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Aleks [24]
3 years ago
15

Change into60f into c​

Physics
1 answer:
Reil [10]3 years ago
8 0

Answer:

15.6 C

Explanation:

Here we are given a temperature in Fahrenheit (F). We want to convert it into Celsius degrees (C).

The conversion can be done by using the following equation:

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

where

T(C) is the temperature in Celsius degrees

T(F) is the temperature in Fahrenheit degrees

In this problem, we have

T(F) = 60 F

Therefore, by applying the equation,

T(C)=\frac{5}{9}(60-32)=15.6 C

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Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would
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Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

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2.The frost line in the solar nebula lies between Mars and Jupiter. It is the distance where it was cold enough for hydrogen compounds to condense into ices. Frost line: Explain how temperature differences led to the formation of two distinct types of planets.

3. The frost line is the point moving away from the Sun where it is cool enough for hydrogen compounds to freeze. Since the solar nebula was hotter near the center of the disk, hydrogen compounds such as water stayed gaseous in the inner solar system. Outside of the frost line, they froze.

4.The solar nebula flattened into a rotating disk. As gas became dense and hot, then it spins faster and pulled towards the center whereby the sun is formed. Solar nebula they collapse where the protostellar disk rotates. In the center of of nebula, there is a fusion begins and then sun is being formed

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4 0
3 years ago
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

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4 0
3 years ago
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at
Ghella [55]

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

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Let initial speed of Tyrone is u

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mu=(M+m)v

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irina [24]
A. Weathering I think
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