Answer:
53.63 μA
Explanation:
radius of solenoid, r = 6 cm
Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2
n = 17 turns / cm = 1700 /m
di / dt = 5 A/s
The magnetic field due to the solenoid is given by
B = μ0 n i
dB / dt = μ0 n di / dt
The rate of change in magnetic flux linked with the solenoid =
Area of coil x dB/dt
= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt
= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4
The induced emf is given by the rate of change in magnetic flux linked with the coil.
e = 2.145 x 10^-4 V
i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA
Answer:
0.24 kgm²
Explanation:
= length of the bat = 81.3 cm = 0.813 m
= mass of the bat = 0.96 kg
= distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m
= Period of oscillation = 1.35 sec
= moment of inertia of the bat
Period of oscillation is given as
= 0.24 kgm²
Answer:
a) f = 615.2 Hz b) f = 307.6 Hz
Explanation:
The speed in a wave on a string is
v = √ T / μ
also the speed a wave must meet the relationship
v = λ f
Let's use these expressions in our problem, for the initial conditions
v = √ T₀ /μ
√ (T₀/ μ) = λ₀ f₀
now it indicates that the tension is doubled
T = 2T₀
√ (T /μ) = λ f
√( 2To /μ) = λ f
√2 √ T₀ /μ = λ f
we substitute
√2 (λ₀ f₀) = λ f
if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change
λ₀ = λ
f = f₀ √2
f = 435 √ 2
f = 615.2 Hz
b) The tension is cut in half
T = T₀ / 2
√ (T₀ / 2muy) = f = λ f
√ (T₀ / μ) 1 /√2 = λ f
fo / √2 = f
f = 435 / √2
f = 307.6 Hz
Traslate
La velocidad en una onda en una cuerda es
v = √ T/μ
ademas la velocidad una onda debe cumplir la relación
v= λ f
Usemos estas expresión en nuestro problema, para las condiciones iniciales
v= √ To/μ
√ ( T₀/μ) = λ₀ f₀
ahora nos indica que la tensión se duplica
T = 2T₀
√ ( T/μ) = λf
√ ) 2T₀/μ = λ f
√ 2 √ T₀/μ = λ f
substituimos
√2 ( λ₀ f₀) = λ f
si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia
λ₀ = λ
f = f₀ √2
f = 435 √2
f = 615,2 Hz
b) La tension se reduce a la mitad
T = T₀/2
RA ( T₀/2μ) = λ f
Ra(T₀/μ) 1/ra 2 = λ f
fo /√ 2 = f
f = 435/√2
f = 307,6 Hz
Answer:
The order is 2>4>3>1 (TE)
Explanation:
Look up attached file
