Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.
Answer:
Changing oil.
Explanation:
You need to regularly check and change your car’s oil to ensure smooth running of the vehicle and to prolong the lifespan of its engine.
Answer:

Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

Where
is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

The horizontal and vertical distances are, respectively:


The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

Let's solve for 

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:


Or equivalently:

Given Vo=37 m/s and R=70 m


And

Answer:
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