Improvements in muscular strength will not affect muscular endurance.

4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s

melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

$V_o=Vrms*\frac{1}{12}\\V_o=10Vrms$

the peak voltage before the bridge rectifier is given by:

$V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V$

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

$V_l=V_{op}-2(0.7)\\V_l=12.74V$

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

$V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V$

Using Ohm's law:

$I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA$

7 0

3 years ago

Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in

Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0

3 years ago

"Consider a 2-kg ball on the end of a 1-m long cord whose mass is negligible. The ball is swung in a horizontal circle and makes

qaws [65]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of

bekas [8.4K]

Answer:

The magnitude of the induced Emf is $0.003371V$

Explanation:

The width of the truck is given as 79inch but we need to convert to meter for consistency, then

The width= 79inch × 0.0254=2.0066 metres.

Now we can calculate the induced Emf using expresion below;

Then the $induced EMF= B L v$

Where B= magnetic field component

L= width

V= velocity

=(40*10^-6) × (42) × (2.0066)

=0.003371V

Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V

8 0

4 years ago

The materials that made up their solar nebula can be categorized into the four general type as follows. Rank these materials fro

Alchen [17]

Answer: Hydrogen and Helium > Hydrogen compounds > Rocks > Metals

Explanation: The solar nebular system contains 98% Light Gases of both Hydrogen and Helium (produced in the Big Bang); followed by 1.4% of Hydrogen Compounds including water, methane, and ammonia; then followed by 0.4% of Rocks like Silicates; and lastly 0.2% of Metals including Iron, Nickel and Aluminium.

4 0

3 years ago