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​

Answer:

0.068kg

Explanation:

In compressed state, a spring 3.00 meters long hanging from the ceiling has a potential energy of 2.0 joules. If the entire potential energy is used to release a ball of mass

Potential energy is expressed using the formula

PE = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height =  3.0m

PE = 2.0Joules

Required

Mass of the ball

Substitute the given values into the formula as shown;

2.0 = m(9.8)(3)

2 = 29.4m

m = 2/29.4

m = 0.068kg

Hence the mass of the ball is 0.068kg

Answer:

8 seconds

Explanation:

as per 1st equation of motion

v=u+at,

u= initial velocity, v= final velocity , t=time, a=acceleration

since it starts from rest , u=0

v=16m/s

a=2m/s^2

16=0+2t

16=2t

t=8

Answer: the length of the pendulum should be 1.4% longer

Explanation:

Given that;

when its noon, the clock reads 11:50 am,

i.e we have 10 minutes delay ⇒ 10min × 60 = 600secs

we know that in simple pendulum

T = 2π√(l/g)  

d means delay and c means correct;

24hrs = 86400 secs

Now

Td/Tc = (86400-600) / 86400  = 0.993 = [2π√(ld/gc)] / [2π√(ld/gc)] = √(ld/lc)

so ld/lc = 0.98616  

lc = 1.014 ld

lc/l.d = -1 + 1.014 = 0.014 × 100 = 1.4% longer

therefore the length of the pendulum should be 1.4% longer

The correct asnwer is D. The permanent magnets cause the metal loop in the

electromagnet to turn.

In an electric motor, the permanent magnets cause the metal loop in the

electromagnet to turn. So, option (D) is correct.

An electrical device that transforms electrical energy into mechanical energy is an electric motor. The majority of electric motors work by creating force in the form of torque imparted to the motor shaft through the interplay of the magnetic field of the motor and electric current in a wire winding.

In an electric motor, their remains a permanent magnet to produce constant magnetic field. When electric current is passed through the conducting loop, it acts ling electromagnet and begins to rotate. Thus an electric motor works. Hence, option(D) is correct.

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Answer:

OD. The wave diffracted as it passed through the opening in the  barrier.

Explanation:

A progressive wave (i.e waves in motion) has the capacity to bend around an obstacle on its path. This is one of the general properties of waves called diffraction. Others are: reflection, refraction, interference. Note that only transverse waves undergo polarization.

Diffraction of waves is the ability of waves to bend around an obstacle on its path during progression.

Thus, the bending of the part of waves as it passes through the barrier implies that the wave diffracted as it passed through the opening in the  barrier.

Answer:

The wave diffrected as it passed through the opening

Explanation:

Answer:

the new frequency is : 302.25Hz

using the formula F2 = [tex]\frac{F1 L1}{L2}[/tex]

Explanation:

frequency of a string is the number of vibrations of a plucked string per second. it is measured in Hertz.

the frequency of a string is inversely proportional to twice the length of the string. which means the longer the string, the smaller the freqency and the higher the string the higher the frequency.

f ∝ 1/2L.

f = k/2L

where f = frequency

                         L = length of string

                         k = constant

k = 2fL is a constant

given data

L1 = 0.62m

f1 = 234Hz

L2 = 0.48m

2f1L1 = 2f2L2

f1L1 = f2L2

f2 = f1L1/L2

f2 = [tex]\frac{234 x 0.62}{0.48}[/tex] = 302.25Hz

in conclusion, the new frequency is 302.25Hz

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Answer:

The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q.

Explanation:

Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.

Answer:

Q = 189000 [J]

Explanation:

The internal energy or heat can be calculated by means of the following expression.

[tex]Q=m*c_{p}*DT[/tex]

where:

Q = internal energy or heat [J]

m = mass = 15 [kg]

Cp = 840 [J/kg*°C]

DT = temperature change = 15 [°C]

[tex]Q = 15*840*15\\Q = 189000 [J][/tex]

Answer:

a. 6602.7 N b. 64.44 km/h

Explanation:

a. Find the force of friction required to keep it from  sliding down the hill.

