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mel-nik [20]
3 years ago
11

A duck floating on a lake oscillates up and down

Physics
1 answer:
slega [8]3 years ago
8 0
<span>The answer is (2) 0.50 Hz. The frequency (f) of oscillation is the number of oscillations (n) per time (t) in seconds: f = n/t. A duck floating on a lake oscillates up and down 5.0 times (n = 5.0) during a 10.-second interval (t = 10.0 s). So, the frequency of duck's oscillations is: f = 5.0/10.0 s = 0.50 1/s = 0.50 Hz.Hope I helped! :) Cheers!</span>
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4 years ago
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Figure 15-1 shows how a(n) ____________________ breeze works.
Firlakuza [10]
<h2>Right answer: Sea breeze </h2>

The sea breeze is formed because during the day the surface of the land on the coast tends to warm up before and more than the surface of the sea. This difference in temperature between these two air masses means that on a sunny day the land warms up much more than the ocean causing a small area of ​​low pressure.

Then, the air rises as the land warms it and the colder air located on the surface of the sea forms a high pressure zone that makes this air mass tend to occupy the space left by the warmer air that has ascended on the coast. Therefore, the mass of air of a high pressure on the ocean always tends to move towards the zone of low pressure located on the coast.

It is important to note that the <u>sea breeze blows perpendicularly to the coast</u> and that the best breezes are formed in the spring and summer seasons because during the spring the water temperature is still cold and during the summer the sun produces high temperatures over the land in the coast.


<h2>So, <u>the greater the temperature contrast </u>between the land and the sea, <u>the greater the force of the wind generated</u>.</h2>
8 0
4 years ago
If the sun's mass is about average, how many stars are there in the milky way galaxy? the mass of the sun is of the order of 103
ra1l [238]
Just assume that the sun has the average mass of all the stars
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4 0
4 years ago
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Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce
ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

F=k\frac{q_{1}q_{2}  }{r^{2} }

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}  

4 0
4 years ago
10. The electron dot diagram for the element Ne would have ​
zmey [24]

Answer:

2,8

Explanation:

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7 0
3 years ago
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