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2 years ago

13

Describe the shape of the light waves that would be created from a single uncovered light bulb

2 answers:

aliya0001 [1]2 years ago

4 0

The shape of the wave will be like that of a sine curve, and it will repeat periodically over a time interval.
Its wavelength will be between 400 nm and 700 nm.

dem82 [27]2 years ago

3 0

The light emitted by the bulb is a visible light spectrum where a monochromatic ray is subdivided into multiple rays that have different colors.

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A student stands on a scale, and the scale reads 85 N. What is being measured by the scale

Dmitry [639]

Force. Force is measured by newtons, probably because he came up with the equations to calculate force.

Hope this helps!

5 0

3 years ago

the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.

asambeis [7]

$The\ area\ of\ a\ circle\ = \pi r^2 \\ 725\ cm^2 = 3.14r^2 \\
230.57 cm^2 = r^2 \\ \sqrt{230.57\ cm^2} = r \\ \boxed{r = 15.18\ cm}$

7 0

2 years ago

A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

$\frac{3}{(r + a)^2} = \frac{2}{r^2}$

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

6 0

3 years ago

In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7

jasenka [17]

Explanation:

Given that,

Current of clothes dryer, I = 16 A

Voltage, V = 240 V

Time, t = 45 min = 2700 s

Current of personal computer, I' = 2.7 A

Voltage, V' = 120 C

Energy used by clothes dryer is given by :

$E=I^2Rt\\\\E=IVt\\\\E=16\times 240\times 2700\\E=1.03\times 10^7\ J$

Let t' is the time for this computer to "surf" the internet. Again using formula of energy used as :

$E=I'V't'\\\\t'=\dfrac{E}{I'V'}\\\\t'=\dfrac{1.03\times 10^7}{2.7\times 120}\\\\t'=31790.12\ s\\\\t'=8.83\ h$

So, for 8.83 hours you could use this computer to "surf" the internet.

6 0

2 years ago

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘C and 1.00 atm. The molar mass of oxygen is 32

VashaNatasha [74]

1) 5765 mol

First of all, we need to find the volume of the gas, which corresponds to the volume of the room:

$V=7.00 m\cdot 8.00 m\cdot 2.50 m=140 m^3$

Now we can fidn the number of moles of the gas by using the ideal gas equation:

$pV=nRT$

where

$p=1.00 atm=1.01\cdot 10^5 Pa$ is the gas pressure

$V=140 m^3$ is the gas volume

n is the number of moles

R is the gas constant

$T=22.0^{\circ}+273=295 K$ is the gas temperature

Solving for n,

$n=\frac{pV}{RT}=\frac{(1.01\cdot 10^5 Pa)(140 m^3)}{(8.314 J/molK)(295 K)}=5765 mol$

2) 184 kg

The mass of one mole is equal to the molar mass of the oxygen:

$M_m = 32.0 g/mol$

so if we have n moles, the mass of the n moles will be given by

$m : n = 32.0 g/mol : 1 mol$

since n = 5765 mol, we find

$m=\frac{(5765 mol)(32.0 g/mol)}{1 mol}=1.84\cdot 10^5 g=184 kg$

3 0

3 years ago