The frictional force f equals the component of the weight of the car, W perpendicular to the inclined hill = Wcosθ times the coefficient of static friction, μ = 0.45.

Since f = μN = μWcosθ = μmgcosθ where m = mass of car = 1500 kg, g = acceleration due to gravity = 9.8 m/s² and θ = angle of incline of hill = 3.5°

So, f = μmgcosθ

= 0.45 × 1500 kg × 9.8 m/s²cos3.5°

= 6615cos3.5°

= 6602.7 N

b. Determine the speed of the car after it emerges from the mud (in km/h)

Since the car drops a vertical height of 14 m, its potential energy decreases by mgh and its kinetic energy increases by mgh where m =mass of car and h = height drop = 14 m. So its kinetic energy increase is ΔK = mgh = 1500 kg × 9.8 m/s² × 14 m = 205800 J

Since it has an initial velocity of u = 50 km/h = 50 km/h 1000m/3600 s = 13.89 m/s, its initial kinetic energy is K = 1/2mu² = 1/2 × 1500 kg × (13.89 m/s)² = 144699.08 J.

Its new kinetic energy after the drop is thus K' = K + ΔK = 144699.08 J + 205800 J = 350499.08 J

Let v be its velocity after the drop, since K' = 1/2mv²,

v = √(2K'/m) = √(2 × 350499.08 J/1500 kg) = √(700998.16 J/1500 kg) = √(467.332 J/kg) = 21.62 m/s

Now, from work kinetic energy principles, the kinetic energy change in the car is the work done on car by friction

So, ΔK" = -fd = -μmgd

Let v' be the velocity of the car after emerging from the mud and moving a distance d = 30.0 m.

So, 1/2m(v'² - v²) = -μmgd

v'² - v² = -2μgd

v'² = v² - 2μgd

Substituting the values of the variables, we have

v'² = (21.62 m/s)² - 2 × 0.25 × 9.8 m/s² × 30.0 m

v'² = 467.42 m²/s² - 147 m²/s²

v'² = 320.42 m²/s²

taking square root of both sides, we have

v' = √(320.42 m²/s²)

= 17.9 m/s

Converting v to km/h we have v' = 17.9 m/s × 3600 s/h × 1 km/1000 m = 64.44 km/h.

So, the car emerges from the mud with a speed of 64.44 km/h

Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

Explanation:

Given that;

friction force of ground box = 68 N

student of 7th grade = n

Whitmore can apply a force of 25 N

every other 7th grade student can apply a force of 6 N.

now

friction force = forced applied by whitmore + total force ny 7th grade student

we substitute

68 = 25 + 6n

6n = 68 - 25

6n = 43

n = 43/6

n = 7.17

Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

Answer:

Explanation:

The impulse of an object can be found by using the formula

impulse = force × time

From the question we have

impulse= 23 × 3.1

We have the final answer as

Hope this helps you

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.

Answer:

The value is  [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]

Explanation:

From the question we are told that

   The measure voltage is  [tex]E_{rms} = 1.0 \ n V = 1.0 *10^{-9} \ V[/tex]

Generally the peak voltage  is   mathematically represented as

       [tex]E_{max} = \sqrt{2} * E_{rms}[/tex]

=>    [tex]E_{max} = \sqrt{2} * 1.0 *10^{-9}[/tex]

=>    [tex]E_{max} = 1.4142 *10^{-9} \ volts[/tex]

Generally the potential difference between the two extremes is mathematically represented as

       [tex]V = 2 * E_{max}[/tex]

=>    [tex]V = 2 * 1.4142 *10^{-9}[/tex]

=>    [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]

Answer:

2.01

Explanation:

First, we need to find the centripetal acceleration.

