Answer:
In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.
The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4
Granite: 2.70 × 10 32.70 × 10 3
Lead: 1.13 × 10 41.13 × 10 4
Iron: 7.86 × 10 37.86 × 10 3
Oak: 7.10 × 10 27.10 × 10 2
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel
across a 30 V potential difference to a light bulb.
a. Calculate the current delivered through the light bulb in the two cases.
b. Draw the circuit connection that will achieve the brightest light bulb.
Explanation:
Given that,
Two resistors 4.5 Ω and 2.3 Ω .
Potential difference = 30 V
When they are in series, the current through each resistor remains the same. First find the equivalent resistance.
R' = 4.5 + 2.3
= 6.8 Ω
Current,
[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]
So, the current through both lightbulb is the same i.e. 4.41 A.
When they are in parallel, the current divides.
Current flowing in 4.5 resistor,
[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]
Current flowing in 2.3 ohm resistor,
[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]
In parallel combination, are brighter than bulbs in series.
How does an airpump work?
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Answer:
The correct answer is - low pitch
Explanation:
Now for the case it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
if frequency increases then pitch will be increase as well as pitch depends on frequency.
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.
As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)
Answer:
Explanation:
The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .
v² = u² + 2 g H .
v² = 0 + 2 x 9.8 x 4.7 m
= 92.12
v = 9.6 m /s
The fish's final velocity will have two components
vertical component = 9.6 m/s downwards
Horizontal component = 2.6 m /s .
Resultant velocity = √ ( 9.6² + 2.6² )
= √ ( 92.16 + 6.76 )
= 9.9 m /s
Answer:
The speed of fish at the time of hitting the surface is 9.95 m/s.
Explanation:
Horizontal speed, u = 2.6 m/s
height, h = 4.7 m
Let the vertical velocity at the time of hitting to water is v.
Use third equation of motion
[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]
The net velocity with which the fish strikes to the water is
[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.
1. How many revolutions do the kids make before the constant operational speed is reached ?
2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.
3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?
Answer:
we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer
Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field
Answer:
Option (2) is correct.
The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Explanation:
An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.
So, the correct option is (2).
The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.
The given problem is based on the concept and fundamentals of electromagnetic waves. The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.
In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.
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you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes
I guess that we want to find how much money you get each week.
We know that the job pays $8.60 per hour.
We know that you work 20 hours per week.
Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:
20*$8.60 = $172.
Now we know that your employer witholds:
10% + 7.65% + 5% = 22.65%
Then your employer withholds 22.65% of your gross pay.
if the 100% of your gross pay is $172
Then the 22.65% will be:
(22.65%/100%)*$172 = 0.2265*$172 = $38.96
This means that your employer withholds $38.96 of your weekly gross pay.
Then each week you get:
$172 - $38.96 = $133.04
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A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please
Answer:
Force A newton Law first law
F = M.A which Force in 0 N as you Questions Above
Force B
Newton Law 3
Action = -Reaction
Hope you can explain this formula as you want to scribe to explaining
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge
Answer:
I think it is of science is it true na i knew it bro dont take tension
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.
Answer:
A. 16 J
B. 16 J
C. 8 N
Explanation:
A. Determination of the kinetic energy.
Mass (m) = 0.5 Kg
Velocity (v) =. 8 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 8²
KE = ½ × 0.5 × 64
KE = 0.5 × 32
KE = 16 J
B. Determination of the Workdone by the force.
Kinetic energy (KE) = 16 J
Workdone =.?
Workdone and kinetic energy has the same unit of measurement. Thus,
Workdone = kinetic energy
Workdone = 16 J
C. Determination of the force.
Workdone (Wd) = 16 J
Displacement (s) = 2 m
Force (F) =?
Wd = F × s
16 = F × 2
Divide both side by 2
F = 16 / 2
F = 8 N
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A
Complete Question
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2
Answer:
[tex]h=1614m[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]
Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]
Density [tex]\rho=1.20kg/m^2[/tex]
Generally the equation for Height climbed is mathematically given by
[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]
[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]
[tex]h=1614m[/tex]
A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram
Answer:
b
Explanation:
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement, [tex]\theta=12\ rad[/tex]
Final angular speed, [tex]\omega_f=17\ rad/s[/tex]
The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]
Put all the values,
[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]
So, the initial angular speed of the CD is equal to 14.73 rad/s.
Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.
Answer:
thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.
Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?
a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features
Answer:
c). Easier setup
Explanation:
As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Answer:
13.3 pounds.
Explanation:
For a spring of constant K, with an attached object of mass M, the period can be written as:
T = 2*π*√(M/K)
Where π = 3.14
First, we know that the period is 4 seconds, then we have:
4s = (2*π)*√(M/K)
We know that if the mass is reduced by 10lb, the period becomes 2s.
