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3 years ago

Red light has a wavelength of 500 nm. What is the frequency of red light?

Physics

1 answer:

snow_tiger [21]3 years ago

3 0

Answer:

$0.6 \times 10 {}^{15}$

Explanation:

By using relation,

Speed of light = frequency × wavelength (in m)

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A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh we need to know h

h = 50 sin 10 = 8.68 meters

then: PE = 20 * 9.81 * 8.68 = 1703.49 j

KE = 1/2 m v^2 = 1703 .49

v = 13 m/s

7 0

2 years ago

An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do

ipn [44]

Answer:

(a) ΔU=747J

(b) γ=1.3

Explanation:

For (a) change in internal energy

According to first law of thermodynamics the change in internal energy is given as

ΔU=Q-W

Substitute the given values

ΔU=970J-223J

ΔU=747J

For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

Where Cp is the molar heat capacity at constant pressure

Cv is the molar heat capacity at constant volume

To calculate γ we first need to find Cp and Cv

For Cp

As we know

Q=nCpΔT

Cp=(Q/nΔT)

$C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K$

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

$C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\$

γ=Cp/Cv

γ=[(37J/mol.K) / (28.687J/mol.K)]

γ=1.3

4 0

3 years ago

Which law states that for any closed loop, the sum of the length segments times the magnetic field in the direction of the segme

dedylja [7]

That's Ampere's Law. ( C ).
The magnetic permeability is the proportionality constant.

3 0

3 years ago

Read 2 more answers

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$