Answer:
Inverse multiplexer or demultiplexer
Explanation:
An inverse multiplexer or demultiplexer behaves in an opposite way to the multiplexer. It collects data from a single source and breaks it down into several smaller outputs.
<em>A transponder performs the function of transmitting and receiving data in digital communication.</em>
<em>A multiplexer collects data from several analog or digital input signals and forwards them to a single output line</em>
<em>codec is used to decode and decode digital data streams</em>
<em>intelligent terminals are simply terminals that have processing power just like a smart keyboard.</em>
Answer:
0.1 mm
Explanation:
60 sheets = 6 mm
We want 1 sheet so we divide both sides by 60 :
1 sheet = 6/60 mm
1 sheet = 1/10 mm
1 sheet is 0.1 mm
Hope this helped and brainliest please
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
![v=15 m/s](https://tex.z-dn.net/?f=v%3D15%20m%2Fs)
- radius of the hill:
![r=100 m](https://tex.z-dn.net/?f=r%3D100%20m)
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,
![m \frac{v^2}{r}](https://tex.z-dn.net/?f=m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
, so we can write:
![mg-N=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg-N%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
(1)
By rearranging the equation and substituting the numbers, we find N:
![N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%28975%20kg%29%289.81%20m%2Fs%5E2%29-%28975%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D7371%20N%20%20)
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
![N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N](https://tex.z-dn.net/?f=N%3Dmg-m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%2862%20kg%29%289.81%20m%2Fs%5E2%29-%2862%20kg%29%20%5Cfrac%7B%2815%20m%2Fs%29%5E2%7D%7B100%20m%7D%3D469%20N%20)
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
![mg=m \frac{v^2}{r}](https://tex.z-dn.net/?f=mg%3Dm%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
from which we find
Answer:
C
Explanation:
There are two forces on the table: weight and normal force. Newton's second law:
∑F = ma
N - mg = 0
N = mg
N = (23.5 kg) (9.81 m/s²)
N = 230 N
Your answer is in the picture