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gregori [183]
3 years ago
11

Which gas in earths atmosphere protects earth by reflecting ultraviolet radiation

Physics
1 answer:
lana66690 [7]3 years ago
4 0
It's the ozone layer (O3)
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Hi, I was wondering what is the use of a temperature sensor since the sensor wouldn't be fully immersed in the hydrogen.
777dan777 [17]

Answer:

Within our homes, temperature sensors are used in many electrical appliances, from our refrigerators and freezers to help regulate and maintain cold temperatures as well as within stoves and ovens to ensure that they heat to the required levels for cooking, air confectioners/heaters.

Explanation:

A temperature sensor is a device used to measure temperature. This can be air temperature, liquid temperature or the temperature of solid matter. There are different types of temperature sensors available and they each use different technologies and principles to take the temperature measurement.

6 0
3 years ago
A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac
Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

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What unit is used for the value of G in Newton's Law of Universal<br> Gravitation?<br> *
Helga [31]

Answer:

In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.

Explanation:

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3 years ago
The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
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