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3 years ago

What function do the wind-driven currents play in the transport of heat energy in the oceans?

1 answer:

7 0

Wind driven currents create in the equatorial and warm zones of the earth’s oceans. The currents then transmit the warm water north, where the heat in the water is misplaced to the colder atmosphere. In the instance of the Gulf Stream, this transmission of heat to the air warms up England and Europe.
As the water cools, its' density upsurges because of chilling and increasing salinity. This cold water drops to the bottom of the ocean, where it is then stimulated by another current back south, similar to a train creating a big circuit.

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Light traveling through air encounter a second medium which slows the light to 2.7 x 10^8. What is the index of the medium?

kozerog [31]

Answer:

1.11

Explanation:

The index of the medium can be calculated using below formula

V= c/ n ............eqn(1)

Where V= velocity of the light is reduced to while traveling through the second medium= 2.7 x 10^8 m/s

n= index of the medium

c= speed of light= 3 x 10^8 m/s

Substitute for the values in eqn(1)

2.7 x 10^8 = (3 x 10^8 m/s)/ n

Making " n" subject of the formula, we have

n= (3 x 10^8 )/(2.7 x 10^8)

n= 1.11

Hence, the index of the medium is 1.11

5 0

3 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0

3 years ago

The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr

Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V= $A_1V_1+ A_2V_2+A_3V_3$

where suffix 1,2,3 refers to the three pipes.

=27L/min+16L/min+11 L/min

=54L/min

b) A x V=54L/min => $\frac{\pi }{4} d^2$ x v

d= 2 cm

$\frac{\pi }{4} d^2$ x v = 54

v= $\frac{4}{\pi }$ x $\frac{54}{2^2}$

-> $A_1$ x $V_1$ =27L/min => $\frac{\pi }{4} d_1^2$ x $v_1$

$d_1$ = 1.3cm

$\frac{\pi }{4} d^2$ x $v_1$ = 27

$v_1$ = $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$

Next is to find the ratio of speed i.e $\frac{v}{v_1}$

$\frac{4}{\pi }$ x $\frac{54}{2^2}$ / $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$ => $\frac{54}{27}$ $\frac{1.3^2}{2^2}$

$\frac{v}{v_1}$ = 0.845

8 0

3 years ago

A vector has an x component of 25.0 units and a y

sammy [17]

Here is the answer to your question

8 0

3 years ago

What happened to the maximum height of consecutive swings

GarryVolchara [31]

Answer:

we need more info

Explanation:

3 0

3 years ago