9. In Millikan's oil drop experiment an oil
drop is held stationary by p.d) of 400 V,
if another drop of double the radius bu
carries same charge is to be held
stationary, the p.d) required is
a) 800 V
b) 1600 V
c) 3200 V
d) 400 V
The strona​

Answers

Answer 1
From the figure, F
e

and F
g

represents electric and gravitational forces respectively.
We can observe that,

V∝
q
m



and in turn m∝r
3

given V = 400V is initial voltage and let V
1


substituting these values in the respective equations we get

V
V
1



=
r
3
(q)
(2r)
3
q



solving the above eqaution will give us
V
1

=8×V=3200 V
So, the answer is option (C).

Related Questions

A nuclear power plant is being built to generate electricity for a large
metropolitan area. Which process will it use?
O A. Nuclear decay of uranium-238
B. Chemical decay of wood
C. Nuclear fission of uranium-235
O D. Combustion of fossil fuels

Answers

Answer:

C. Nuclear fission of uranium-235

Answer:

C. Nuclear fission of uranium-235

Explanation:

Give thanks to above me

Question 2: Start-Up
Now you will use another Tracker experiment to examine the displacement, velocity, and acceleration of a car in the x (horizontal) direction. To
begin your investigation, open the Tracker Car Start-up experiment.
Part A
Play () the video. At the end, rewind (H) and step forward (►) one frame ata time to observe the step-by-step changes in position. Based on
your observations, describe the car's displacement (distance and direction from the starting place) over time.

Answers

Answer:

The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.

Explanation:

Answer:  The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.

Explanation: edmentum sample answer

difference between incident ray and refracted ray ​

Answers

Answer:

** incident ray.

Incident ray - the ray of light falling on the surface AB is called the incident ray

reflected ray.

** Reflected ray - the incident ray bouncing back in the same medium after striking the reflecting surface is called reflected ray.

Two ice skater are at rest, andy and brenda. andy has a mass of 62.5 kg. they push off each other. after, andy moves 1.59 m/s east, while brenda moves 2.22 m/s west. what is brenda's mass?
Unit=kg

Answers

Answer: 44.76 kg

Explanation: There are two options for formulas to use here.

0=(m1)(V1f)+(m2)(V2f)

Or

(m1)(V1f) = —(m2)(V2f)

Both formulas should give you an answer of 44.763514, which can be rounded to the specifications of whatever your curriculum or teacher wants. Most likely 44.76.

Blessings to all of you in the Name of Jesus Christ our Savior!

What is albedo? What on Earth has high albedo?

Answers

Answer:

The fraction of incident light or radiation reflected by a surface or body, commonly expressed as a percentage.

2.

biology The whitish inner portion of the rind of citrus fruits that is a source of pectin, commonly referred to as the pith.

Explanation:

Given a force of 100N and an acceleration of 5 m/s2 , what is the mass?

Answers

Answer:

20kg

Explanation:

Given parameters:

Force  = 100N

Acceleration  = 5m/s²

Unknown:

Mass  = ?

Solution:

According to Newton's second law of motion:

 Force  = mass x acceleration;

 So;

          Mass  = [tex]\frac{force }{acceleration}[/tex]  

          Mass  = [tex]\frac{100}{5}[/tex]   = 20kg

in what condition do acceleration and deaccelration

Answers

Im not sure what are you asking about. But in physic/ mechanic usually when the object against gravity, its deceleration but when the object follow the gravity its acceleration

Also usually when the object move to the left side the speed will be negative (-) this is probably because of deceleration? since deceleration = -acceleration.

During the pitching motion, a baseball pitcher exerted an average horizontal force of 90 N against the 0.1 kg baseball while moving it through a horizontal displacement of 2.0 m before release. (1) what was the amount of work performed by the pitcher on the baseball (2) If the velocity at the start of the pitching motion was zero, at release the ball was traveling horizontally at which velocity

Answers

Answer:

(1) 180 J

(2) 60 m/s

Explanation:

(1) From the question,

Amount of work performed by the pitcher on the baseball = Force × distance.

W = F×d............... Equation 1

Given: F = 90 N,  and d = 2.0 m.

Substitute into equation 1

W = 90×2

W = 180 Joules.

(2)

F = ma........... Equation 2

Where a = acceleration of the ball.

a = F/m

Given: F = 90 N, m = 0.1 kg.

therefore,

a = 90/0.1

a = 900 m/s².

Using,

v² = u² + 2as ............ Equation 3

Where u = 0 m/s, a = 900 m/s², s  = 2 m.

substitute into equation 3

v² = 0² + 2×900×2

v² = 3600

v = √3600

v = 60 m/s

Determine the wavelength of a wave with a frequency of 100hz and a velocity of 5m/s.

