Answer:
Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?
Explanation:
solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).
Answer:
7 / 1
Explanation:
The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength
Ax / Ay = λx / λy = 1 / 7
λy / λx = 7 / 1
Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
Answer:
it is True as the operational definition of electric current.
Explanation:
The definition of electric current is
I = dQ / dt
By convention the direction of the current is the direction in which a positive charge flows.
The initial expression is the derivative that is the change of the load in the unit of time and this occurs in a given cross-sectional cable.
The proposed definition is the same as this, so it is True as the operational definition of electric current.
