All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>
Answer:
The coefficient of performance for the cycle is 2.33.
Explanation:
Given that,
Output energy 
Work done 
We need to calculate the coefficient of performance
Using formula of the coefficient of performance
We need to calculate the 
Put the value into the formula
Now put the value of into the formula of COP
Hence, The coefficient of performance for the cycle is 2.33.
Answer: Use this F=Ma.
Explanation: So your answer will be
F=1 Kg+9.8 ms-2
So the answer will be
F=9.8N
How'd I do this?
I just used Newton's second law of motion.
I'll also put the derivation just in case.
Applied force α (Not its alpha, proportionality symbol) change in momentum
Δp α p final- p initial
Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)
or then
F α m(v-u)/t
So, as we know v=final velocity & u= initial velocity and v-u/t =a.
So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.
So, F=<em>k</em>ma (where k is the proportionality constant)
<em>k</em> is 1 so you can ignore it.
So, our equation becomes F=ma
Communication circuit <em>(D)</em> is becoming more common in residential electrical design and construction.
LAN Ethernet cables, outlets, and even hubs and bridges, are being built into the walls of new homes, along with the usual electrical outlet wiring, to give the owner the networking infrastructure and internet access that everybody needs now ... without stringing a mess of cables on the floor and through doors all over the house.
Answer:
b. Static > sliding > rolling friction.
Explanation:
Static friction is greater than sliding friction. It takes more force to get an object to start sliding than to keep it sliding.
Sliding friction is greater than rolling friction. There are fewer points of contact for a round surface compared to a flat one.
