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3 years ago

My brother john has never learned this lesson is that a sentence

Engineering

1 answer:

oee [108]3 years ago

6 0

Answer:

Yes

Explanation:

You just need to add a period behind the word "lesson", and make it My brother john has never learned this lesson.

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Saul is testing an installation and discovers a short circuit. What's causing this?

sergey [27]

A high voltage!! Hope this helps

6 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago

Two bodies have heat capacities (at constant volume) c, = a and c2 = bT and are thermally isolated from the rest of the universe

prisoha [69]

Answer:

Explanation:

The answer to the above question is given in attached files.

8 0

3 years ago

The first thing you are going to create is a logical one bit full adder in continuous assignment verilog. You can only use logic

IgorC [24]

Answer:

See Explaination

Explanation:

// use the `timescale directive which u have used in ur testbench here

module FA1(a,b,cin,s,cout);

input a,b,cin;

output s,cout;

wire s1,c1,c2;

assign s1= #4 a ^ b;

assign s= #4 s1 ^ cin;

assign c1= #2 a & b;

assign c2= #2 s1 & cin;

assign cout= #3 c1 | c2;

endmodule

5 0

3 years ago

A civil engineer is performing a tensile test on a brick using a tension machine. Immediately after he adds the first weight, th

ki77a [65]

A high elastic modulus of brick

7 0

3 years ago