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4 years ago

Where is visible light located on the electromagnetic spectrum and why?

1 answer:

5 0

Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic ... The different types of radiation are defined by the the amount of energy found in the photons.

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A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at

Liono4ka [1.6K]

Answer:

(a) the runner's kinetic energy at the given instant is 308 J

(b) the kinetic energy increased by a factor of 4.

Explanation:

Given;

mass of the runner, m = 64.1 kg

speed of the runner, u = 3.10 m/s

(a) the kinetic energy of the runner at this instant is calculated as;

$K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J$

(b) when the runner doubles his speed, his final kinetic energy is calculated as;

$K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J$

the change in the kinetic energy is calculated as;

$\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4$

Thus, the kinetic energy increased by a factor of 4.

4 0

3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

4 years ago

PHYSICS PLEASE HELP EASY!!<br> I put C and it was incorrect; I'm so confused :((

devlian [24]

Answer:

the answer is A i took this class and it was hard but once you know it you will be a master

Explanation:

8 0

3 years ago

GRAPH INCLUDED PLEASE HELP

swat32

Answer:

10 days

Explanation:

The half-life of a radioactive sample is the time taken for half of the sample to decay.

In the diagram, the half-life corresponds to the time after which the % of cobalt-57 has halved. We can observe the following:

At t=10 days, the % of Co remaining is approximately 45%

At t=20 days, the % of Co remaining is approximately 22%

This means that the sample of cobalt-57 has halved in 10 days, so the half-life of cobalt-57 is 10 days.

5 0

4 years ago

A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed

Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0

3 years ago