Answer:
The average velocity of the sled is vavg = s/t.
Explanation:
Hi there!
The average velocity is calculated as the traveled distance over time:
vavg = Δx/Δt
Where:
vavg = average velocity.
Δx = traveled distance.
Δt = elapsed time.
We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:
vavg = s/t
Have a nice day!
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
a) for y=0.200m, r1=0.200m and r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.200m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D228391.8N%2FC)
b) for y=0.600m, r1=0.600m, r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.600m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D-59345.91N%2FC)
According to Einstein's special theory of relativity, the speed of the light in a vacuum is the same no matter the speed with which an observer travels. So the answer should be A) 0,1c (1/10 the speed of light)
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)
• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)
•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
