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kozerog [31]
3 years ago
8

What is the energy in joules of a mole of photons associated with red light of wavelength 7.00 × 102 nm?

Physics
1 answer:
konstantin123 [22]3 years ago
4 0
<span>The energy of a single photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength. Plugging the values in gives E = 6.63E-34 x 3.00E8 / 700E-9 = 2.84E-19 Joules Now one mole of substance is equivalent to 6.02E23 particles, so one mole of these photons will be: 2.84E-19 x 6.02E23 = 1.71E5 Joules</span>
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A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

\theta=274rev

Explanation:

From the question we are told that:

Angular velocity \omega=510rpm

Mass m=40.kg

Diameter d 75=>0.75m

Off Time t=40.0s

Oscillation at Power off N=210

Generally the equation for Angular displacement is mathematically given by

 \theta_{\infty}=\frac{w+w_0}{t}t

 w=\frac{2*\theta_{\infty}}{t}-w_0

 w=\frac{28210}{40*(\frac{1}{60})}-510

 w=120rpm

Generally the equation for Time to come to rest is mathematically given by

 t=(\frac{\omega_0}{\omega_0-\omega})t

 t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})

 t=0.87min

Therefore Angular displacement is

 \theta =(\frac{120+510}{2})0.87

 \theta=274rev

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A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

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Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

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Which three quantities can be used to calculate acceleration?
PtichkaEL [24]
D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
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