Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as
![s_x=v_x_0 t](https://tex.z-dn.net/?f=s_x%3Dv_x_0%20t)
![t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bs_x%7D%7Bv_x%7D%5C%5Ct%3D%5Cfrac%7B2%20%5Ctimes%2010%5E%7B-2%7D%7D%7B1.1%20%5Ctimes%206%7D%5C%5Ct%3D1.82%20%5Ctimes%2010%5E%7B-8%7D%20s)
Now for the vertical distance
vy_o=0
than the equation of motion becomes
![s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2](https://tex.z-dn.net/?f=s_y%3Dv_y_0%20t%2B%5Cfrac%7B1%7D%7B2%7D%20at%5E2%5C%5Cs_y%3D%5Cfrac%7B1%7D%7B2%7D%20at%5E2%5C%5C0.5%20%5Ctimes%2010%5E%7B-2%7D%3D%5Cfrac%7B1%7D%7B2%7D%20a%281.82%20%5Ctimes%2010%5E%7B-8%7D%29%5E2%5C%5Ca%3D3.02%20%5Ctimes%2010%5E%7B13%7D%20m%2Fs%5E2)
Now using this acceleration the value of electric field is calculated as
![E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BF%7D%7Bq%7D%5C%5CE%3D%5Cfrac%7Bma%7D%7Bq%7D%5C%5CE%3D%5Cfrac%7Bm_ea%7D%7Bq_e%7D%5C%5C)
Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation
![E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.1%5Ctimes%2010%5E%7B-31%7D%20%5Ctimes%203.02%20%5Ctimes%2010%5E%7B13%7D%20%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D%5C%5CE%3D171.76%20N%2FC)
So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Answer:
1. The final velocity of the truck is 15 m/s
2. The distance travelled by the truck is 37.5 m
Explanation:
1. Determination of the final velocity
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Final velocity (v) =?
The final velocity of the truck can be obtained as follow:
v = u + at
v = 0 + (3 × 5)
v = 0 + 15
v = 15 m/s
Therefore, the final velocity of the truck is 15 m/s
2. Determination of the distance travelled
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Distance (s) =?
The distance travelled by the truck can be obtained as follow:
s = ut + ½at²
s = (0 × 5) + (½ × 3 × 5²)
s = 0 + (½ × 3 × 25)
s = 0 + 37.5
s = 37.5 m
Therefore, the distance travelled by the truck is 37.5 m
Answer:
5.791244495 KNm
Explanation:
The height h is given by,
Potential energy, PE is given by
PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as
and h is already given hence substituting 77 Kg for m we obtain
PE=21.6567095 KNm
We also know that Kinetic energy is given by
where v is the velocity and substituting v for 20.3 we obtain
KE=15.865465 KNm
Friction work is the difference between PE and KE hence
Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm
Answer:
33333.35 kg
Explanation:
I got it right on Acellus, rounded to 33300 sigfigs