Answer:
The Answer is 0.019998c
Explanation:
Please see the attached Picture for the answer.
Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D-sin%28-%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%2Bsin%28%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
Since 
where;
L = length
Q = charge
λ = density of the charge;
then substituting
for λ, we have :
![E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%28%5Cfrac%7BQ%7D%7BL%7D%29%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
![E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2Q%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20E_oLR%7D)
substituting our given parameter; we have:
![E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%286.26%2A10%5E%7B-12%7DC%29%5Bsin%5Cfrac%7B2.59rad%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FN.m%5E2%29%280.235431%29%280.0909%29%7D)
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C
Answer:

Given:
Initial speed (u) = 22 m/s
Final speed (v) = 0 m/s (Rest)
Time taken (t) = 4 seconds
To Find:
Distance travelled by car (s)
Explanation:
From equation of motion of object moving with uniform acceleration in straight line we have:

By substituting value of v, u & t in the equation we get:


Distance travelled by car (s) = 44 m
it moves toward the truck because increased air movement between the car and the truck decreases pressure.
Hope this helped :) <3
Answer:
<h2>i. 9J and </h2><h2>ii. -9J</h2>
Explanation:
Please see attached the diagram and also the FBD for your reference.
Step one:
given data
Weight w1=20N
Distance s1=75cm= 0.75m
Weight w2=12N
Distance s2=75cm =0.75m
Clearly, we are asked to find the work done by gravity and the tension in the string.
we know that work is defined as force times the distance traveled
W=F*D
and to know the force we need the acceleration, seeing that the system is stationary the acceleration due to gravity 9.81m/s^2 is not needed here
i . Work done by gravity = w2*s2=12*0.75=9J
ii. The tension is equal but opposite = -9J