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2 years ago

14

1 question, brainliest for whoever gets it right.

1 answer:

Kaylis [27]2 years ago

6 0

Answer:

im pretty sure the answer is c please mark me brainliest

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Which are examples of transverse waves? Check all that apply. earthquake P-waves earthquake S-waves radio waves sound visible li

Mariulka [41]

Earthquake S - Waves are examples of transverse waves. The correct option among all the options that are given in the question is the second option. Other good examples of transverse waves are an oscillating string and light waves. A wave is a kind of disturbance that or an oscillation that travels through space.

8 0

2 years ago

Read 2 more answers

How many electrons do the elements of the most stable group have in their outer energy levels

Wittaler [7]

I think it might be 8 electrons

6 0

2 years ago

Read 2 more answers

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0

2 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago

Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0 kg cart repels one way

JulijaS [17]

Answer:

V = -0.3 m/sec.

Explanation:

5.0 x 0.12 + 2.0 x v = 0. Which means that V = -0.3 m/sec.

The -ve sign shows it moves in the opposite direction.

8 0

2 years ago