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Anika [276]
2 years ago
8

A dolphin in seawater at a temperature of 25°C emits a sound wave directed toward the ocean floor 154 m below. How much time pas

ses before it hears an echo? (The speed of sound in ocean water is 1533 m/s.)
Physics
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

0.2s

Explanation:

SO for the dolphin to hear its echo, the sound wave must travel a distance twice as much as the displacement between the dolphin and the ocean floor. So d = 154 * 2 = 308 m

Since the speed of sound in ocean floor is v = 1533m/s we can find out the time by dividing the distance d by the speed of sound

t = d / v = 308 / 1533 = 0.2s

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Explain why the top of the loop cannot be the same height as (or higher than) the top of the first hill. Assume the roller coast
Ivahew [28]

Answer:

By conservation of energy, it can climb up to a height equal to that it went down before. However, due to the friction in the machines, the total mechanical energy of the roller coaster will decrease. As a result, the first "hill" of many roller coasters are the highest, but the followings will have decreasing heights.

Explanation:

7 0
3 years ago
Read 2 more answers
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
3 years ago
Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =
nekit [7.7K]

Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

\vec a=(ax,ay)\ ,\  \vec b=(bx,by)

The sum of the vectors is:

\vec a+\vec b=(ax+bx,ay+by)

The difference between the vectors is:

\vec a-\vec b=(ax-bx,ay-by)

The magnitude of \vec a is:

|\vec a|=\sqrt{ax^2+ay^2}

The angle \vec a makes with the horizontal positive direction is:

\displaystyle \tan\theta=\frac{ay}{ax}\\

The question provides the vectors:

\vec a=(2,6)

\vec b=(2,22)

\vec d=\vec a-\vec b

Calculate:

\vec d=(2,6)-(2,22)=(0,-16)

The magnitude of \vec d is:

|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16

The angle is calculated by:

\displaystyle \tan\theta=\frac{-16}{0}

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0
3 years ago
Fluorescence occurs when an object absorbs ____________________ and immediately releases the energy as visible light.
soldier1979 [14.2K]

Answer:

The answer should be light or other electromagnetic radiation

Explanation:

Such as x-rays or other things like that.

<em>-Hope This Helps!</em>

<em>-Justin:)</em>

5 0
2 years ago
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh
dem82 [27]

Answer:

Explanation:

a )

Reaction force of the ground

R = mg

= 160 N

Maximum friction force possible

= μ x R

= μ x 160

= .4 x 160

= 64 N .

b )

 160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,

Taking moment about top point of ladder

160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3

240 + 444 + 4f = 2700

f = 504 N

c )

Let x be the required distance.  

Taking moment about top point of ladder

160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4  = 900 x 3  ( .4 x 900 is the maximum friction possible )

240 + 444 x + 1440  = 2700

x = 2.3  m

so man can go upto 2.3 at which maximum friction acts .

8 0
3 years ago
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