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4 years ago

Which wave is a body wave? a. P wave b. L wave c. B wave d. T wave

2 answers:

8 0

<span><span>P-waves are a type of body wave, called seismic
waves in seismology, that travel through a continuum and are the first waves from an earthquake to arrive at a seismograph.</span></span>

ivanzaharov [21]4 years ago

4 0

P And S waves are body waves because each wave shakes the ground in different ways Hope This Helps:)

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Which is the 2nd closest star to earth??

IgorLugansk [536]

Answer:

Proxima Centauri

Explanation:

U 2 can help me by marking as brainliest........

8 0

3 years ago

Read 2 more answers

Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00

MariettaO [177]

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

$sin\theta=\frac {\lambda}{d}\times m$ ....1

The location of the bright fringe is determined by :

$y=L\times tan\theta$

Where, L is the distance between the slit and the screen.

For small angle , $sin\theta=tan\theta$

So,

Formula becomes:

$y=L\times sin\theta$

Using 1, we get:

$y=L\times \frac {\lambda}{d}\times m$

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

$0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1$

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>

7 0

3 years ago

A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 f

Rashid [163]

Answer:

0.28 ft

Explanation:

We are given that

Strength=m= $36ft^2/s$

Distance between source and sink=15 ft

Distance between the sink of the source and origin= $a=\frac{15}{2}$ ft

Uniform velocity, U=12 ft/s

We have to find the length of the oval.

Formula to find the half length of the body

$\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}$

Where a=Distance between sink of source and origin

U=Uniform velocity

m=Strength

l=Half length

Using the formula

$\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}$

$l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}$

$l=0.14$

Length of oval= $2l=2(0.14)=0.28 ft$

8 0

3 years ago

What is true about the velocity of sound waves in solids compared to air?

ivolga24 [154]

<h2>The correct option is (A)</h2>

Explanation:

Travels faster in solids because the particles are closer together.

The reason for the above statement is -

(i) Sound waves travel through the medium (solid, Liquid, and gas), more denser is the medium more is the speed of sound.

(ii) In solid, the particles are arranged very closely and tightly packed therefore the sound waves in solid have the fastest speed or travels fastest in comparison to liquid and gas.

(iii) Slower than solid is the travel of sound waves in liquid than in gas(least dense, particles are far apart.

4 0

3 years ago

Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visi

gregori [183]

Answer:

1.013 s

Explanation:

You can solve this problem using the equations for constant acceleration motion. The velocity at the bottom of the window can be found using this expression:

$(x-x_o) = \frac{1}{2} at^2 + v_ot = -\frac{1}{2}gt^2 + v_ot$

the gravity is negative as it opposes the movement.

$(x-x_o) = -\frac{1}{2}gt^2 + v_ot\\v_o = \frac{(x-x_o)+\frac{1}{2}gt^2}{t} = \frac{1.19m+\frac{1}{2} 9.81m/s^2(0.2s)^2}{0.2s} = 6.931 m/s$

Now, the time elapsed before the ball reappears is 2 times the time that it takes for the ball to go from the bottom of the window, reach maximum height, and reach again the bottom of the window, minus 2 times the time that it takes for the ball to travel from the top to the bottom of the window. The time that takes to the ball to reach maximum height, or in other words, to time that takes for the velocity of the ball to go from vo to 0m/s:

$t_1 = \frac{0m/s - v_o}{g} = \frac{-6.931 m/s}{-9.81m/s^2} = 0.706 s$

Then:

$t = 2t_1 - 2*0.2s = 2*(0.706s) - 0.4s = 1.013 s$

7 0

3 years ago