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olga2289 [7]
2 years ago
12

The temperature of a body of water influences _____

Physics
2 answers:
Dvinal [7]2 years ago
8 0

Answer

The temperature of a body of water influences<u> by surrounding temprature or the temprature of air which blows upon it.</u>

<u>Explanation</u>

If the temperature of air which blow upon the surface of water is high that time there will be transfer of heat from air to surface water due to which temperature of water body will rise. Similarly if the temperature of air which blow upon the surface of water is low that time there will be transfer of heat from  water bodies to surrounding air due to which temperature of water body will fall.

Maslowich2 years ago
5 0
The temperature of the air above it
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The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug
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Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0
3 years ago
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What is the difference between regional metamorphism and contact metamorphism?
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A current runs from the left to the right in a long, straight wire. How does the magnetic field at point X compare with the magn
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2 years ago
A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu
ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb

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3 years ago
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