Answer:
For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive
Explanation:
Solution
Given that:
From the question stated, Anna drew a diagram to compare forces that are strong and weak.
Now,
We are to find which labels are grouped in areas marked as X, Y, Z respectively.
Thus,
For X, Y, Z it is marked as:
X: Always attractive or attractive only
Y: Very small range
Z: Repulsive and attractive
Answer:
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
Explanation:
Electrolytes are substances that, when are dissolved in solution, separates into electrical positive charges (cations) and electrical negative charges (anions) which are known as ions.
These ions have an adequate capacity to conduct particle charges and, therefore electricity.
Sodium, calcium, phosphate and potassium, are examples of electrolytes.
<u>Hence, the correct answer is:</u>
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
I hope it helps you!
Answer:
Same direction: t=234s; d=6.175Km
Opposite direction: t=27.53s; d=0.73Km
Explanation:
If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be
(<em>the train must see the car advancing at a lower speed</em>), where
is the speed of the automobile and
the speed of the train.
So we have
.
So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:
![t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BL%7D%7Bv_%7BAT%7D%7D%20%3D%20%5Cfrac%7B1.3Km%7D%7B20Km%2Fh%7D%20%3D%200.065h%3D234s)
And in that time the car would have traveled (<em>relative to the ground</em>):
![d=v_At=(95Km/h)(0.065h)=6.175Km](https://tex.z-dn.net/?f=d%3Dv_At%3D%2895Km%2Fh%29%280.065h%29%3D6.175Km)
If they are traveling in opposite directions, <u>we have to do all the same</u> but using
(<em>the train must see the car advancing at a faster speed</em>), so repeating the process:
![v_{AT}=(95km/h)+(75Km/h)=170Km/h](https://tex.z-dn.net/?f=v_%7BAT%7D%3D%2895km%2Fh%29%2B%2875Km%2Fh%29%3D170Km%2Fh)
![t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BL%7D%7Bv_%7BAT%7D%7D%20%3D%20%5Cfrac%7B1.3Km%7D%7B170Km%2Fh%7D%20%3D%200.00765h%3D27.53s)
![d=v_At=(95Km/h)(0.00765h)=0.73Km](https://tex.z-dn.net/?f=d%3Dv_At%3D%2895Km%2Fh%29%280.00765h%29%3D0.73Km)
Answer:
This reduces the average force applied during the landing process/ or you can say it reduces the impact your body takes.
Explanation:
Answer: 1.28 sec
Explanation:
Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:
V = ∆x / ∆t
Solving for ∆t, we have:
∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec