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yawa3891 [41]
3 years ago
10

Wind is the result of ___ currents.

Physics
2 answers:
xz_007 [3.2K]3 years ago
8 0

Answer:

Convection Currents. Hope this helps:)

Crank3 years ago
6 0
The word is Conventional.
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An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?
Sunny_sXe [5.5K]
Answer is 6.84 approx
reason:-
            (2.78^2+6.25^2)^1/2=6.84 approx
4 0
3 years ago
Desde lo alto de un acantilado de 140 m, se lanza verticalmente un objeto hacia abajo con velocidad de 3m/s. Entonces la magnitu
grin007 [14]

Answer:

54

Explanation:

d = vo*t + ½*g*t²

d = 3*3 + ½*10*3²

d = 9 + 45

d = 54 m

entonces el objeto tiene 54 m de desplazamiento

7 0
3 years ago
Density is a measure of an object's _____ per unit _____.
ladessa [460]
<span>Density is a measure of an object's </span>mass per unit of volume

Which means that it shows how much mass is contained within a volume of something.
7 0
3 years ago
The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning
sammy [17]

Let the rise in temperature be 5^0C

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the  coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = 1.7*10^{-5}  ^0C^{-1}, and Δt = 5^0C

So expansion in length = 93*1.7*10^{-5}*5 = 0.007905 m = 0.79*10^{-3}m

So order of magnitude in change in length = -3


3 0
3 years ago
Read 2 more answers
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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