Of the options presented in the problem, SAN would be the the exemption of the 4 which is not a network configuration.
LAN or Local Area Network, MAN or Metropolitan Area Network, and WAN or Wide Area Network are examples of a network configuration and are arranged in this solution by their capacity area if they were to be ranked. SAN or Storage Area Network only deals with storage devices and aren't that much connected with the other three presented in the problem.
Answer:
2.44 m/s due East
Explanation:
From the question given above, the following data were obtained:
Mass of 1st car (m₁) = 4 Kg
Velocity of 1st car (u₁) = 3 m/s
Mass of 2nd car (m₂) = 5 Kg
Velocity of 2nd car (u₂) = 2 m/s
Final velocity (v) =?
The final velocity can be obtained as follow:
v(m₁ + m₂) = m₁u₁ + m₂u₂
v(4 + 5) = (4×3) + (5×2)
9v = 12 + 10
9v = 22
Divide both side by 9
v = 22/9
v = 2.44 m/s
Thus, the final velocity is 2.44 m/s.
Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.
Answer:
"Offgassing"
Explanation:
According to my research on Kinesiology, I can say that based on the information provided within the question the process being described is known as "Offgassing". In other words this process is defined as when something gives off or releases a chemical, especially a harmful one, in the form of a gas into the air..
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
