The element in an incandescent lightbulb that releases light energy is

Fe + Br2 --> FeBr3

eimsori [14]

Explanation:

1
2
1
1
3
4
No
2Fe+3Br2-->2FeBr3

6 0

2 years ago

The following table lists the work functions of a few commonmetals, measured in electron volts.

steposvetlana [31]

Answer:

Lithium

Explanation:

The equation for the photoelectric effect is

$\frac{hc}{\lambda}= \phi + K_{max}$

where

$\frac{hc}{\lambda}$ is the energy of the incident photon, with

h being the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

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

$K_{max}$ is the maximum kinetic energy of the emitted photoelectrons

In this problem, we have

$\lambda= 190 nm = 1.9\cdot 10^{-7}m$ is the wavelength of the incident photon

$K_{max}=4.0 eV$ is the maximum kinetic energy of the electrons

First of all we can find the energy of the incident photon

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.90\cdot 10^{-7} m}=1.05\cdot 10^{-18} J$

Converting into electronvolts,

$E=\frac{1.05\cdot 10^{-18} J}{1.6\cdot 10^{-19} J/eV}=6.6 eV$

So now we can re-arrange the equation of the photoelectric effect to find the work function of the metal

$\phi = E-K_{max}=6.6 eV - 4.0 eV=2.6 eV$

So the metal is most likely Lithium, which has a work function of 2.5 eV.

3 0

3 years ago

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

2 years ago

When designing a user interface, the most important information should be placed in the ______ of the screen.

musickatia [10]

Answer:

upper-left corner

Explanation:

Most vital information are positioned in a place where users can view them clearly and without obstruction.

3 0

3 years ago

A hockey player uses her hockey stick to exert a force of 6.81 N on a stationary hockey puck. The hockey puck has a mass of 165

Anna007 [38]

Answer:

41.3 m/s^2 option (e)

Explanation:

force, F = 6.81 N

mass, m = 165 g = 0.165 kg

Let a be the acceleration of the puck.

Use newtons' second law

Force = mass x acceleration

6.81 = 0.165 x a

a = 41.27 m/s^2

a = 41.3 m/s^2

Thus, the acceleration of the puck is 41.3 m/s^2.

5 0

3 years ago