Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = 
Q' = + 42 nC = 
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
Now,
Electric field at the mid-point due to charge Q' is given by:
Now,
The net Electric field is given by:
Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
The photon energy is the light frequency times Planck's constant. ... Planck's constant is usually called "h". Its value is about 6.626*10^-34 Joules/ Hz. A Joule is a unit of energy, and a Hertz is a unit of frequency, so when you multiply a frequency by h you get an energy.
Answer:
1) The medium b has higher optical density.
2) If the optical densities of both A and B are same.
Explanation:
- The refractive index of a material determines the number of times the velocity of light in that material is slower than the velocity of light in a vacuum.
- The higher refractive index means slower propagation of the light in that medium.
- The refractive index of the medium determines the optical density of the medium.
- If the material has a higher refractive index, then its optical density is high.
- If the material has a low refractive index, then its optical density is low.
- The light bends towards the normal while refracting through the medium from a lower optical density medium to the higher optical density medium.
- If both the mediums has the same optical density, the direction of propagation doesn't change.
