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2 years ago

How does the Coriolis effect impact ocean currents in the Northern and Southern Hemispheres?

Physics

1 answer:

Strike441 [17]2 years ago

3 0

The Coriolis effect describes the pattern of deflection taken by objects not firmly connected to the ground as they travel long distances around the Earth.
The key to the Coriolis effect lies in Earth’s rotation. Specifically, Earth rotates faster at the Equator than it does at the poles. Earth is wider at the Equator, s

The Coriolis effect bends the direction of surface currents to the right in the Northern Hemisphere and left in the Southern Hemisphere.

Let’s pretend you’re standing at the Equator and you want to throw a ball to your friend in the middle of North America. If you throw the ball in a straight line, it will appear to land to the right of your friend because he’s moving slower and has not caught up.

Now let’s pretend you’re standing at the North Pole. When you throw the ball to your friend, it will again to appear to land to the right of him. But this time, it’s because he’s moving faster than you are and has moved ahead of the ball.

Everywhere you play global-scale "catch" in the Northern Hemisphere, the ball will deflect to the right.

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Light traveling from water to a gemstone strikes the surface at an angle of 80.0º and has an angle of refraction of 15.2º . (a)

7nadin3 [17]

Answer:

The the speed of light in the gemstone is $v= 0.599*10^8 m/s$

The unreasonable thing about this is that the speed of ligth in the gemstone is too low

the speed is 19% of the speed of light which is very low

One main unreasonable or inconsistent factor is that the assumption that the differnce between the angle of incidence and angle of refraction is very large

Explanation:

From the question we are told that

The angle of incidence $i = 80^o$

The angle of refraction $r = 15.2^o$

From Snell's law we have ,

$n_1 sin \theta_1 = n_2 sin \theta_2$

Where $n_1$ is the refractive index of the first medium (water) with a constant value of $n_1 = 1..333$

$n_2$ is the refractive index of the second medium (gem stone)

$\theta_1$ is the angle between the beam and perpendicular surface of the first medium

$\theta_2$ is the angle between the beam an the perpendicular surface of the second medium

Making the $n_2$ the subject of the formula

$n_2 = n_1 \frac{sin \theta_1}{sin \theta_2}$

$= (1.333)(\frac{sin (80.0)}{sin 15.2} )$

$= 5.007$

Generally refractive index of a material is mathematically represented

$n = \frac{c}{v}$

Where c is the speed light

v is the speed of light observed in a medium

Making v the subject

$v = \frac{c}{n}$

substituting value for gem stone

$v = \frac{3.0*10^8}{5.007}$

$v= 0.599*10^8 m/s$

7 0

3 years ago

this is how i made the map , i’m not sure where to put the symbols but can someone help me now answer the last few questions i p

N76 [4]

Answer:

hi my name is bill im here to help no problem 1+1=2

8 0

2 years ago

E) Thermal energy is released during

vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

$Q_{net,in} - W_{net, out} = \Delta U$ (1)

Where:

$Q_{net, in}$ - Net input heat, measured in joules.

$W_{net, out}$ - Net output work, measured in joules.

$\Delta U$ - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then $\Delta U < 0$ , which is caused by three scenarios:

(i) $Q_{net,in} < 0$ , $W_{net, out} < 0$ , $|Q_{net,in}|>|W_{net,out}|$

(ii) $Q_{net, in} > 0$ , $W_{net,out} > 0$ , $|Q_{net,in}|$

(iii) $Q_{net,in}< 0$ , $W_{net, out}>0$

In the case $Q_{net,in} > 0$ , $W_{net, out}$ , the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

$Q_{net, in} = \Delta U$ , $Q_{net, in} < 0$ (2)

If $Q_{net, in} < 0$ , then $\Delta U < 0$ . Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0

2 years ago

A gas is compressed in a cylinder from a volume of 20.0 l to 2.0 l by a constant pressure of 10.0 atm. calculate the amount of w

jok3333 [9.3K]

First, we need to convert the pressure in SI units. Keeping in mind that $1 atm = 1.01 \cdot 10^5 Pa$ :
$p=10.0 atm =1.01 \cdot 10^6 Pa$

The initial and final volume of the gas are (keeping in mind that $1.0 L = 0.001 m^3$ ):
$V_i = 20.0 L=0.020 m^3$
$V_f = 2.0 L=0.002 m^3$

So, the work done on the gas by the surrounding is
$W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J$
And the final positive sign means that this work corresponds to an increase in internal energy of the gas.

8 0

3 years ago

What type of infant temperament is generally associated with better adjustment in adulthood?

Rashid [163]

Saying no and not throwing fits and manners.

8 0

2 years ago