All categories

2 years ago

Which of the following is an example of mechanical waves in nature?

1 answer:

8 0

The correct answer is D. sound

A mechanical wave is one where oscillations of matter occur and energy is transferred through a medium, which is what occurs with sound waves.

You might be interested in

a person is sitting on the last bench can see clearly see things written on book but cannot see them distinctly on board. what t

marysya [2.9K]

Answer:

I think concave

Explanation:

7 0

2 years ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

<u>From the Coulomb's Law the force between the charges is given as:</u>

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

<em>Now force:</em>

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

7 0

3 years ago

Read 2 more answers

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago

A light wave traveling through medium 1 strikes a boundary with medium 2 at at a 45 degree angle. the light then enters the seco

Alona [7]

Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.

So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal

here it can be said that medium 2 will be rarer then medium 1

So here the possible options are

1. Water

Air

2. Diamond

Air

So in above two options medium 1 is denser and medium 2 is rarer

8 0

2 years ago

A rock is dropped from a garage roof from rest. the roof is 6.0 m from the ground. determine the velocity of the rock as it hits

Dmitrij [34]

From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

V(Final velocity)=? (m/sec)

h(height)= 6.0 m

u(Initial velocity)=0 m/sec

g(gravitational acceleration)=9.81 m/s²

Newton's third equation of motion:

$\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\ v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec$

Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

#SPJ1

7 0

2 years ago