All categories

2 years ago

2.

Physics

2 answers:

Zinaida [17]2 years ago

5 0

Answer:

c. Force is not necessary for maintaining motion

I believe this is the correct answer, hope this helps :)

Explanation:

enot [183]2 years ago

3 0

C is the right answer

You might be interested in

A ball is at rest at the top of a hill until a boy kicked it with his foot. What is the force that causes motion in this scenari

vazorg [7]

The boy’s foot causes the motion. His foot is the one that causes the ball to roll down the hill.

3 0

2 years ago

The ____ of an object is a measure of the amount of matter it contains. on the other hand ____ is a measure of the gravitational

Phoenix [80]

The Gravitational Force of an object is a measure of the amount of matter it contains. on the other hand __Matter__ is a measure of the gravitational force on an object. I hope it helps :)

7 0

3 years ago

The planet Mars has much less gravitational force that Earth. What would happen if you went to Mars?

vovangra [49]

gravitational force of planet exerted on its object near the surface is known as weight

so here we know that gravitational force of mars is much less than the gravitational force of Earth

So on the surface of mars the Weight of objects must be much less than the weight of object on surface of Earth

so here correct answer must be

D. Your weight would decrease.

6 0

3 years ago

How do you think festival dances can help you improve your fitness?

Irina18 [472]

Festival dances can help improve your fitness because your moving and exercising your body without even knowing for hours therefore it helps with weight loss and your health in general.

7 0

2 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal. 

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the frequency has an inverse relation with the wavelength

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago