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
aivan3 [116]
3 years ago
15

which of these changes does a submarine encounter as it returns from the bottom of the ocean to the surface of the ocean

Physics
1 answer:
fiasKO [112]3 years ago
7 0

a.the amount of sunlight increases.

Explanation:

As a submarine rises to the surface, the change it encounters that is true from the given options is that the amount of sunlight increases.

The bottom of the ocean is dark and receives little to no sunlight due to the scattering of the rays by ocean water.

  • As the submarine rises, the volume of water column on it decreases and the pressure on it decreases too.
  • Also, the temperature rises steadily to the surface.

learn more:

Heat and temperature brainly.com/question/914750

#learnwithBrainly

You might be interested in
What is the reaction force to a foot pushing down on the floor?
Helga [31]

Answer:

C. The floor pushing back against the foot​

Explanation:

5 0
2 years ago
Read 2 more answers
A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust
bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0
2 years ago
A bullet with a mass ????b=13.5mb=13.5 g is fired into a block of wood at velocity ????b=253vb=253 m/s. The block is attached to
Lorico [155]

Answer:

450 grams

Explanation:

Given:

mass of the bullet, m = 13.5 g = 0.0135 kg

velocity of the bullet, v = 253 m/s

spring constant of the spring, k = 205 N/m

Compression of the spring, x = 35.0 cm = 0.35 m

Now, the kinetic energy of the moving system (bullet + board) will tend to move the spring

thus,

let velocity of the system, V

now, applying the concept of conservation of momentum, we have

mv = (M + m)V

where,

M is the mass of the block

thus,

V = mv/(M + m)

now,

the kinetic energy of the system = (1/2)(M + m)V²

or

the kinetic energy of the system = (1/2)(M + m)(mv/(m + M))²

Energy gained by the spring = (1/2)kx²

now,

equating both the energies, we get

(1/2)(M + m)(mv/(m + M))² = (1/2)kx²

or

(mv)²/(m + M) = kx²

on substituting the values, we get

(0.0135 × 253)²/(0.0135 + M) = 205 × (0.35)²

or

11.66/(0.0135 + M) = 25.1125

or

M = 0.450 kg = 450 grams

8 0
3 years ago
What is the relationship between radio waves and the visible spectrum
evablogger [386]

Answer:

They both are part of electromagnetic radiation.

Radio waves have longer wavelength than visible waves.

Radio waves have lower frequency than visible waves.

Explanation:

8 0
2 years ago
A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?
andriy [413]

Answer:

v = 22.54 mph.

Explanation:

Given that,

Distance moved, d = 200 m

Time, t = 19.8 s

We need to find the runner's average speed.

We know that,

1 mile = 1609.34 m

200 m = 0.124 miles

19.8 seconds = 0.0055 h

So,

Speed = distance/time

v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph

So, the runner's average speed is 22.54 mph.

4 0
2 years ago
Other questions:
  • The rate at which work is done is also the definition of which of the following?
    7·1 answer
  • What units are used in the periodic table for atomic mass?
    14·2 answers
  • Computer science <br> Need answers please ^^
    8·1 answer
  • A race car travels a circular track at an average rate of 135 mi/hr. The radius of the track is 0.450 miles. What is the centrip
    7·2 answers
  • The movement of the plates of the Earth are responsible for exposing rocks to weathering.
    15·1 answer
  • Plants and trees grow nearly everywhere. this is one of the advantages of what energy?
    7·2 answers
  • a car traveling at 40m/s starts to decelerate steadily. it comes to a complete stop in 12 seconds. what is the acceleration?
    11·1 answer
  • A ball is rolling across the floor. Why does the ball come to a stop?
    9·2 answers
  • If the distance between two charged particles is doubled, the force between them changes by a factor of
    12·1 answer
  • On Earth, the gravitational field strength is 10 N/kg. Calculate E for a 4 kg bowling ball that is being
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!