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3 years ago

Wladimir Köppen’s climate system was developed _____.

Physics

1 answer:

son4ous [18]3 years ago

5 0

Answer:

From 1846 to 1940 it would be about 1950

Explanation:

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Object 1 and 2 attract each other with a electrostatic force of 36.0 units. If the charge of object 1 is changed to four times t

Mrac [35]

The answer would be 6

8 0

2 years ago

An alternator provides 20 A of peak current at 100 V peak AC to a resistive electric heater. How much heating power is delivered

Liono4ka [1.6K]

-- The resistance of the heater is (volts/current) = 5 ohms

-- The heating (RMS) value of a sinusoidal AC is V(peak)/√2 . For this particular alternator, V(peak)=100V, so the heating (RMS) equivalent is 70.71 V.

-- The heating power delivered to the electric heater is (E²/R).

Power = (100/√2)² / 5

Power = 5,000 / 5

<u>Power = 1,000 watts </u>

6 0

3 years ago

Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer

Reptile [31]

Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

$I' = I_{0}Cos^{2}\theta$ .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

$I'' = I'Cos^{2}\left ( 90-\theta \right )$

So,

$I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )$

$I'' = I_{0}Cos^{2}\theta Sin^{2}\theta$

$I'' = \frac{I_{0}}{4}Sin^{2}2\theta$

(b)

Now differentiate with respect to θ.

$I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta$

$I'' = \frac{I_{0}}{2}\times Sin 4\theta$

7 0

3 years ago

A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki

Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

$\tau=I*\alpha$

given

$\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?$

$\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2$

the angular velocity is

$\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s$

$KE= 1/2*I* \omega^2$

$KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J$

4 0

2 years ago

Test populations are studied. Population one is found to obey the differential equation dy1/da=o.2y1 and the population two obey

oksano4ka [1.4K]

Answer:

Population 1 indicates growth while Population 2 indicates a declining population

Explanation:

Here, using the given rate of change of the population, we want to determine which of the two is growing and which is declining

From the rate of change of both, we can determine this. Looking at the differential equation for the first one, we can see that it is of positive value. Looking at the differential equation for the second one. we can see it is of negative value

While a positive change rate indicates growth, a negative change rate will indicate otherwise

Hence, we can conclude that the one with a negative rate change will indicate a declining population

8 0

3 years ago