For Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter
a. it discards the frame without sending an error message to the network layer
b. it sends a NACK (not acknowledged frame) to the sending host
c. it delivers the frame to the network layer, and lets the network layer decide what to do
d. it discards the frame and sends an error message to the network layer
Answer:
Option A
Explanation:
The nodal address has to match the signal message address for it to function well but if the it doesn't match the nodal receiver address, it disregards it.
Answer:
Explanation:
The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

Where:

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since
, so:

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

Where in this case:

Therefore:

<u>Answer:</u>
2N/cm
<u>Step-by-step explanation:</u>
According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

where,
is the force which is stretching or compressing the spring,
is the spring constant; and
is the distance the spring is stretched.
Substituting the given values to find the elastic constant
to get:




Therefore, the elastic constant is 2 Newton/cm.
Answer:
a) 627840000 W
b) 0.9233
Explanation:
The rate at which energy flows through the dam for the generation of electricity is proportional to the volulme flow rate times the potential energy per unit volume.
Mathematically,
P = Q(ρgh)
Where, ρ is water density, g is accelaration due to gravity, h is height through which water falls over the turbine and Q is the discharge.
a) The power can be calculated as
P = 640(1000×9.81×221)
P = 627840000 W
b) The ratio = Calculated power / avarage power
= 62784×10^4/680×10^6
= 0.9233