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stira [4]
3 years ago
13

Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with

a frequency of 125 kHz. What is the wavelength of this sound, in meters?While remaining stationary, the dolphin emits a sound pulse and receives an echo after 0.220 s. How far away, in meters, is the reflecting object from the dolphin?
Physics
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

wavelength =  0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 *  0.220 s/2

x= 162.8 m

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Natali5045456 [20]

Answer:

True

Explanation:

With the increase in temperature hypothalamus fails and heatstroke occurs due to this failure. Hypothalamus is the region of our brain that act as a thermostat. It co-ordinates our physiological response to excessive heat. When the person’s temperature reaches to 104 degrees then it causes heatstroke. This heatstroke is very sudden and can kill person. Hence, we can conclude that when person’s temperature reaches to 104 degrees chances of survival decreases dramatically.

8 0
3 years ago
In avarage,How many times do a child breathe in a
ollegr [7]
On an approximate scale, A child breaths 20 times a minute as compared to only 12 to 16 in resting phase of an Adult.

So, In 60 minutes (1 hour), They breathe = 20 * 60 = 1200
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In short, Your Answer would be: 28,800

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5 0
3 years ago
Read 2 more answers
We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one
Len [333]

Answer:

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

3 0
3 years ago
Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2
Ugo [173]

Answer:

+1 m/s^2

Explanation:

Acceleration is given by

a=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in velocity

\Delta t is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

\Delta t = -3 -(-4) = 1 s

\Delta v = 2 - (1) = +1 m/s

So, the acceleration is

a=\frac{+1}{1}=+1 m/s^2

4 0
3 years ago
Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated
Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>
  • Gravitation is the force of attraction between any two bodies.
  • Every body in the universe attracts every other body with a force.
  • This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
  • Mathematically we can expressed it as,

                        F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2

<h3>How to solve the problem?</h3>
  • Here, we have given with the data's,

                      M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m

  • Thus, the force of attraction between these two bodies will be,

               F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

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6 0
1 year ago
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