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Alla [95]
1 year ago
6

Of the following approximate conversions of celsius into fahrenheit, which is most accurate? a. 24°c almost-equals 75.9°f b. 2°c

almost-equals 35.8°f c. 34°c almost-equals 92.7°f d. 13°c almost-equals 55.0°f
Physics
1 answer:
Softa [21]1 year ago
4 0

The most accurate among the options is 13°c almost-equals 55.0°f, when you round down 55.4, you get 55.0

Hence, option D) is the most accurate.

<h3>What is Scale of Temperature?</h3>

Scale of temperature is simply a method used in calibrating the physical quantity of temperature in metrology.

Conversion of Celsius into Fahrenheit, Formula is expressed as;

Degree = (0 × 9/5) + 32 = Fahrenheit

First we check each option;

A) 24°c almost-equals 75.9°f

24°C = (24 × 9/5) + 32 = 75.2°F

B) 2°C almost-equals 35.8°f

2°C = (2 × 9/5) + 32 = 35.6°F

C) 34°C almost-equals 92.7°f

34°C = (34 × 9/5) + 32 = 93.2°F

D) 13°c almost-equals 55.0°f

13°C = (13 × 9/5) + 32 = 55.4°F

The most accurate among the options is 13°c almost-equals 55.0°f, when you round down 55.4, you get 55.0

Hence, option D) is the most accurate.

Learn more about temperature scales here: brainly.com/question/88395

#SPJ4

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Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
. Friction is a rubbing force that ___________ a spinning yo-yo.
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A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

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i hope i have been useful buddy.

good luck ♥️♥️

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If any part of a meteoroid survives the fall through the atmosphere and lands on Earth, it is called a meteorite.

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