Then the new mass of the object will be: (M - 10lb)
Then the period equation becomes:
2s = (2*π)*√((M-10lb)/K)
So we have two equations:
4s = (2*π)*√(M/K)
2s = (2*π)*√((M-10lb)/K)
We want to solve this for M.
First, we need to isolate K in one of the equations.
Let's isolate K in the first one:
4s = (2*π)*√(M/K)
(4s/2*π) = √(M/K)
(2s/π)^2 = M/K
K = M/(2s/π)^2 = M*(π/2s)^2
Now we can replace it in the other equation.
2s = (2*π)*√((M-10lb)/K)
First, let's simplify the equation:
2s/(2*π) = √((M-10lb)/K)
1s/π = √((M-10lb)/K)
(1s/π)^2 = ((M-10lb)/K
K*(1s/π)^2 = M - 10lb
Now we can use the equation: K = M*(π/2s)^2
then we get:
K*(1s/π)^2 = M - 10lb
(M*(π/2s)^2)*(1s/π)^2 = M - 10lb
M/4 = M - 10lb
10lb = M - M/4
10lb = (3/4)*M
10lb*(4/3) = M
13.3 lb = M
What is the maximum wavelength, in nm, of light that can eject an electron from a metal with Φ =4.50 x 10–19 J?
[tex]4.4×10^{-7}\:\text{m}[/tex]
Explanation:
The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:
[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]
or
[tex]\dfrac{hc}{\lambda} = \phi[/tex]
Solving for the wavelength [tex]\lambda[/tex],
[tex]\lambda = \dfrac{hc}{\phi}[/tex]
[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]
[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]
Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.
The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.
To find the wavelength, the given values are,
Energy (E) = 4.50×10¯¹⁹ J
What is wavelength?The distance between two consecutive crests and troughs is called the wavelength of a wave.
Here, for the wavelength,
Energy (E) = 4.50×10¯¹⁹ J
Planck's constant (h) = 6.626×10¯³⁴ Js
Speed of light (v) = 3×10⁸ m/s
The wavelength of the light can be obtained as illustrated below:
E = hv / λ
Cross multiply λ,
E × λ = hv
Divide both sides by E,
λ = hv / E
Substituting all the values,
λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹
λ = 0.000000441733 m
λ = 441.73nm
λ - The maximum wavelength of light.
Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm
Learn more about wavelength,
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The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
[tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex] (you has an mistake in the formula)
[tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]
[tex]\frac{1}{C_{eq1}}[/tex] = 0.1 10⁶
[tex]C_{eq1}[/tex] = 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
[tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]
[tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]
[tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶
[tex]C_{eq2}[/tex] = 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]
U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]
U₄ = 0.5 J
pha của dao động làm hàm
Answer:
pha của dao động là hàm bậc nhất của thời gian.
An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?
105 km/h ≈ 29.2 m/s
60.0 km/h ≈ 16.7 m/s
Before the collision the test car has an acceleration a of
a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²
During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is
a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²
i.e. with magnitude about 19.7 m/s².
Overall, the car has an average acceleration of
a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²
A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.
Answer:
[tex]T=3*10^-3 N/m[/tex]
Explanation:
From the question we are told that:
Radius :
[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]
Work [tex]W=1.5 * 10^{-4}[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=TdA[/tex]
Where
[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]
[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]
[tex]dA=0.050m^2[/tex]
Therefore
[tex]W=TdA[/tex]
[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]
[tex]T=3*10^-3 N/m[/tex]
The reason why a teacher is more important then a farmer
Answer:
A teacher is more important than a famer.
Explanation:
A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.
A certain heating element is made out of Nichrome wire and used with the standard voltage source of V=120 V. Immediately after the voltage is turned on, the current running through the element is measured at I1=1.28 A and its temperature at T1=25°C. As the heating element warms up and reaches its steady-state (operating) temperature, the current becomes I2=1.229 A.
Required:
Find this steady-state temperature T2.
Answer:
T₁ = 232.5 ºC
Explanation:
For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law
V = i R
R = V / i
i₀ = 1.28 A
R₀ = 120 / 1.28
R₀ = 93.75 ohm
i₁ = 1.229 A
R₁ = 120 / 1.229
R₁ = 97.64 or
Resistance in a metal is linear with temperature
ΔR = α R₀ ΔT
where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹
ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]
ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]
ΔT = 2,075 10² C
ΔT = T₁-T₀ = 2,075 10²
T₁ = T₀ + 207.5
T₁ = 25+ 207.5
T₁ = 232.5 ºC
