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3 years ago

What happens to ocean water as it moves from Antarctica to the equator?

Physics

2 answers:

creativ13 [48]3 years ago

7 0

It becomes less dense and rises to the surface

IrinaVladis [17]3 years ago

3 0

Answer:

B. It becomes less dense and rises to the surface.

Explanation:

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Shah did not trust the results of an experiment that she had read about, so she is conducting the experiment herself. She goes t

Hitman42 [59]

The question is incomplete, the complete question is;

Shah did not trust the results of an experiment that she had read about, so she is conducting the experiment herself. She goes through the same set of steps and measures the effect of the amount of sugar on a single organism.

Which best describes what Shah is doing?

replication

repetition

both repetition and replication

neither repetition nor replication

Answer:

replication

Explanation:

Replication in scientific experiments refer to the process of going through all the steps in an experiment in order to obtain the same set of results initially obtained.

All scientific experiments is expected to be replicated under identical experimental conditions. If the results obtained by a scientific investigation cannot be replicated under identical circumstances by other scientists, then the scientific data must be taken with a pinch of salt.

Shah is expected to obtain almost exactly the same data as the original researchers who first published the data.

5 0

2 years ago

Read 2 more answers

Which process binds together sediment to form new rock?

vekshin1

Answer:

its cementation i got it correct

Explanation:

6 0

3 years ago

A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s

valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

m = 20kg

F = 50 Newtons (kg*m/sec^2)

acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

7 0

2 years ago

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

3 0

2 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

From the Fourier's law of conduction:

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

Now calculating the equivalent thermal resistance for conductivity using electrical analogy:

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

For plywood-Styrofoam interface from inside:

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&For Styrofoam-plywood interface from inside:

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

2 years ago