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Akimi4 [234]
3 years ago
12

A power electronics package has been designed to handle 400W of power dissipation. Find out the change in output power handling

capability of the power electronics package when efficiency is increased from 89% to 94%. Discuss briefly the importance of efficiency base on output power capability calculated.
Physics
1 answer:
shtirl [24]3 years ago
6 0

Answer:

Explanation:

First we must find the real power that is dissipating for each of the two cases, this is achieved by multiplying the nominal power by the efficiency

P=Poη

P=power

Po=nominal power

η=efficiency

POWER NUMBER 1

P1=400*0.89=356W

POWER NUMBER 2

P2=400*0.94=376W

then we find the power change by subtracting both results

ΔP=P2-P1=376-356=20W

always ensure that the efficiency in any system is maximized, because it is always required that the amount of energy delivered by the system is maximum with the same input power.

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Answer:3

Explanation:

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Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.
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Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

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3 years ago
A chair of weight 70.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =
Annette [7]

Answer:

94.67 N

Explanation:

Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.

At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.  

\begin{array}{l}\\\Sigma {F_y} = 0\\\\N - W - F\sin \theta = 0\\\\N = W + F\sin \theta \\\end{array}

Substituting 70 N for W, 41 N for F and \theta for 37 degrees

N=70+41sin37=94.67441595  N and rounding off to 2 decimal places

N=94.67 N

​

6 0
3 years ago
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

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