Answer:
B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.
Explanation:
In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential
so according to the given situation, the option b is correct
And, the rest of the options are wrong
<span>The rule of inertia states that an object in motion will stay in motion unless another force has acted upon it. Because the person doesn't have their seatbelt on, they will keep moving. But if they were wearing a seatbelt, that would work as the force that is supposed to stop the person from flying forward.
Hope this helps :)
Please give brainliest</span>
Answer:
gfhvgtrtjrgvfjrrgfrfftuyrisnhdvfcgfridkjhsybvvtfvjfcgvwjfccegvghcvgrcgvrekgvrkgvkvvrvkvfgkerruuyti
To answer this question, you must remember the equation:
a²+b²= c²
(6.4)² + (12)²= (12.2)²
<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84
This cannot be a triangle
hope this helps
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)