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
Leto [7]
3 years ago
8

What are some ways scientists can study the seafloor?

Physics
2 answers:
TiliK225 [7]3 years ago
4 0

Answer: A device records the time it takes sound waves to travel from the surface to the ocean floor and back again. Sound waves travel through water at a known speed. Once scientists know the travel time of the wave, they can calculate the distance to the ocean floor.

Explanation:

musickatia [10]3 years ago
3 0

Answer:

Sonar can be used to measure how deep the ocean is. A device records the time it takes sound waves to travel from the surface to the ocean floor and back again. Sound waves travel through water at a known speed. Once scientists know the travel time of the wave, they can calculate the distance to the ocean floor.

Explanation:

You might be interested in
What structure on the south african coast has a range of 63 km and releases flashes every 30 seconds?
I am Lyosha [343]

Answer:

Cape Point lighthouse

Explanation:

The lighthouse is located South of Cape Town, at a point where boats turn to go around the tip of Africa.

The new lighthouse has the most powerful light of all the lighthouses in South Africa, with a power of 10 megacandelas.  That's why it's being visible for up to 63 km around.

It's a very important lighthouse helping to conduct the traffic around dangerous waters where 2 oceans meet.

7 0
4 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 55.0° above the horizontal. It explodes on a mountainside 42.
GuDViN [60]

Answer:

x=7227

y=1678

(7227,1678)

Explanation:

Ok, check the picture I attached you, because of the problem don't give us aditional information, let's asume that the projectile is fired from an initial position x=0 and y=0. Now let's use projectile motion equations, but firs let's find the initial velocity components in x-axis and y-axis:

v_ox=v_o*cos(\theta_o)=300*cos(55)=172.0729309m/s

v_oy=v_o*sin(\theta_o)=300*sin(55)=245.7456133m/s

Now, let's find the x coordinate with this equation:

x-x_o=x-0=x=v_ox*t=172.0729309*42=7227.063098m

Finally asumming a gravity constant g=9.8, let's find the y coordinate with the next equation:

y-y_o=y-0=y=v_oy*t-\frac{1}{2}*g*t^{2} =(245.7456133*42)-\frac{42^{2} *9.8}{2}

y=10321.31576-8643.6=1677.71576m

8 0
3 years ago
Newton's First Law of Motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon
atroni [7]
This is sometimes called the Law of Inertia
3 0
3 years ago
A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the
Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = 9.40\times 10^{4}\ Pa

Pressure in the smaller pipe, P' = 2.80\times 10^{4}\ Pa

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'

18^{2}v = 9^v'

v' = 4v

Now,

By using Bernoulli's eqn:

P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'

where

h = h'

\rho = 10^{3}\ kg/m^{3}

9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}

6.6\times 10^{4}  = \frac{1}{2}\rho 15v^{2}

v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s

Now, the rate of flow is given by:

\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av

\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s

3 0
3 years ago
Read 2 more answers
A piece of gum becomes stuck upon a skateboard's wheel. What is the centripetal acceleration of the piece of gum if the wheel's
Vilka [71]

Answer:

a

Explanation:

<em>a_c</em><em> </em><em>=</em><em> </em><em>v</em><em>_t</em><em>^</em><em>2</em><em>/</em><em>r</em>

<em>a_c</em><em> </em><em>=</em><em> </em>(0.5)^2/0.03

<em>a_c</em><em> </em><em>=</em><em> </em>8.33 m/s^2

7 0
3 years ago
Other questions:
  • !!31 Points and BRAINLIEST
    10·2 answers
  • When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th
    7·1 answer
  • How are weak nuclear forces and gravity alike
    10·1 answer
  • how will you connect three resistors of 2 ohm centrioles and 5 ohms respectively so as obtain of the result and the resistance o
    8·1 answer
  • A 300g ball and a 100g ball are dropped from a tower. Both balls are the same size and their is no air resistance. Which one hit
    15·1 answer
  • The neurons of giant squids, for example, consist of axons with very large radii, which allows the squid to react very quickly w
    15·1 answer
  • a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.
    12·1 answer
  • Which sentence avoids using a pronoun reference error?
    5·2 answers
  • I NEED MAJOR HELP ON THIS AS WELL PLEASE SOMEONE HELP ME
    14·1 answer
  • An object is moving at a velocity of 23 m/s. It accelerates to a velocity of 85 m/s over a time of 2.0 s. What acceleration did
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!