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Basile [38]
3 years ago
9

Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below. How much time el

apses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately 1530 m/s.)
Physics
1 answer:
umka2103 [35]3 years ago
4 0

Answer:

Time, t = 0.104 seconds

Explanation:

Frequency of the click of the Dolphin, f = 55.3 kHz

A dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below, d = 80 m

The speed of sound in seawater is, v = 1530 m/s

Once the sound is send and reflects, the total distance covered by it is 2d such that,

t=\dfrac{2d}{v}\\\\t=\dfrac{2\times 80}{1530}\\\\t=0.104\ s

So, the time elapses before the dolphin hears the echoes of the clicks is 0.104 seconds.

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Answer:

pressure=force/area

=10/2

5

Explanation:

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an airplane is traveling at an altitude of 31,360. a box of supplies is driped from the cargo hold how long will it take to reac
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Answer: 1,600 seconds

Explanation:

31,360/9.8 = 3,200.

Then divide 3,200/2 = 1,600

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A Chef is looking for a new frying pan that will allow her to cook food quickly at temperatures above 500°C. Use the data provid
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Answer:

Explanation:

1st one

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2 years ago
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a
Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

p_i = m u = (8 kg)(2 m/s)=16 kg m/s

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

p_f = p_{fB}+p_{fS}

where p_{fS} is the momentum of the spring. For the conservation of momentum,

p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s


b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s


c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

\Delta p=F \Delta t

Since we know \Delta t=0.5 s, we can find the magnitude of the force:

F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.


d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

5 0
3 years ago
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