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Alexeev081 [22]
3 years ago
14

An electromagnetic wave with a peak magnetic field component of 3.20 × 10−7 T carries what average power per unit area? (μ0 = 4π

× 10−7 T⋅m/A, ε0 = 8.85 × 10−12 C2/N⋅m2 and c = 3.00 × 108 m/s)
Physics
2 answers:
Oksanka [162]3 years ago
6 0

Answer:

The average power per unit area is 12.22 W/m²

Explanation:

Given that,

Peak magnetic field B=3.20\times10^{-7}\ T

We need to calculate the peak electric field

Using formula of electric filed

E_{peak}=B_{peak}\times c

Put the value into the formula

E_{peak}=3.20\times10^{-7}\times3\times10^{8}

E_{peak}=96\ N/C

We need to calculate the average power per unit area

Using formula of average power

I=\dfrac{E_{max}^2}{2\mu_{0}c}

Put the value into the formula

I=\dfrac{96^2}{2\times4\pi\times10^{-7}\times3\times10^{8}}

I=12.22\ W/m^2

Hence, The average power per unit area is 12.22 W/m²

IRISSAK [1]3 years ago
4 0

Answer:

Explanation:

Average power per unit area =

I avg = c B₀² / 2μ₀

= 3 X 10⁸ X (3.2 X 10⁻⁷)² / 2 X 4π × 10⁻⁷

= 1.22 X 10

12 . 2 W / m²

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8 0
3 years ago
A series RC circuit contains a 1,000 ohm resistor and a 0.025 microfarad capacitor. What is the time constant of this circuit?
stiks02 [169]

The RC time constant, also called tau, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads), i.e.

\tau = RC

Here,

R = Resistance

C = Capacitance

Replacing we have that

\tau = (1000)(0.025*10^{6})

\tau = 25*10^{-6}

\tau = 25\mu s

Therefore the time constant of this circuit is \tau = 25\mu s

5 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
Ask Your Teacher The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s
adell [148]

Answer:

\theta = 20\ rev

Explanation:

Case 1

Given,

initial angular speed = 0 rev/s

final angular speed = 2 rev/s

time, t = 7 s

angular acceleration of the washer

\alpha = \dfrac{\omega_f-\omega_i}{t}

\alpha = \dfrac{2-0}{7}

\alpha = 0.286\ rev/s^2

using equation of rotational motion

\omega_f^2 = \omega_i^2 + 2 \alpha \theta_1

2^2 = 0^2 + 2\times 0.286 \times \theta_1

\theta_1 = 7 rev

Case 2

Given,

initial angular speed = 2 rev/s

final angular speed = 0 rev/s

time, t = 12 s

angular acceleration of the washer

\alpha = \dfrac{\omega_f-\omega_i}{t}

\alpha = \dfrac{0-2}{13}

\alpha = -0.154\ rev/s^2

using equation of rotational motion

\omega_f^2 = \omega_i^2 + 2 \alpha \theta_2

0^2 = 2^2 + 2\times (-0.154) \times \theta_2

\theta_2 = 13 rev

total revolution in this case

\theta = \theta_1 + \theta_2

\theta =7 +13

\theta = 20\ rev

total revolution of the washer is equal to 20 rev.

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3 years ago
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