Answers

0.05 I think not sure

We are given:

Frequency of the wave = 100 Hz

Velocity of the wave = 5 m/s

Finding the Wavelength:

We know the relation between the wavelength and frequency is:

u = νλ    [where ν is the frequency, u is the speed and λ is the wavelength]

5 = 100*(λ)

λ = 5/100         [dividing both sides by 100]

λ = 1/20

λ = 0.05 m

Hence, the wavelength is 0.05 m

what belongs in the center section

Answers

Answer:

The second one I think

Explanation:

B

what is force?answer in one line.​

Answers

any interaction that, when unopposed, will change the motion of an object

A block of wood of length L = 21.0 cm, width w = 9.53 cm, and height h = 5.92 cm is just barely immersed in water by placing a mass m on the top of the block. The density of the wood is rho = 0.390 g/cm^3. The value of m is:______

a. 0.72 kg
b. 1.2 kg
c. 1.6 kg
d. 7.1 kg
e. 0.36 kg

Answers

Answer:

0.462kg

Explanation:

Density = Mass/Volume

Given

Density of the wood = 0.390 g/cm^3

Volume of the wood = Length * Width * Height

Volume = 21 * 9.53 * 5.92

Volume = 1,184.7696cm³

Get the mass m;

mass = density * volume

mass = 0.390 * 1,184.7696

mass = 462.060144g

The mass in kg is 0.462kg

A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have in order to hit the wall with the same magnitude of momentum as the ball?

Answers

Answer:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

The ball momentum is

[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]

As there is a similar momentum

So,

[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

The arrow strikes a deer in the woods with the speed of 55 m/sec at an angle of 315 degrees. Calculate the Horizontal and vertical components of the arrow’s velocity.

Answers

Answer:

100 m

Explanation:

In which reaction are the atoms of elements rearranged?
A. A neutron fuses with hydrogen-1 to form hydrogen-2.
B. Uranium-235 breaks apart into cesium, rubidium, and two
neutrons.
C. A neutron fuses with uranium-238 to form uranium-239.
D. Hydrogen combines with oxygen to form water.

Answers

Answer:

D. Hydrogen combines with oxygen to form water.

Explanation:

Hydrogen combining with oxygen to form water is a typical example of chemical reaction.

During a chemical reaction, atoms of elements are rearranged. Most chemical reactions obey the law of conservation of mass which states that "matter is neither created nor destroyed in a chemical reaction but atoms are simply rearranged".

The other choices given are nuclear reactions. In such reactions, atoms are not rearranged but are simply destroyed and made in the process.

Answer:

D.) Hydrogen combines

Explanation:

BE HAPPY

a car accelerates from rest at 2 m/s. what is the speed after 8 sec?

Answers

Answer:

16m/s

Explanation:

[tex]v_{f}=v_{i}+at[/tex]

[tex]v_{f}=0+2\cdot8[/tex]

[tex]v_{f}=16\ \frac{m}{s}[/tex]

Therefore,  the speed after 8 seconds is 16m/s

Through which material would you expect sound waves to move fastest? O A. Iron O B. Water O c. Air O D. Milk​

Answers

Answer:

Solids

Is this what your looking for, It might tell you the answer?

A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power. How do their rates of photon emission compare

Answers

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

During an isothermal process one mole of a monoatomic gas did 3000 J of work on its surrounding. The final volume and pressure of the gas are 25 L and 1 atm, respectively. What was the initial volume of the gas

Answers

Answer:  The initial volume of the gas is 7.72 L

Explanation:

For an isothermal process the temperature is constant.  

[tex]PV=nRT[/tex]

as P = pressure = 1 atm ,

V = Volume =  25 L

n = moles

R= gas constant

T = temperature

[tex]PV=1atm\times 25L[/tex]

[tex]nRT=25Latm=25\times 101.3J=2532.5J[/tex]     (1Latm=101.3 J)

For isothermal reaction :

[tex]w=-2.303nRT\log\frac{V_2}{V_1}[/tex]

where , w = work done by  system = -ve

n = moles = 1

[tex]V_2[/tex] = final volume = 25 L

[tex]V_1[/tex] = initial volume = ?

[tex]-3000J=-2.303\times 2532.5\log \frac{25}{V_1}[/tex]

[tex]V_1=7.72L[/tex]

Thus initial volume of the gas is 7.72 L

charge of one of electron is 1.6022×10*-19 coulomb. what is the total charge on 1 mole of electrons​

Answers

There are

[tex]6.023 \times {10}^{23} [/tex]

electrons in a mole. So charge of 1 mole electron is

[tex]6 \times {10}^{23} \times ( - 1.6022) \times {10 }^{ - 19} [/tex]

Can we use a hydrometer to
measure the density of milk?​

Answers

Answer:

yes

Explanation:

i hope this helps not sure im right

a 0.8 ,^3 insulated rigid tank contains 1.54 kg of carbon dioxide at 100 kPa. Now paddle wheel work is done on the system until pressure in the tank rises to 135 kpa assuming the ideal gas model and negligible kinetic and potential energy effects determine the paddle wheel work dine during the process and the energychange during this process

Answers

Answer:

The answer is "[tex]W= 100.44 \ KJ\ \ \ \W_{win}=7.23 \ KJ[/tex]"

Explanation:

Please find the complete question in the attached file.

