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2 years ago

15

An increase of 180 degrees in Fahrenheit is equal to an increase of how many degrees in celsius?

1 answer:

andriy [413]2 years ago

7 0

Answer:

it is 100 degree which is boiling point in water

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Why can't a hot air balloon go to higher layers?

Elena L [17]

If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.

5 0

2 years ago

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

notka56 [123]

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

4 0

2 years ago

STatiana [176]

15 degrees because a glass of water won't do anything to a bath tub of 15 degree water

4 0

2 years ago

An object at 27°C has its temperature increased to 37°C.

Firlakuza [10]

Answer:

b. 14

Explanation:

$T_{i}$ = Initial temperature = 27 °C = 27 + 273 = 300 K

$T_{f}$ = Final temperature = 37 °C = 37 + 273 = 310 K

$P_{i}$ = Initial Power radiated by the object

$P_{f}$ = Final Power radiated by the object

We know that the power radiated is directly proportional to fourth power of the temperature. hence

$\frac{P_{f}}{P_{i}} = \frac{T_{f}^{4} }{T_{i}^{4} }\\\frac{P_{f}}{P_{i}} = \frac{(310)^{4} }{(300)^{4} }\\\frac{P_{f}}{P_{i}} = 1.14\\P_{f} = (1.14) P_{i}$

Percentage increase in power is given as

$\frac{(P_{f} - P_{i})\times100}{P_{i}} \\\frac{((1.14) P_{i} - P_{i})\times100}{P_{i}} \\14$

3 0

3 years ago

What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

$V_{e}=\sqrt{\frac{2GM}{R}}$

Where:

$V_{e}$ is the escape velocity

$G=6.67(10)^{-11} Nm^{2}/kg^{2}$ is the Universal Gravitational constant

$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

$V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}$

Finally:

$V_{e}=3633.86 m/s \approx 3.63 km/s$

7 0

2 years ago