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3 years ago

How many Joules of energy are required to run a 100W light bulb for one day?

Physics

1 answer:

Vlada [557]3 years ago

4 0

The following are the answers to the questions presented:

a. The joules of energy required to run a 100W light bulb for one day is 8640000J
b. The amount of coals that has to be burned to light that light bulb for one day is 0.96kg

The solution would be like this for this specific problem:

P=W/s→W=Pt=100W1day 24h/1day 3600s/1h=8640000J

W=30/100wm→m=100W/30w=100×8640000J/30×30×10in thepowerof6 J/kg=0.96kg

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values

marishachu [46]

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

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If a ball is thrown horizontally with a speed of 65 mph, how far will it fall while traveling 90 ft of horizontal distance?

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Answer:

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Explanation:

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3 years ago

If the total time is 120 seconds and the total distance is 320 meters. Calculate

RUDIKE [14]

Answer:

S=D/T

320/120= 2.66 miles

Explanation:

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3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

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3 years ago

You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with

Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

Explanation:

For this exercise we use the constructor equation or Gaussian equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

m = $\frac{h'}{h}$ = - $\frac{q}{p}$

h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

$\frac{1}{q} = \frac{1}{f } - \frac{1}{p}$

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

h ’= h q / p

where p has a high value and is the same for all systems

h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

f = 12 cm h ’= fo 12 cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0

3 years ago