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3 years ago

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Calculate the wavelength λ and the frequency f of the photons that have an energy of E photon = 2.32 × 10 − 19 J. Ephoton=2.32×1

0−19 J. Use c = 3.00 × 10 8 m/s c=3.00×108 m/s for the speed of light in a vacuum.

1 answer:

klasskru [66]3 years ago

4 0

Answer:

Frequency =3.5×10^14hertz

Wavelength = 8.6×20^-7m

Explanation:

Energy of photon is expressed as;

E = hf where;

h is the Planck constant

f is the frequency of the photons

From the formula f = E/h

Given E = 2.32 × 10^ − 19 J

h = 6.63×10^-34Js

f = 2.32×10^-19/6.63×10^-34

f = 0.35×10^(-19+34)

f = 0.35 × 10^15

f = 3.5×10^14hertz

To get the wavelength, we will use the relationship:

f = c/λ

λ = c/f

Given c = 3×10^8m/s

f = 3.5×10^14hertz

λ = 3×10^8/3.5×10^14

λ = 0.86×10^-6m

λ = 8.6×20^-7m

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Answer:

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Explanation:

Power: This can be defined as the rate at which work is done or energy is used up.

The expression for power is given as,

P = E/t

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Where P = power, W = Work, t = time, m = mass, h = height, g = acceleration due to gravity.

Hans' power

P = mgh/t

Given: m = 100 kg, h = 2 m, g = 9.8 m/s², t = 3 s

Substitute into equation 1

P = 100(9.8)(2)/3

P = 653.33 W.

Frans' power

P' = mgh/t

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P' = 9800/20

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from the above,

since P>P'

Hence, Hans is more powerful

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3 years ago

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2 years ago

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

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Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

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3 years ago

regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

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3 years ago