Answer:
μ ≈ 0.238
a ≈ 0.441 m/s²
Explanation:
a) F = ma (a = 0 constant velocity)
Fcosθ - μN = 0
μ = Fcosθ / N
μ = Fcosθ / (mg + Fsinθ)
μ = 310cos20 / (1120 + 310sin20)
μ = 0.2376007... ≈ 0.238
b)
Fcosθ - μN = ma
a = (Fcosθ - μN) / m
a = (Fcosθ - μ(mg - Fsinθ) / m
a = (310cos20 - 0.2376(1120 - 310sin20) / (1120/9.81)
a = 0.441308... ≈ 0.441 m/s²
Brass is an alloy, a combination of two metals, and aluminum is a pure metal
Answer:
![\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%203.6%20%5C%20ns%2F%20km%7D)
Explanation:
From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.
Hence, the material dispersion is ![\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )](https://tex.z-dn.net/?f=%5Cdfrac%20%7Bd%20%5Ctau%20_%7Bmat%7D%7D%7Bd%20%5Clambda%20%7D%20%5Csimeq%20%2880%20%5C%20ps%20%2F%20%28nm.km%29%20%5C%20%29)
Now, using the pulse spread formula:
![\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%20%5Cdfrac%7Bd%20%5Ctau%20_%7Bmat%7D%20%7D%7Bd%20%5Clambda%7D%20%5Csigma%20%5Clambda)
![\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%20%2880%20%5C%20ps%2F%20%28%20m.km%29%20%5C%20%29%20%20%5Ctimes%20%2845%20%5C%20nm%29)
Thus, the pulse spreading as a result of material dispersion is:![\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Csigma_%7Bmat%7D%7D%7BL%7D%20%3D%203.6%20%5C%20ns%2F%20km%7D)
Answer:
the car’s final displacement is 60 m
Explanation:
Given;
initail velocity of the car, u = 0
acceleration of the car, a = 0.8 m/s²
time of motion, t = 10 s
The first displacement of the car:
![x_1 = ut + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} (0.8)(10)^2\\\\x_1 = 40 \ m](https://tex.z-dn.net/?f=x_1%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2%5C%5C%5C%5Cx_1%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%280.8%29%2810%29%5E2%5C%5C%5C%5Cx_1%20%3D%2040%20%5C%20m)
The second displacement of the car;
acceleration, a = 0.4 m/s², time of motion, t = 10 s
![x_2 = ut + \frac{1}{2} at^2\\\\x_2 = 0 + \frac{1}{2} (0.4)(10)^2\\\\x_2 = 20 \ m](https://tex.z-dn.net/?f=x_2%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2%5C%5C%5C%5Cx_2%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%280.4%29%2810%29%5E2%5C%5C%5C%5Cx_2%20%3D%2020%20%5C%20m)
The final displacement of the car;
x = x₁ + x₂
x = 40 m + 20 m
x = 60 m
Therefore, the car’s final displacement is 60 m
Answer:
A: All of the above
Explanation:
The instantaneous speed of an object is simply the current seed of the object at any given time. The SI unit is m/S and it is a vector quantity.
Therefore, according to the given options, they all have SI units that are consistent with distance and time which makes them all an example of instantaneous speed.