All categories

4 years ago

A sound wave traveling downward with a speed of about 4,000 m/s suddenly slows to 1,500 m/s not far below the Earth’s surface.

Physics

2 answers:

WINSTONCH [101]4 years ago

5 0

Speed of sound wave depends on the elasticity of medium

It is given by formula

$v = \sqrt{\frac{E}{\rho}}$

now if we are given that initially sound wave is travelling with speed 4000 m/s

Then its speed becomes 1500 m/s

so here since speed is decreased by large value

so here elasticity must decreased by large amount

A. a different kind of rock

B. an underground lake

C. an open cavity

D. a pile of dinosaur bones

So in above all options the density is changed or decreased by large value if the medium is changed to solid to liquid

So the correct answer will be

B. an underground lake 

alexira [117]4 years ago

4 0

Answer: An underground lake

Hope that helped :D

You might be interested in

What rule does static electricity follow

Luba_88 [7]

Static electricity works when there is an imbalance of positive and negative charges, and thus causes a little spark.

3 0

3 years ago

a spotted lizard runs at 3m/s at top speed. a girl wants to catch the lizard to keep as a pet. where should the girl place her c

Ann [662]

The acceleration due to earth's gravity is -9.8 m/s [dn] I thought... I'm assuming this is a projectile motion question asking for the range.

Break each kinematic quantity into their x and y components.

x y

v₁ = 3 m/s 0

v₂ = 3 m/s ?

Δd = ? -1.5

Δt = ? ?

a = 0 -10 m/s²

So the variable we are trying to find is Δdx (x component of displacement). We need to use a kinematic equation to do so. However, we obviously don't have enough given to find Δdx. This means we need to find something first, something we can use. How about Δt? Δt can be applied to both the x and y components. We have enough information in the y component list to find Δt. We can use this formula and solve for Δt.

Δdy = v₁y ( Δt ) + 1/2 ( ay ) ( Δt )²

Δdy = 1/2 ( ay ) ( Δt )² <- the first term cancels out since v₁y = 0.

2Δdy = ay ( Δt )²

2Δdy / ay = ( Δt )²

√ 2Δdy / ay = Δt

√ 2(-1.5 m/s) / -9.8 m/s² = Δt

√ -3.0 m/s / -9.8 m/s² = Δt

√ 0.306122449 s² = Δt

0.5532833352 s = Δt

Now, we can use this newly found quantity to solve for Δdx using the x component values using the appropriate kinematic equation.

Δdx = ( v₁x + v₂x / 2) ( Δt )

Δdx = ( ( 3.0 m/s + 3.0 m/s ) / 2 ) ( 0.5532833352 s )

Δdx = ( 6.0 m/s / 2 ) ( 0.5532833352 s )

Δdx = ( 3.0 m / s )( 0.5532833352 s )

Δdx = 1.659850006 m

Therefore, the girl should place her cage 1.7 m away from the platform to catch the lizard.

This solution assumes that the acceleration due to gravity is -10 m/s² [dn] and not -9.8 m/s² [dn]. If you need -9.8 m/s² [dn], then just substitute it into my solution. This was a pain to type lol

6 0

4 years ago

A triangle has angles measuring 10°, 40°, and 130° with side lengths of 23 cm, 84 cm, and 100 cm. Classify this triangle by its

Maurinko [17]

For the lengths of its sides
As a scalene triangle (from the Greek σκαληνός "unequal"), if all its sides have different lengths (in a scalene triangle there are no two angles having the same measure)
For its angles
obtuse triangle: if one of its interior angles is obtuse (greater than 90 °); the other two are acute (less than 90 °).

5 0

3 years ago

Differentiate between sound waves and seismic waves?

algol13

The only real difference is that common seismic waves travel through the ground and sound waves travel through the air. If you had a pipe attached to granite and you were listening to it, you might detect both.

7 0

3 years ago

Read 2 more answers

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

2 years ago