We're given that the merry go round rotates 1 revolution in 2.09 seconds. Converting to rpm, we know that it rotates 30 revolution per minute

Now this speed gotten in rpm will be converted to m/s, to ease the calculation

30 rpm = πdN/60 m/s

30 rpm = (3.142 * 4 * 30)/60

30 rpm = 377.04/60

30 rpm = 6.284 m/s

a(c) = v²/r

a(c) = 6.284²/2

a(c) = 39.49 / 2

a(c) = 19.74 m/s²

F = ma

F = 50 * 19.74

F = 987 N

Also, Normal Force, F(n) =

F(n) = mg

F(n) = 50 * 9.81

F(n) = 490.5

We then use this to find the coefficient of static friction, μ

μ = F/F(n)

μ = 987 / 490.5

μ = 2.01

Given :

A pinball machine launches a .045 kg ball at a speed of 9.2 m/s.

To Find :

The potential energy of the spring just before it launched the pinball.

Solution :

We know, their is no external force applied on system.

It means that kinetic energy will remains constant.

Initial Energy = Final Energy

[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\0 + P.E_i= \dfrac{mv^2}{2}+ 0\\\\P.E_i = \dfrac{0.045 \times 9.2^2}{2}\ J\\\\P.E_i = 1.9044\ J[/tex]

Therefore, the potential energy of the spring just before it launched the pinball os 1.9044 J.

Answer:

Wonho, Seonghwa, and Bangchan

Explanation:

Answer:

namjoon,seonghwa,bangchan

Explanation:

Answer:

i am!

Explanation:

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

W = Work (J)

F = Force / weight (N)

s = distance (m)

Work done in physics is the product of force and displacement. The displacement for the object is 36 m and force acts on it is 10 N. Then the work done is 360 J.

Work done is the dot product of force acting on a body and the resultant displacement. When a force applied on an object results in a displacement from its position, the force is said to be work done.

Work done is a vector quantity thus, characterised by a magnitude and direction.  The common unit of work done is joule.

Given that force applied on the weight = 10 N

displacement occurred = 40 m - 4 m = 36 m

Work done = F . ds

ds = 36 m and f = 10 N

Then W = 10 N × 36m

             = 360 J.

Therefore, the work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth is 360 J.

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The question is incomplete. The complete question is :

First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:

The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number

What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.

Solution :

a). Given :

[tex]$Q_1 = 42 \ J$[/tex] , [tex]$Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J $[/tex]

Net heat transfer

[tex]$Q= Q_1+Q_2$[/tex]

   = 42 + (-22)

   = 20 J

Total work

[tex]$W= W_1+W_2$[/tex]

   = 6 + (-6)

   = 0 J    

∵ ΔU = Q - W

       = 20 - 0

        = 20 J

This is the net change in the internal energy of the system.

b). ΔU = Q + W

           = (-140) + (-165)

           = -305 J

c). ΔU = Q + W

           = (-440) + (645)

           = 205 J

d). ΔU = Q + W

           = (-450) + (690)

           = 240 J

Answer:

The value is  [tex]\beta_f = 84.95 \ dB[/tex]  

Explanation:

From the question we are told that  

    The intensity level  of the shout of a single person is [tex]\beta = 48.1 \ dB[/tex]

     The number of fans  is  [tex]n = 4841[/tex]

Gnerally intensity level is mathematically represented as

                     [tex]\beta = 10 log * \frac{I}{I_o }[/tex]

Here [tex]I_o[/tex] is the minimum intensity of sound human ear can pick and the value is  

          [tex]I_o = 1 * 10^{-12} \ W/m ^2[/tex]

when  [tex]\beta = 48.1 \ dB[/tex]

          [tex]48.1 = 10 log * \frac{I}{ 1 * 10^{-12}}[/tex]

    =>   [tex]4.81 = log ( \frac{ I}{ 1 * 10^{-12}} )[/tex]

taking antilog of  both sides

        [tex]64565.42 = \frac{I}{ 1 *10^{-12}}[/tex]

=>     [tex]I = 6.457 *10^{-8} \ W/m^2[/tex]

