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3 years ago

Which is colder, 0°C or 20°F?

Physics

2 answers:

Allushta [10]3 years ago

8 0

20 degrees F because 0 degrees celcius is around 32 degrees F

Mashutka [201]3 years ago

7 0

In order to compare the two temperatures, we need to convert 20°F into Celsius. The formula we need to use is:

$T(^{\circ}C)=\frac{5}{9}(T(^{\circ}F)-32)$

Substituting 20°F, we find

$T(^{\circ}C)=\frac{5}{9}(20^{\circ}F-32) =-6.7^{\circ}C$

And this is less than 0°C, so the answer is

20°F is colder than 0°C.

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A type O star is likely to appear _____. yellow, blue, green or red

Murljashka [212]

A type O star is likely to appear blue.

5 0

3 years ago

Read 2 more answers

A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and

vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

$U=mgh$

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

$h=4-4cos(30\°)=0.53m$

Then, the potential energy is:

$U=(55kg)(9.8m/s^2)(0.53m)=285.67J$

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

$U'=(m_1+m_2)gh'=285.67\ J$

From the previous equation you can get h':

$h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m$

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

$h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°$

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0

3 years ago

From t=0 onwards what happens to the voltage across the inductor annd the curreent through the inductor relative to their values

gregori [183]

From t=0 onwards I changes slowly and V changes abruptly across the inductor.

At time t=0, the voltage across the inductor equalises the battery voltage; nevertheless, Lenz's Law states that this induced EMF will always be opposed to the polarity of the battery. The voltage across the inductor is equivalent to the voltage of a battery because the inductor at time zero behaves like a second battery of the same voltage linked in reverse.

Because current can never be zero, voltage across the inductor decreases with time. If it did, there would be no back EMF to stop the current from flowing through the inductor because the magnetic field would not be changing. As a result, the inductor will become less of an open circuit as the current increases over time. The inductor will essentially behave like a resistor.

Learn more about inductor here:

brainly.com/question/15893850

#SPJ4

3 0

1 year ago

The intensity of sunlight at the Earth is about 1367 W/m^2. How many times further away would the Sun have to be from the Earth

grandymaker [24]

The intensity of sunlight on the Earth is about 1367 W/m^2. The earth has to be 131 the distance farther away

This is further explained below.

<h3>What is the distance?</h3>

Generally, bet the power emitted by the sun is P. so. at the distance of earth intensely where $r_E$ is the distance of the earth from the sun. were $I_1=1362 \frac{w}{m^2}$

Therefore

p=4 $\pi 1367 r_E^2$

let at distance $\gamma r_E$$ is a positive number) the intensity is the same as the intensity at 10m away from a 100w bulb. So.

In conclusion,

$\begin{aligned}&\frac{P}{4 \pi\left(\gamma r_{\varepsilon}\right)^2}=\frac{100}{4 \pi 10^2}\\\\&\text { or, }\left(\gamma r_E\right)^2=P=4 \pi 1367 r_E^2\\\\&\therefore \gamma=\sqrt{4 \pi \times 1367}\\\\&=131.06\\\\ }\end{aligned}$