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Likurg_2 [28]
2 years ago
13

Describe how the boiling point of water on top of a mountain would be different from its boiling point at sea level.

Physics
1 answer:
Sonbull [250]2 years ago
7 0

Answer:

At elevated altitudes, any cooking that involves boiling or steaming generally requires compensation for lower temperatures because the boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure.

Explanation:

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3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?
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The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

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How can a cyclist minimize friction as he or she rides? A. by increasing the weight of the bike B. by increasing the tread on hi
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Compare the strengths of UV light and microwaves (in Hz). Which type of light is more powerful and how do you know?
TEA [102]

Answer:

UV light is more powerful as it has greater energy.

Explanation:

The energy propagated by electromagnetic waves ( light ) through vacuum or medium is known as electromagnetic radiation.

The frequency/wavelength range of electromagnetic radiation is known as electromagnetic spectrum. The electromagnetic spectrum ranging from gamma ray to radio waves.

Frequency range of UV light = ( 8 x 10¹⁴ to 3 x 10¹⁶ ) Hz

Frequency range of Microwaves = ( 300 x 10⁶ to 300 x 10⁹ ) Hz

Ratio of UV light to Microwaves = (\frac{8\times10^{14} }{300\times10^{6} } to \frac{3\times10^{16} }{300\times10^{9} } )

                                                     = ( 2.66 x 10⁶ to 1 x 10⁸ )

Energy of electromagnetic radiation is given by the relation:

E = hν

Here h is plank's constant and ν is frequency.

UV light is more powerful than Microwaves as frequency of UV light is greater than frequency of microwaves. Thus, by the above equation, the energy of UV light is more than energy of Microwaves.  

5 0
3 years ago
A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
3 years ago
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