From of the ideal gas relation that initial and the last temperatures were determined:  

[tex]T_1 = \frac{P_1 V}{m R}[/tex]

    [tex]= \frac{100 \times 0.8}{1.54 \times 0.1889} \\\\ = 275 \ K[/tex]

[tex]T_2 = \frac{P_2 V}{m R}[/tex]

    [tex]= \frac{135 \times 0.8}{1.54 \times 0.1889} \\\\ = 317 \ K[/tex]

In the initial and final states, the internal energies for given temperatures are described from A-20 by means of intelmpolation and divided by the carlxon molar mass.  

[tex]u_1 = 141.56 \frac{KJ}{kg}\\\\u_2 = 206.78 \frac{KJ}{kg}[/tex]

The real job is just the difference between internal energies:  

[tex]W = m(u_2 - u_1) \\\\[/tex]

    [tex]= 1.54(206.78 -141.56) \ kJ \\\\ =100.44 \ kJ[/tex]

In the initial and final states, the zero entries are as determined as internal energies:

[tex]S_1^{\circ} =4.788 \frac{KJ}{kg K}\\\\S_2^{\circ} =5.0478 \frac{KJ}{kg K}[/tex]

From its energy increase, the minimum work required is determined:

[tex]W_{min} = m(u_2-u_1 - T_0(s_2 -S_1))\\\\=W-mT_0(S_2^{\circ}- S_1^{\circ} -R \In \frac{P_2}{P_1})\\\\= 100.44kJ -1.54 \times 298( 5.0478-4.788-0.1889 \In \frac{135}{100})\\\\=7.23\ KJ\\[/tex]

Investigator Campbell has bullets that were collected from the crime scene and puts them under her comparison microscope. What other item would she need to examine?


The victim's clothing with the gunshot hole

Test bullets fired from the suspected gun

A bullet that has been fired into a wall or hard surface

The barrel of the suspected gun

Answers

Answer:

Test bullets fired from the suspected gun

Explanation:

This is the correct answer!! I took the test!!

from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa at point B. Friction between the water and the pipe walls is negligible. What is the rate of discharge at point B

Answers

Answer:

1.8 m/s

Explanation:

Here is the complete question

The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.

What is the rate of discharge at point B?

Solution

Using Bernoulli's equation,

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)

(h₂ - h₁) = 5 m

Substituting the values of the variables, we have

700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)

36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)

- 13000 Pa = 500 kg/m³(v₂² - v₁²)

(v₂² - v₁²) = - 13000 Pa/500 kg/m³

(v₂² - v₁²) = -26 m²/s²

By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B

πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m

d₁²v₁ = d₂²v₂

v₁  = v₂(d₂/d₁)²

= v₂(0.15/0.05)²

= v₂(3)²

= 9v₂

So, substituting v₁ = 9v₂ into the above equation, we have

(v₂² - v₁²) = -26 m²/s²

v₂² - 9v₂² = -26 m²/s²

- 8v₂² = -26 m²/s²

v₂² = -26 m²/s² ÷ (-8)

v₂² = 3.25 m²/s²

taking square root of both sides,

v₂ = √(3.25 m²/s²)

= 1.8 m/s

So, the rate of discharge at point B is 1.8 m/s

which statement about force is incorrect

Answers

Answer:

What are the options?

Explanation:

A microscope using ultraviolet light is used to study bacteria. If the aperture diameter is 1.5 cm and it is desired to distinguish features with an angular size 0.036 arc seconds, what maximum wavelength can be used

Answers

Given :

A microscope using ultraviolet light is used to study bacteria. If the aperture diameter is 1.5 cm.

Angular size is 0.036 arc seconds.

To Find :

The maximum wavelength that can be used.

Solution :

Converting given angle into radians :

[tex]\theta = \dfrac{0.036}{3600}\times \dfrac{\pi}{180}\\\\\theta = 1.745\times 10^{-7}\ radians[/tex]

Now, we know maximum wavelength is given by :

[tex]\lambda = \dfrac{\theta \times D}{1.22}\\\\\lambda = \dfrac{1.745\times 10^{-7} \times 0.015}{1.22}\ m\\\\\lambda =2.145 \times 10^{-7}\ m[/tex]

Hence, this is the required solution.

A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:

Answers

Answer: Induced emf is given by:

ε= Bvl

On putting the values we get

=5×10

−5

×1.50×2

=0.15mV

Explanation:

Hope these helped have a wonderful Christmas break ❄️

Can a physical guantity have unit but is dimensionless ?

Answers

Ans: A quantity cannot have units until it has dimensions. A quantity that has dimensions must have units. Yes a quantity have units but still be dimensionless for example,unit of angle is radian ,but it is a dimensionless quantity. A dimensionless quantity may have unit.

Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT

Answers

Answer:

2.753*10^-11N

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

Hence the gravitational force between the asteroid is 2.753*10^-11N

a 60N at an angle of 30°from
horizontal​

Answers

Explanation:

Force, F = 60 N

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find horizontal and vertical component of the force.

Horizontal component,

[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]

Vertical component,

[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]

So, the horizontal and vertical component are 51.96 N and 30 N respectively.

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