What is the common trait of all main-sequence stars?.

In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe

sweet-ann [11.9K]

Answer:

Mass of the oil drop, $m=3.01\times 10^{-15}\ kg$

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

$qE=mg$

$m=\dfrac{qE}{g}$

$q=6e$ , e is the charge on electron

E is the electric field, $E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m$

$m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}$

$m=3.01\times 10^{-15}\ kg$

So, the mass of the oil drop is $3.01\times 10^{-15}\ kg$ . Hence, this is the required solution.

5 0

2 years ago

A 900 kg car travelling at 12 m/s due east collides with a 600 kg car travelling at 24 m/s due north. As a result of the collisi

katrin [286]

Calculate force of each car:

P1 = 900 kg x 12m/s = 10,800

P2= 600 x 24 = 14,400

Degree of travel = arctan(14,300/10800)

Degree of travel = 53.1 N of E

8 0

2 years ago

A cyclist is traveling 10 m/s with an acceleration of 6 m/s2. How fast is the cyclist traveling at the end of 20 seconds

krok68 [10]

Answer:

<em>The cyclist is traveling at 130 m/s</em>

Explanation:

<u>Constant Acceleration Motion </u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at$

The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.

Apply the above equation:

$v_f=10+6\cdot 20$

$v_f=10+120$

$v_f=130\ m/s$

The cyclist is traveling at 130 m/s

7 0

2 years ago

On a road trip, a driver achieved an average speed of (48.0+A) km/h for the first 86.0 km and an average speed of (43.0-B) km/h

Bond [772]

A = 15, B = 1
48 + 15 = 63 for the first 86km
43 - 1 = 42 for the remaining 54 km
time = distance/speed
t1 = 86/63 = 1.3651hours
t2 = 54/42 = 1.2857hours
speed = distance/time
s = 140/2.6508
average speed = 52.8km/h

4 0

3 years ago

Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. If titanium is irradiated with light of

butalik [34]

Answer:

a) $1.59(10)^{-19} J$

b) $2.34(10)^{12} electrons$

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.

<u>This is what Einstein proposed: </u>

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

$E=\frac{hc}{\lambda}$ (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 $\Phi=6.94(10)^{-19} J$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a) Maximum possible kinetic energy of the emitted electrons ( $K$ )</h3>

From (1) we can know the energy of one photon of 233 nm light:

$E=\frac{hc}{\lambda}$

Where:

$h=6.63(10)^{-34}J.s$ is the Planck constant

$\lambda=233 (10)^{-9} m$ is the wavelength

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

$E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s}$ (3)

$E=8.53(10)^{-19} J$ (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

$8.53(10)^{-19} J=6.94(10)^{-19} J+K$ (5)

Finding $K$ :

$K=1.59(10)^{-19} J$ (5) This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is $2 \mu J=2(10)^{-6} J$ </h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

$\frac{E_{burst}}{E}$

Where $E_{burst}=2(10)^{-6} J$ is the energy of the burst of light

Hence:

$\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons$ This is the maximum number of electrons that can be freed by the burst of light.

4 0

2 years ago