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Alisiya [41]
3 years ago
10

Are ocean currents always cold

Engineering
1 answer:
swat323 years ago
8 0

Answer:

The surface ocean currents have a strong effect on Earth's climate. ... However, these areas do not constantly get warmer and warmer, because the ocean currents and winds transport the heat from the lower latitudes near the equator to higher latitudes near the poles.

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A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender? select
Nataly_w [17]

Answer:

Option D

H_o equals 0.5 H_a less than 0.5

Explanation:

Information given

N=144

X=66

P=\frac {x}{n}=66/144= 0.458333\approx 0.46

The hypothesis

Null hypothesis

Probability, p =0.5

H_o: p =0.5

Alternative hypothesis

Probability, p<0.5

H_a: p

5 0
4 years ago
An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?
kifflom [539]

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = \frac{NE}{t}

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = \frac{NE}{t} = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}          [t = 1 s for per second]

or

550 × 10³ = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

7 0
3 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
3 years ago
You and your design group are competing for a multidisciplinary concept project. Your firm is the lead group in the design profe
sasho [114]

Answer: the answer for the question is a

Explanation:

4 0
3 years ago
How do I find v0 using Kirchhoff's laws and Ohm's law? The answer is 25V but I'm confused about how to get that.
telo118 [61]

Answer:

  25 V

Explanation:

It is convenient to use Kirchoff's current law (KCL), which tells you the sum of currents into a node is zero. The node of interest is the top left node.

The currents into it are ...

  20 mA + (-5 -Vo)/(2kΩ) -(Vo/(5kΩ)) = 0

  20 mA -2.5 mA = Vo(1/(2kΩ) +1/(5kΩ)) . . . . add the opposite of Vo terms

  (17.5 mA)(10/7 kΩ) = Vo = 25 . . . volts . . . . divide by the coefficient of Vo

_____

You will notice that the equation resolves to what you would get if you drew the Norton equivalent of the voltage source with its 2k impedance. You have two current sources, one of +20 mA, and one of -2.5 mA supplying current to a load of 2k║5k = (10/7)kΩ. KCL tells you the total current into the node is equal to the current through that load (out of the node).

5 0
3 years ago
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