Generally  the intensity for the whole fans is mathematically represented as  

         [tex]I_f = n * I[/tex]

=>      [tex]I_f = 4841 * 6.457 *10^{-8 }[/tex]

=>      [tex]I_f = 0.0003126 \ W/m^2[/tex]

Gnerally the intensity level for the whole fans is mathematically represented as    

               [tex]\beta_f = 10 log [ \frac{I_f }{I_o } ][/tex]

=>             [tex]\beta_f = 10 log [ \frac{ 0.0003126 }{1*10^{-12}}[/tex]

=>             [tex]\beta_f = 84.95 \ dB[/tex]  

Explanation:

The attatched figure shows the vector diagram for a force that has magnitude of 60 N and it is acting at an angle of 30° from the horizontal​.

When it is resolved, the horizontal and vertical components are given by :

[tex]F_x=F\cos\theta\\\\=60\times \cos30\\\\=51.96\ N[/tex]

And

[tex]F_y=F\sin\theta\\\\=60\times \sin30\\\\=30\ N[/tex]

Hence, this is the required solution.

Answer: Tidal range

Explanation:

Tides are considering the rise and fall of sea levels and there are two types of it which are called high tide and low tide. The difference between high tide and low tide is called the tidal range.

The tidal range is not constant and it is considering height difference. It can change and it is depending on the locations of the Sun or the Moon.

Answer:

v = 35 km/s

Explanation:

Given that,

Orbital radius of Venus, [tex]r=10.8\times 10^{10}\ m=10.8\times 10^{10}\times 10^{-3}=10.8\times 10^{7}\ km[/tex]

The rotation period is, [tex]T = 224.7\ \text{days}=19414080\ s[/tex]

We need to find the orbital speed of the Venus. The formula for the orbital speed is given by :

[tex]v=\dfrac{2\pi r}{T}\\\\v=\dfrac{2\pi \times 10.8\times 10^{7}}{19414080}\\\\=34.95\ km/s[/tex]

or

v = 35 km/s

So, the orbital speed of Venus is 35 km/s.

Answer:

0.00129rad/s

Explanation:

The angular velocity is expressed as;

v = wr

w is the angular velocity

r is the radius

Given

v =  20,000 mph

r = 4300mi

Get w;

w = v/r

w = 20000* 0.44704/4300*1609.34

w = 8940.8/6,920,162

w = 0.00129rad/s

Hence the angular velocity generated is 0.00129rad/s

Answer:

the answer is B

Explanation:

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 11760 × 0.5

We have the final answer as

Hope this helps you

Answer:

[tex]v_=-8.5\times 10^{-4}\ m/s[/tex]

Explanation:

Given that,

Mass of a wrench, m₁ = 4 g = 0.004 kg

Speed of wrench, v₁ = -15 m/s

Mass of the Astronaut, m₂ = 70 kg

We need to find Astronaut's velocity. Let it is v₂. Using the conservation of linear momentum to find it.

[tex]m_1v_1=m_2v_2\\\\v_2=\dfrac{m_1v_1}{m_2}\\\\v_2=\dfrac{0.004\times (-15)}{70}\\\\=-8.5\times 10^{-4}\ m/s[/tex]

So, the speed of Astronaut is [tex]8.5\times 10^{-4}\ m/s[/tex].

Answer:

See the answers and the explanation below.

Explanation:

To solve these questions we must understand that speed is the relationship between the displacement and the time that the displacement lasts.

1. 300 [km], in 5 [hr]

d = displacement [km]

[tex]v = d/t[/tex]

where:

v = velocity [km/h]

t = time [hr]

[tex]v = 300/5\\v = 60 [km/h][/tex]

In one hour : [tex]d = 60*1\\d = 60 [km][/tex]

b. Car B 100 [km] in 2 [hr]

[tex]v =100/2\\v = 50 [km/h]\\[/tex]

C. The car A has the greatest average speed.

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED