What does flowing electrical charge produce?

Which of the following would be used in luminosity calculations?

Daniel [21]

Answer:

1) joule

2) $kgm^{2}/s^{2}$

3) $10\%$

Explanation:

1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second ( $\frac{J}{s}$ ), which is equal to Watts ( $W$ ).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work $W$ is expressed as force $F$ multiplied by the distane $d$ :

$W=F.d$

Where force has units of $kgm/s^{2}$ and distance units of $m$ .

If we input the units we will have:

$W=(kgm/s^{2})(m)$

$W=kgm^{2}/s^{2}$ This is 1Joule ( $1 J$ ) in the SI system, which is also equal to $1 Nm$

3) The formula to calculate the percent error is:

$\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%$

Where:

$V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2}$ is the experimental value

$V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2}$ is the accepted value

$\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%$

$\% error=10.04\% \approx 10\%$ This is the percent error

8 0

3 years ago

Electric fish generate current with biological cells called electroplaques, which are physiological emf devices. The electroplaq

Harlamova29_29 [7]

Answer:

current in water = 0.924 A

Explanation:

Let the current in each row be i.

Thus, current in water is contributed by each row and total current in water becomes 140i.

We are given;

emf of each electroplaque = 0.15 V

Number of electroplaques = 5000

internal resistance = 0.25 Ω

resistance = 800Ω

Applying Kirchoff's Voltage Law to row and water, we have;

5000E − (5000r)i − 800(140i) = 0

Rearranging;

5000E = (5000r)i + 800(140i)

Plugging in the relevant values;

5000 x 0.15 = i((5000 x 0.25) + 112,000)

750 = 113,250i

i = 750/113,250

i = 0.0066 A

Recall earlier, the current in water is 140i.

Thus, current in water = 140 x 0.0066

= 0.924

5 0

3 years ago

Which of the following is a physical property of wood

Black_prince [1.1K]

The density, hard, strong, and rough.

3 0

3 years ago

Read 2 more answers

Why bananas are curved

Vinil7 [7]

Because they built:different

6 0

3 years ago

Read 2 more answers

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

$O=(0,0)$

$A=(0,-b)$

$B=(h,0)$

We need to calculate the position vector of AB

$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

$u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}$

We need to calculate the force acting along the edge

$\hat{F}=F(u_{AB})$

$\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})$

We need to calculate the net moment

$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0

3 years ago