In a transistor, collector current is controlled by Base Collector Emitter

Which of the following statements about gravitational potential energy is not true?

nevsk [136]

Answer:

B and D

Explanation

Gravitational potential energy does not relate to graphs and it is about a moving object so b and d would not be correct

3 0

3 years ago

Car A is accelerating in the direction of its motion at the rate of 3 ft /sec2. Car B is rounding a curve of 440-ft radius at a

Katena32 [7]

Answer:

Incomplete question

Check attachment for the diagram of the problem.

Explanation:

The acceleration of the car A is given as

a=3ft/s²

Car B is rounding a curve of radius

r=440ft

Car B is moving at constant speed of Vb=30mi/hr.

Car A reach a speed of 45mi/hr

Note, 1 mile = 5280ft

And 1 hour= 3600s

Then

Va=45mi/hr=45×5280/3600

Va=66ft/s

Also,

Vb=30mi/hour=30×5280/3600

Vb=44ft/s

Now,

a. Let write the relative velocity of car B, relative to car A

Vb = Va + Vb/a

Then,

Using triangle rule, because vectors cannot be added automatically

Vb/a²= Vb²+Va²-2Va•VbCosθ

From the given graphical question the angle between Va and Vb is 60°.

Vb/a²=44²+66² - 2•44•66Cos60

Vb/a²=1936+ 4356 - 5808Cos60

Vb/a² = 3388

Vb/a = √3388

Vb/a = 58.21 ft/s

The direction is given as

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Va/SinA = Vb/SinB = (Vb/a)/SinC

66/SinA = 44/SinB = 58.21/Sin60

Then, to get B

44/SinB = 58.21/Sin60

44Sin60/58.21 = SinB

0.6546 = SinB

B=arcsin(0.6546)

B=40.89°

b. The acceleration of Car B due to Car A.

Let write the relative acceleration of car B, relative to car A.

Let Aa be acceleration of car A

Ab be the acceleration of car B.

Ab = Aa + Ab/a

Given the acceleration of car A

Aa=3ft/s²

Then to get the acceleration of car B, using the tangential acceleration formular

a = v²/r

Ab = Vb²/r

Ab = 44²/440

Ab = 4.4ft/s²

Using cosine rule again as above

Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ

Ab/a²= 3²+4.4²- 2•3•4.4•Cos30

Ab/a²= 9+19.36 - 22.863

Ab/a² = 5.497

Ab/a = √5.497

Ab/a = 2.34ft/s²

To get the direction using Sine rule again, as done above

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Aa/SinA = Ab/SinB = (Ab/a)/SinC

3/SinA = 4.4/SinB = 2.34/Sin30

Then, to get B

4.4/SinB = 2.34/Sin30

4.4Sin30/2.34 = SinB

0.9402 = SinB

B=arcsin(0.9402)

B=70.1°

Since B is obtuse, the other solution for Sine is given as

B= nπ - θ. , when n=1

B=180-70.1

B=109.92°

4 0

3 years ago

What is the maximum kinetic energy (in eV) of a photoelectron when a surface, whose work function is 5.0 eV, is illuminated by p

Aleksandr [31]

Answer: -1.898 ev (this means the photoelectric effect cannot occur)

Explanation:

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

If the light is 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

$E=h.\nu$ (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 surface.

In this case $\Phi=5 eV$

So, applying equation (1) in this problem:

$E=h \nu$

Where:

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

$\nu$ is the frequency

Now, the frequency has an inverse relation with the wavelength $\lambda=400(10)^{-9}m$ :

$\nu=\frac{c}{\lambda}$ (3) Being $c=3(10)^{8}m/s$ is the speed of light in vacuum

$\nu=\frac{3(10)^{8} m/s}{400(10)^{-9}m}$ (4)

$\nu=7.5(10)^{14} Hz$ (5)

Substituting (5) in (1):

$E=(4.136(10)^{-15}eV.s)(7.5(10)^{14} Hz)$ (6)

$E=3.102 eV$ (7)

Now, substituting (7) in (2):

$3.102 eV=5 eV+K$ (8)

Finding $K$ :

$K=-1.898 eV$ (9) Since a negative kinetic energy is not physically possible, the only explanation is: the conditions for a photoelectric effect were not met, hence the photoelectric effect cannot occur.

To understand it better:

The main condition for the occurrence of the photoelectric effect is that the energy incident photon $E$ must be greater than the work function $\Phi$

$E>\Phi$