Answer:
E = 1.04*10⁻¹ N/C
Explanation:
Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:
![vf^{2} -vo^{2} = 2*a*x](https://tex.z-dn.net/?f=vf%5E%7B2%7D%20-vo%5E%7B2%7D%20%3D%202%2Aa%2Ax)
As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.
Replacing this values in the equation above, we can solve for a, as follows:
![a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bvo%5E%7B2%7D%2Amp%7D%7B2%2Ax%7D%20%3D%20%5Cfrac%7B%282.00e3m%2Fs%29%5E%7B2%7D%7D%7B2%2A0.2m%7D%20%3D%201e7%20m%2Fs2)
According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:
F = mp*a = q*E
For a proton, we have the following values:
mp = 1.67*10⁻²⁷ kg
q = e = 1.6*10⁻¹⁹ C
So, we can solve for E (in magnitude) , as follows:
![E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bmp%2Aa%7D%7Be%7D%20%3D%5Cfrac%7B1.67e-27kg%2A1e7m%2Fs2%7D%7B1.6e-19C%7D%20%3D%201.04e-1%20N%2FC)
⇒ E = 1.04*10⁻¹ N/C
Answer:
63N, 45N
Explanation:
Given:
Tension 1 = T1
We have the following free body equations:
F - T1 = 100*0.3 -(I)
T1 - T2 = 60*0.3 -(II)
T2 = 150*0.3 -(III)
Adding the three above equations, we have
F - T1 + T1 - T2 + T2 = (100*0.3) + (60*0.3) + (150*0.3)
Hence, F = 93N
Hence, substituting into equation I we have
T1 = Force by bar on lighter cart = 93 - (100*0.3) = 63N
Hence, force that heavier cart exerts on bar
T2 = 150*0.3 = 45N
Answer: The amplitude is 0. (assuming that the amplitude ot both initial waves is the same)
Explanation:
When two monochromatic light waves of the same wavelength and same amplitude undergo destructive interference, means that the peak of one of the waves coincides with the trough of the other, so the waves "cancel" each other in that point in space.
Then if two light waves undergo destructive interference, the amplitude of the resultant wave in that particular point is 0.
The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude
![W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N](https://tex.z-dn.net/?f=W%3Dmg%20%3D%20%282.5%20kg%29%289.81%20m%2Fs%5E2%29%3D24.5%20N)
the force of the wind F, acting horizontally, with intensity
![F=12 N](https://tex.z-dn.net/?f=F%3D12%20N)
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
![T \cos \alpha -F=0](https://tex.z-dn.net/?f=T%20%5Ccos%20%5Calpha%20-F%3D0)
![T \sin \alpha -W=](https://tex.z-dn.net/?f=T%20%5Csin%20%5Calpha%20-W%3D)
By dividing the second equation by the first one, we get
![\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04](https://tex.z-dn.net/?f=%5Ctan%20%5Calpha%20%3D%20%20%5Cfrac%7BW%7D%7BF%7D%3D%20%5Cfrac%7B24.5%20N%7D%7B12%20N%7D%3D2.04%20%20)
From which we find
![\alpha = 63.8 ^{\circ}](https://tex.z-dn.net/?f=%5Calpha%20%3D%2063.8%20%5E%7B%5Ccirc%7D)
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope: