Unpolarized light is passed through an optical filter that is oriented in the vertical direction. What is the polarization of th
1 answer:
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Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
t = 1.4 second
Answer:
The intensity of the net electric field will:
Explanation:
Here we need first find the electric field due to the first charge at the midway point.
The electric field equation is given by:
Where:
- k is Coulomb's constant
- q(1) is 20.00 μC or 20*10⁻⁶ C
- d is the distance from q1 to the midpoint (d=10.0 cm)
So, we will have:
The direction of E1 is to the right of the midpoint.
Now, the second electric field is:
The direction of E2 is to the right of the midpoint because the second charge is negative.
Finally, the intensity of the net electric field will:
I hope it helps you!
Answer:
a) 10.8 m
b) 24.3 m/s
Explanation:
a)
- In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
- The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
- We can find this distance simply applying the definition of average velocity, as follows:
- The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
- We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):
- where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
- Solving for Δx, we get:
- So, the total distance traveled was the sum of (1) and (3):
- Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
- Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.
b)
- We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
- The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
- ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
- Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:
- Δx₂, is the distance traveled while decelerating, and can be obtained using (2):
- Solving for Δx₂, we get:
- Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
- Taking the positive root in the quadratic formula, we get the following value for vomax:
- v₀max = 24.3 m/s.