1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
n200080 [17]
4 years ago
11

Sonar is used to locate objects underwater by measuring the time it takes an ultrasound sound pulse to reflect back to the recei

ver. An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic sound that travels 1530 m/s in seawater. Find the depth of the water if the time delay of the echo to the ocean floor and back is 8 seconds.
Physics
1 answer:
Romashka-Z-Leto [24]4 years ago
8 0

Answer:

The floor of the ocean is 6120 m deep.

Explanation:

In order to find the depth of the ocean we need to use the speed of the ultrasonic sound 1530 m/s and the time it takes for the echo to comeback. Since the wave is transmitted by the vessel goes to the bottom of the ocean and comeback, it travels the distance between the vessel and the floor two times, so we can divide the time by two. We then have:

D = V*t/2 = 1530*8/2 = 1530*4 = 6120 m

You might be interested in
I need help with this one <br><br> Coulombs law <br><br> 100 points and brainliest
Ganezh [65]

Answer:

my dad is a physics professor so when he comes home i will tell him to answer the question for you

3 0
3 years ago
Read 2 more answers
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0
3 years ago
Which exoplanet has the most Earth-like orbit?​
Digiron [165]

Answer: Kepler-452b

I think

hope it helps

4 0
3 years ago
Calculate the work done by a 47 n force pushing a 0.025 kg pencil 0.25 m against a force of 23.
inna [77]
Fnet=fa-fr
=47-23
=24n
W=F∆Xcos°
=24n×0.025cos0
=0.6J
8 0
3 years ago
If the inside of a container has a surface area of 20cm², what will be the pressure on each square inch of the container if 117.
egoroff_w [7]

Answer:

Explanation:

given :-

force applied = 117.6N

Area = 20 m^2

Pressure =?

solution:-

Pressure = Force / Area

= 117.6 N / 20 m^2

= 5.88 Pascals

Hope it helps :D!

Explanation:

6 0
2 years ago
Other questions:
  • 50 POINTS! A supplier wants to make a profit by buying metal by weight at one altitude and selling it at the same price per poun
    11·2 answers
  • During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)
    5·2 answers
  • A Car is travelling at 36ft/s change to miles per hour
    10·1 answer
  • after the car leaves the platform , gravity causes it to accelerate downward at a rate of 9.8 m/s2. what is the gravitational fo
    6·1 answer
  • A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc
    14·1 answer
  • A 214 g aluminum cup holds and is in thermal equilibrium with 892 g of water at 76°C. The combination of cup and water is cooled
    13·2 answers
  • Why is the mechanical advantage of using a single pulley always 1? Assume there’s no friction. a) The input force is in a differ
    13·2 answers
  • I really need some help with this. The question is in the photo
    7·2 answers
  • Uh... can you help me with this all? or you can only the one you know.​
    14·1 answer
  • A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!