Which of the following drivers has the right-of-way?

Add a capacitor, C2 = 6 pF, in parallel with R3, and a capacitor, C3 = 2 pF in parallel with R4. Use PSpice to plot the magnitud

Ganezh [65]

Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember

6 0

3 years ago

Calculate the theoretical density of FCC iron (eg. austenitic stainless steel). The lattice parameter for FCC iron is 0.357 nm a

Ann [662]

Answer: 12.4 feet

Explanation:

If there is a smooth transition and there is no change in slopes, energy considerations can be used

The cube has a kinetic energy of

ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2

At the highest point when there is a gain in potential energy

pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

If there is no loss in energies,

pe = ke

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

h = 2000 /322 = 6.211 (ft)

= h / sin(30) = 12.4 ft

8 0

3 years ago

III. During January, at a location in Alaska winds at −27°C can be observed. However, several meters below ground the temperatur

Naya [18.7K]

Answer:

Not possible.

Explanation:

According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:

$\eta=1-\frac{T_{cold}}{T_{hot}}$

Where

$T_{hot}$ and $T_{cold}$ are temperature (in Kelvin) of heat source and heatsink respectively

In our case (I will be using K = 273+°C) :

$\eta=1-\frac{-27+273}{14+273}\\=0.1428$

In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.

6 0

3 years ago

A velocity field is given by V=(2x)i + (yt)j m/s, where x and y are in meters and t is in seconds. Find the equation of the stre

lesya692 [45]

Answer:

Equation of the streamline V = 4i - 4j m/s

Unit Vector n = (4i+4j)/4√2

Explanation:

Parameters

x=2 and y=-1

V=(2x)i + (yt)j m/s

Substituting (x=2) and (y=-1) into the velocity field V

Therefore V = 4i - tj where t=4s

Equation of the streamline

V = 4i - 4j m/s

Unit vector normal to the streamline n

Note V.n=0 and the velocity only have x- and y - components

Therefore

V.n=(4i - 4j).(nₓi+nyj)=0 or 4nₓ - 4ny = 0

The unit vector requires that nₓ^2+ny^2=1

Therefore n = (4i+4j)/4√2

3 0

3 years ago

Read 2 more answers

A particle moves along a straight line such that its acceleration is a=(4t^2-2) m/s, where t is in seconds. When t = 0, the part

klasskru [66]

Answer with Explanations:

We are given:

a(t)=4*t^2-2............................(1)

where t= time in seconds, and a(t) = acceleration as a function of time.

and

x(0)=-2 .................................(2)

x(2) = -20 ............................(3)

where x(t) = distance travelled as a function of time.

Need to find x(4).

Solution:

From (1), we express x(t) by integrating, twice.

velocity = v(t) = integral of (1) with respect to t

v(t) = 4t^3/3 - 2t + k1 ................(4)

where k1 is a constant, to be determined.

Integrate (4) to find the displacement x(t) = integral of (4).

x(t) = integral of v(t) with respect to t

= (t^4)/3 - t^2 + (k1)t + k2 .............(5) where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 ......................(6)

substitute (6) in (3)

x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20

16/3 -4 + 2k1 -2 = -20

2k1 = -20-16/3 +4 +2 = -58/3

=>

k1 = -29/3 ....................................(7)

Thus substituting (6) and (7) in (5), we get

x(t) = (t^4)/3 - t^2 - 29t/3 + 2 ..............(8)

which, by putting t=4 in (8)

x(4) = (4^4)/3 - (4^2 - 29*4/3 +2

= 86/3, or

= 28 2/3, or

= 28.67 (to two places of decimal)

4 0

3 years ago