Explanation:
dnndndndndndndndndnndmfnfnf
Answer: 1766.667 Ω = 1.767kΩ
Explanation:
V=iR
where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).
3mA = 0.003 A
Rearranging the equation, we get
R=V/i
Now we are solving for resistance. Plug in 0.003 A and 5.3 V.
R = 5.3 / 0.003
= 1766.6667 Ω
= 1.7666667 kΩ
The 6s are repeating so round off to whichever value you need for exactness.
Answer:
a) 254.6 GPa
b) 140.86 GPa
Explanation:
a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;
Ec(u) = EmVm + EpVp
To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,
Vm + Vp = 1
Vm = 1 - 0.63
Vm = 0.37
In the first equation,
Where
Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,
The modulus of elasticity upper-bound is,
Ec(u) = EmVm + EpVp
Ec(u) = (68 x 0.37) + (380 x 0.63)
Ec(u) = 254.6 GPa.
b) Considering the express of rule of mixtures for lower bound;
Ec(l) = (EmEp)/(VmEp + VpEm)
Substituting values into the equation,
Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)
Ec(l) = 25840/183.44
Ec(l) = 140.86 GPa
Answer:
b. spark plugs
Explanation:
Diesel engines are characterized in that the mechanism that activates the explosion of fuel is high pressures, so when the piston reaches the top of the cylinder, the pressure of the air-diessel mixture is so high that it causes the explosion, this It is what generates the power in a diesel engine.
