Evaluate tan(249). O A. 1.36 B. 0.45 C. 0.41 D. 0.91

Surface currents are driven by _____. wind density salinity temperature

sweet-ann [11.9K]

The question says, what drives surface current. The correct option is A, that is wind. There are two different current system in the ocean, they are deep circulation and surface circulation. Surface current is majorly driven by the wind. The wind is capable of moving the top 400 meters of the ocean, thereby creating ocean surface current. The pattern of the surface current is determined by the direction of the wind, forces from the earth rotation and the position of the landform that interact with the current.

7 0

3 years ago

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A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t

QveST [7]

Explanation:

(a) First, we will calculate the number of moles as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

= $\frac{\text{100 g}}{4 g/mol}$

= 25 mol

According to the ideal gas equation,

PV = nRT

or, $(P_{2} - P_{1})V = nR (T_{2} - T_{1})$

$(6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T$

$\Delta T$ = 336.17 K

Hence, temperature change will be 336.17 K.

(b) The total amount of heat required for this process will be calculated as follows.

q = $mC \Delta T$

= $100 g \times 5.193 J/g K \times 336.17 K$

= 174573.081 J/K

or, = 174.57 kJ/K (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

3 0

3 years ago

Consider the electronic elements that are cooled by forced convection in Problem 6.31. The cooling system is designed and tested

Stella [2.4K]

Answer:

surface temperature of the chip located 120 mm Ts=42.5°C

surface temperature of the chip in Mexico Ts=46.9°C

Explanation:

from the energy balance equation we have to:

q=E=30W

from Newton´s law:

Ts=Tα+(q/(h*A)), where A=l^2

N=h/k=0.04*(Vl/V)^0.85*Pr^1/3

data given:

l=0.12 m

v=10 m/s

k=0.0269 W/(m*K)

Pr=0.703

Replacing:

h=0.04*(0.0269/0.12)*(10*0.12)/((16*69x10^-6))^0.85*(0.703^1/3) = 107 W/m^2*K

The surface temperature at sea level is equal to:

Ts=25+(30x10^-3/107*0.004^2)=42.5°C

h=0.04*(0.0269/0.12)*((10*0.12)/(21*81x10^-6))^0.85*(0.705^1/3)=85.32 W/(m*K)

the surface temperature at Mexico City is equal to:

Ts=25+(30x10^-3/85.32*0.004^2)=46.9°C

8 0

3 years ago

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

$2 n t = (m + \dfrac{1}{2})\lambda$

At m = 0 for the minimum thickness, so

$2\times 1.45 \times t = (0 + 0.5)\times 600$

$2.9 \times t =300$

$t =\dfrac{300}{2.9}$

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0

3 years ago

In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0

3 years ago