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2 years ago

Water heating over a stove

Physics

2 answers:

enot [183]2 years ago

8 0

Answer:

Use a pot??

Explanation:

Leto [7]2 years ago

8 0

ahh yes

Y. E. S. W A. T. E. R. A. N. D. S T. ON. V. E

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By what angle should the second polarized sheet be rotated relative to the first to reduce the transmitted intensity to one-half

algol [13]

Answer:

θ = 45º

Explanation:

The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz

I = I₀ cos² θ

in the problem they ask us

I = ½ I₀

let's look for the angles

½ I₀ = I₀ cos² θ

cos θ = √ ½ = 0.707

θ = cos 0.707

θ = 45º

6 0

2 years ago

put the following in order of a full lunar month: full moon, new moon, 1st quarter, waxing gibbous, waning gibbous, waxing cresc

ASHA 777 [7]

The same cycle of phases repeats over and over and over and over and over again, so I could start the list with whatever phase I want, and build the list until I get to the same phase I started with.

But the cultures that track their months by the phases of the Moon (traditional Muslim, traditional Jewish, traditional Chinese) all start the new month with the New Moon, so I guess I'll start my list there too.

-- New Moon

-- Waxing Crescent

-- First Quarter

-- Waxing Gibbous

-- Full Moon

-- Waning Gibbous

-- Third Quarter

-- Waning Crescent

-- next New Moon

4 0

2 years ago

A closely wound rectangular coil of 80 turns has dimensions of 25.0 cm by 40.0 cm. The plane of the coil is rotated from a posit

strojnjashka [21]

Answer:

The average induced emf is 50.88 V

Explanation:

Given;

number of turns, N = 80 turns

angle between the plane of the coil and magnetic field, θ = 43°

strength of magnetic field, B = 1.40 T

time to make 80 turns = 0.0700 s

$Emf_{avg} = \frac{N \delta \phi}{dt} = N(\frac{BACos \theta {_f}-BACos \theta {_i}}{dt} )\\\\Emf_{avg} = \frac{NBA(Cos \theta {_f}-Cos \theta {_i})}{dt}$

where;

A is the area of the coil = 0.25 x 0.4 = 0.1 m²

If the plane makes an angle of 43° with a magnetic field to a position perpendicular to the field, then the initial angle θi = 90 - 43 = 47° and the final angle θf with the field = 0

$Emf_{avg} = \frac{NBA(Cos \theta {_f}-Cos \theta {_i})}{dt}\\\\Emf_{avg} = \frac{80*1.4*0.1(Cos(0)-Cos(47))}{0.07} = 50.88 \ V$