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3 years ago

The two main parts of a wave are the

1 answer:

3 0

<span>So we want to know the two main physical quantities that describe the wave. Those are frequency and wavelength. These two quantities are connected via the equation: v=f*L, where v is velocity of a wave, f is frequency of a wave and L is the wavelength of the wave. </span>

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PLease help me WIll give brainliest

melisa1 [442]

Answer:

Explanation:

3 0

2 years ago

Danielle exerts a 14.0 N force to compress a spring by a distance of 8.00 cm. What is the spring constant of this spring

Lady_Fox [76]

Answer:

175 N/m

Explanation:

Given:

Force = F= 14.0 N

Distance = x = 8.00 cm = 0.08 m

To find:

spring constant

Solution:

spring constant is calculated by using Hooke's law:

k = F/x

Putting the values in above formula:

k = 14.0 / 0.08

k = 175 N/m

4 0

3 years ago

To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar

Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

Explanation:

$f = \frac{1}{2\pi*\sqrt{C*L}}$

a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF replacing:

$f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}$

$f = 475.14 Hz$

b. second smallest use the capacitive 6 uF so:

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}$

$f = 1959Hz$

c. second largest and largest oscillation first combination so:

Use 4.2 uF

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{4.2uF*11mH}}$

$f = 2341.53 Hz$

And finally largest oscillation cap in serie so:

$C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}$

$C=2.47 uF$

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}$

$f = 3053.34 Hz$

5 0

3 years ago

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 m2 . At

Anastaziya [24]

Answer:

31.8 × 10⁻⁴ J = 3.18 mJ

Explanation:

We know the intensity I of a wave is I = P/A where P = power and A = area = 0.500 m²

The intensity of an electromagnetic wave is also equal to I = E₀²/μ₀c

where E₀ = maximum electric field strength = √2E where E = rms value of electric field = 0.0200 N/C, μ₀ = 4π × 10⁻⁷ H/m ,c = 3 × 10⁸ m/s

P/A = E₀²/μ₀c = 2E²/μ₀c

P = 2E²A/μ₀c = 2 × (0.02 N/C)² × 0.5 m²/(4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)

= 1.06 × 10⁻⁴ W = 0.106 mW

Since P = E/t where E = Energy and t = time

E = Pt with t = 30 s

E = 1.06 × 10⁻⁴ W × 30 s = 31.8 × 10⁻⁴ J = 3.18 mJ

So the wave carries 3.18 mJ of energy through the window in 30 s

5 0

3 years ago

An object is moving with an initial velocity of 3.3m/s it is subject to a constant acceleration of 3.7 m/s2 for 10 s. How far wi

egoroff_w [7]

Answer:

218m

Explanation:

Given parameters:

Initial velocity = 3.3m/s

acceleration = 3.7m/s²

time = 10s

Unknown:

How far will it travel during the time of acceleration = ?

Solution:

We use of the kinematics equations to solve this problem;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

t is the time

a is the acceleration

So;

S = (3.3x10) + ( $\frac{1}{2}$ x 3.7 x 10²) = 218m

8 0

2 years ago