1) Refraction
2)Reflection
3)Concave
4)Convex
I took the test and got this right so you can believe me :)
Hope this helps
C. The range of a projectile increases with an increase in the angle of launch.
The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body.
P = W = mass x acceleration due to gravity
P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N
Solving for the static friction force (F),
F = P x c
F = (2.94 N) x 0.6 = 1.794 N
Therefore, the maximum force of static friction is 1.794 N.
F=K*X,
F=M*a
M*a=K*X
2.5*9.81=K*0.0276
24.525=K*0.0276
24.525/0.0276=K
K= 888.6 N/m ---- force constant
assuming 2.5 refers to the new extension, just divide F/ 0.025
to get
981N/m
Answer:
a)![E=50.53\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D50.53%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The direction will be negative direction.
b)
The direction will be positive direction.
Explanation:
Given that
q1 = +7.7 µC is at x1 = +3.1 cm
q2 = -19 µC is at x2 = +8.9 cm
We know that electric filed due to a charge given as
![E=K\dfrac{q}{r^2}](https://tex.z-dn.net/?f=E%3DK%5Cdfrac%7Bq%7D%7Br%5E2%7D)
![E_1=K\dfrac{q_1}{r_1^2}](https://tex.z-dn.net/?f=E_1%3DK%5Cdfrac%7Bq_1%7D%7Br_1%5E2%7D)
![E_2=K\dfrac{q_2}{r_2^2}](https://tex.z-dn.net/?f=E_2%3DK%5Cdfrac%7Bq_2%7D%7Br_2%5E2%7D)
Now by putting the va;ues
a)
![E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C](https://tex.z-dn.net/?f=E_1%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B7.7%5Ctimes%2010%5E%7B-6%7D%7D%7B0.031%5E2%7D%5C%20N%2FC)
![E_1=72.11\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_1%3D72.11%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
![E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C](https://tex.z-dn.net/?f=E_2%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B19%5Ctimes%2010%5E%7B-6%7D%7D%7B0.089%5E2%7D%5C%20N%2FC)
![E_2=21.58\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_2%3D21.58%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The net electric field
![E=E_1-E_2](https://tex.z-dn.net/?f=E%3DE_1-E_2)
![E=50.53\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D50.53%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The direction will be negative direction.
As we know that electric filed line emerge from positive charge and concentrated at negative charge.
b)
Now
distance for charge 1 will become =5.5 - 3.1 = 2.4 cm
distance for charge 2 will become =8.9-5.5 = 3.4 cm
![E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C](https://tex.z-dn.net/?f=E_1%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B7.7%5Ctimes%2010%5E%7B-6%7D%7D%7B0.024%5E2%7D%5C%20N%2FC)
![E_1=120.3\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_1%3D120.3%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
![E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C](https://tex.z-dn.net/?f=E_2%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B19%5Ctimes%2010%5E%7B-6%7D%7D%7B0.034%5E2%7D%5C%20N%2FC)
![E_2=147.92\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_2%3D147.92%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The net electric field
![E=E_1+E_2](https://tex.z-dn.net/?f=E%3DE_1%2BE_2)
![E=268.22\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D268.22%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The direction will be positive direction.