Answer:
19.0 m/s
Explanation:
First of all, we can find the acceleration of the object by using Newton's second law:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
where
F = 200 N is the net force applied to the object
m = 50.0 kg is the mass of the object
a is the acceleration
Solving for a,
![a=\frac{F}{m}=\frac{200 N}{50.0 kg}=4.0 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7B200%20N%7D%7B50.0%20kg%7D%3D4.0%20m%2Fs%5E2)
Now we can find the final speed of the object:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where
u = 3.0 m/s is the initial speed
a = 4.0 m/s^2 is the acceleration
t = 4.0 s is the time
Substituting,
![v=3.0 m/s + (4.0 m/s^2)(4.0 s)=19.0 m/s](https://tex.z-dn.net/?f=v%3D3.0%20m%2Fs%20%2B%20%284.0%20m%2Fs%5E2%29%284.0%20s%29%3D19.0%20m%2Fs)
Answer:
A. ![KE_h=\frac{1}{16}\times KE_l](https://tex.z-dn.net/?f=KE_h%3D%5Cfrac%7B1%7D%7B16%7D%5Ctimes%20KE_l)
B. the lighter block travels at a speed 4 times faster than the heavier block.
C. b The heavy block must be pushed 4 times farther than the light block.
Explanation:
- mass of lighter block,
![m](https://tex.z-dn.net/?f=m)
- mass of the heavier block,
![4m](https://tex.z-dn.net/?f=4m)
Given that the blocks are acted upon by equal force.
A.
Then the kinetic energy of the blocks depends on their individual velocity.
<u>And velocity is related to momentum through Newton's second law of motion:</u>
<u />
<u />
<u />
<u />
considering that the time for which the force acts on each mass is equal.
![dv_l=\frac{F.dt}{m}](https://tex.z-dn.net/?f=dv_l%3D%5Cfrac%7BF.dt%7D%7Bm%7D)
For the heavier block:
![dv_{_h} =\frac{F.dt}{4m}](https://tex.z-dn.net/?f=dv_%7B_h%7D%20%3D%5Cfrac%7BF.dt%7D%7B4m%7D)
Therefore:
Kinetic energy of lighter block:
![KE_l=\frac{1}{2}\times m.(\frac{F.dt}{m} )^2](https://tex.z-dn.net/?f=KE_l%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20m.%28%5Cfrac%7BF.dt%7D%7Bm%7D%20%29%5E2)
![KE_l=\frac{1}{2m} \times (F.dt)^2](https://tex.z-dn.net/?f=KE_l%3D%5Cfrac%7B1%7D%7B2m%7D%20%5Ctimes%20%28F.dt%29%5E2)
Kinetic energy of heavier block:
![KE_h=\frac{1}{2} \times m.(\frac{F.dt}{4m} )^2](https://tex.z-dn.net/?f=KE_h%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20m.%28%5Cfrac%7BF.dt%7D%7B4m%7D%20%29%5E2)
![KE_h=\frac{1}{16}\times (\frac{1}{2m} \times (F.dt)^2)](https://tex.z-dn.net/?f=KE_h%3D%5Cfrac%7B1%7D%7B16%7D%5Ctimes%20%28%5Cfrac%7B1%7D%7B2m%7D%20%5Ctimes%20%28F.dt%29%5E2%29)
![KE_h=\frac{1}{16}\times KE_l](https://tex.z-dn.net/?f=KE_h%3D%5Cfrac%7B1%7D%7B16%7D%5Ctimes%20KE_l)
B.
From the above calculations and assumptions we observe that the lighter block travels at a speed 4 times faster than the heavier block.
C.
Since the lighter block is having the speed 4 times more than the heavier block so it must be pushed 4 times farther because the speed is directly proportional to the distance.
The speed of the proton is ![9.79*10^5m/s](https://tex.z-dn.net/?f=9.79%2A10%5E5m%2Fs)
Data;
- potential difference (A) = 3500V
- potential difference (B) = -1500V
<h3>Velocity of The Proton</h3>
The work done to move through a potential velocity 'v' is q
The potential difference 'v' is the difference between A and B
![V = 3500-(-1500) = 5000v](https://tex.z-dn.net/?f=V%20%3D%203500-%28-1500%29%20%3D%205000v)
But the work is converted into kinetic energy of proton.
![q_pV = \frac{1}{2}m_pV^2\\](https://tex.z-dn.net/?f=q_pV%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_pV%5E2%5C%5C)
Let's substitute the values and solve for the velocity
![(1.6*10^-^1^9)*(5000)=\frac{1}{2}(1.67*10^-^2^7)v^2\\v^2 = \frac{2* 1.6*10^-^1^9 * 5000}{1.67*10^-^2^7}\\ v = \sqrt{9.58*10^1^1} \\v = 9.79 * 10^5m/s](https://tex.z-dn.net/?f=%281.6%2A10%5E-%5E1%5E9%29%2A%285000%29%3D%5Cfrac%7B1%7D%7B2%7D%281.67%2A10%5E-%5E2%5E7%29v%5E2%5C%5Cv%5E2%20%3D%20%5Cfrac%7B2%2A%201.6%2A10%5E-%5E1%5E9%20%2A%205000%7D%7B1.67%2A10%5E-%5E2%5E7%7D%5C%5C%20v%20%3D%20%5Csqrt%7B9.58%2A10%5E1%5E1%7D%20%5C%5Cv%20%3D%209.79%20%2A%2010%5E5m%2Fs)
The speed of the proton is ![9.79*10^5m/s](https://tex.z-dn.net/?f=9.79%2A10%5E5m%2Fs)
Learn more about work done on an electron here;
brainly.com/question/13673636