All categories

2 years ago

A blue-light photon has a wavelength of

1 answer:

8 0

Energy = hc/λ, where h is planck's constant, c is the speed of light and λ is the wavelength.

Energy = （6.626 x 10^-34*3.00 x 10^8)/（<span>4.80 × 10^−7）= 4.14*10^-19 J, which is 3.

Normally, when solving these equations, don't forget to check the units are correct (eg. wavelength is in meters) and dont forget to include the units in the calculation.</span>

You might be interested in

ASAP pls answer right if can’t see picture don’t answer

Anna35 [415]

Answer:

Not sure but

F = m* a

32= 5 * a

a= 6.4 m/s^2

6 0

3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

2 years ago

A truck is moving with a certain uniform velocity. It is accelerated uniformly by 0.75 m/s^2. After 20 seconds , the velocity be

scoundrel [369]

Answer:

Vi = 5 m/s

Explanation:

let (a) acceleration = 0.75 m/s²

(t) time = 20 seconds

Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)

find Initial velocity (Vi)

Vf - Vi

a = -----------

Vi = Vf - (a * t) = 20 - (0.75 * 20)

Vi = 5 m/s

5 0

3 years ago

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

4 0

3 years ago

Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

$a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*t+\frac{1}{2} *2*t^{2}$

$t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s$

Check

$t_{2}=8s$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*+\frac{1}{2} *2*8^{2}$

$x_{f}=-48+64\\x_{f}=16$

5 0

3 years ago