Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
Newton's third law of motion
Explanation:
Newton's third law of motion states that:
<em>"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force (reaction force) on object A"</em>
It is important to note that this law is always valid, even when it seems it is not.
Consider for example the gravitational force that the Earth exerts on your body (= your weight). We can say that this is the action force. It may seems that there is no reaction force in this case. However, this is not true: in fact, your body also exerts an equal and opposite force on the Earth, and this is the reaction force. The reason that explains why we don't notice any effect on Earth due to this force is that the mass of the Earth is much larger than your mass, therefore the acceleration produced on the Earth because of the force you apply is negligible.
It is also important to note that the action-reaction pair of forces always act on two different objects, so they never appear in the same free-body diagram.
Learn more about Newton's third law of motion:
brainly.com/question/11411375
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In order to determine the required force to stop the car, proceed as follow:
Calculate the deceleration of the car, by using the following formula:
![v^2=v^2_o-2ax](https://tex.z-dn.net/?f=v%5E2%3Dv%5E2_o-2ax)
where,
v: final speed = 0m/s (the car stops)
vo: initial speed = 36m/s
x: distance traveled = 980m
a: deceleration of the car= ?
Solve the equation above for a, replace the values of the other parameters and simplify:
![\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%3D%5Cfrac%7Bv%5E2_o-v%5E2%7D%7B2x%7D%20%5C%5C%20a%3D%5Cfrac%7B%2836%5Cfrac%7Bm%7D%7Bs%7D%29%5E2-%280%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%7D%7B2%28980m%29%7D%3D0.66%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%5Cend%7Bgathered%7D)
Next, consider that the formula for the force is:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
where,
m: mass of the car = 820 kg
a: deceleration of the car = 0.66m/s^2
Replace the previous values and simplify:
![F=(820kg)(0.66\frac{m}{s^2})=542.20N](https://tex.z-dn.net/?f=F%3D%28820kg%29%280.66%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%3D542.20N)
Hence, the required force to stop the car is 542.20N
<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2
hope this helps</span>