Answer:
The tension on the string is 2.353 N.
Explanation:
Given;
the speed of sound in air, v₀ = 343 m/s
then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s
mass per unit length, m/l = μ = 0.002 kg/m
The speed of sound on the string is given as;
where;
T is the tension on the string
T = (34.3)²(0.002)
T = 2.353 N
Therefore, the tension on the string is 2.353 N.
Answer:
ΔV=0.484mV
Explanation:
The potential difference across the end of conductor that obeys Ohms law:
ΔV=IR
Where I is current
R is resistance
The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A
R=(pL)/A
Given data
Length L=3.87 cm =0.0387m
Diameter d=2.11 cm =0.0211 m
Current I=165 A
Resistivity of aluminum p=2.65×10⁻⁸ ohms
So
ΔV=IR
ΔV=0.484mV
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is
Clock time of computer B is
Effective CPI of computer A is
Effective CPI of computer B is
CPU time of A is
CPU time of B is
Hence Computer A is Faster by
Computer A is 1.41 times faster than the Computer B