A... (2)
From equation 1, B = A/1.2... (3)
To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give
C/B = 3.2A/{A/1.2}
C/B = 3.2A×1.2/A
C/B = 3.2×1.2
C/B = 3.84dB
A square loop whose sides are long is made of copper wire of radius , given the resistivity of copper is . if the magnetic field perpendicular to the loop changes at a constant rate of I = 14.029 mA.
The basic characteristic of a substance that measures how effectively it resists an electric current is called electrical resistance. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. Electrical resistance is a basic property of a material that measures how strongly it resists an electric current. The SI unit for electrical resistance is the ohmmeter.
We use magnetic field as a tool to describe how the magnetic field is distributed in the space around and inside something of a magnetic nature. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. An ohmmeter is a unit of electrical resistance in the SI system.
Learn more about magnetic field here;
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The complete question is :
A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
A is the answer I really don’t know the answer to that but if u can u can help me on my work
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ = 
with the new dough
v = 
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v = 
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v = 
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v = 
v = 1 /√5 v₀
v = 0.447 v₀
