All categories

3 years ago

Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong

Physics

1 answer:

sveta [45]3 years ago

8 0

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI

You might be interested in

Similarity between mass and charge

IrinaK [193]

I think there is only one.
<span>
</span>Both are conserved<span>.</span>

5 0

3 years ago

A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m

mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

5 0

2 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface: