Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...
(C)
Explanation:
[tex]E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})[/tex]
[tex]= 5.3×10^{-18}\:\text{J}[/tex]
Answer:
It's D
Explanation:
It's from alvs
A physics student likes to study while listening to loud music. If electricity costs 12.00$/kWh (kilowatt-hour), how much would it cost the student to run a 220 W stereo system 8.0 hours per day for 10 days of studying?
Answer:
the cost of running the stereo is $211.2
Explanation:
Given;
cost of electricity, = 12.00$/kWh
power consumed by the stereo system, P = 220 W
duration of the power consumption, t = 8 hours
number of days, = 10 days
total time of the power consumption = 8 hours x 10 = 80 hours
power consumed in kW = 220 W / 1000 = 0.22 kW
Energy consumed = 0.22 kW x 80 h = 17.6 kWh
The cost of using 17.6 kWh
= 17.6 x $12
= $211.2
Therefore, the cost of running the stereo is $211.2
Xác định ứng lực trong các thanh BC,
CF và FE của hệ giàn và chỉ rõ các thanh
chịu kéo hay nén. Cho P1=P2=600 lb,
P3=800 lb
Answer:
............................................
Explanation:
a method of reducing friction
Answer:
Lubrication
Explanation:
People oil/lubricate bicycle chains because the chain turns around the cogs and rub together so this help with friction.
Hope this helps :)
Answer:
The method of reducing friction are :
i) In moving parts of machine friction can be reduced by using a ball bearing between the moving surfaces
ii) The bodies of aeroplane ,ship ,boat etc are made streamlined to reduce friction.
iii) Friction can be reduced by polishing rough surfaces. For example : carrom boards are highly polished to reduce friction.
I hope this help you:)
Using the given temperature and pressures, determine: a) the diameter of the water scale model balloon (m), b) the weight of the scale model, c) the specific gravity of the buoyant material such that the model conditions will be similar to the full-scale balloon.
Complete Question
Complete Question is attached below
Answer:
a) [tex]D=0.7[/tex]
b) [tex]W=1787.5N[/tex]
c) [tex]\rho'=998.19kg/m^3[/tex]
Explanation:
From the question we are told that:
Hot air:
Temperature [tex]T_a=360K[/tex]
Pressure [tex]P_a=100kPa[/tex]
Distance [tex]d=12m[/tex]
Weight [tex]W=1400N[/tex]
Water:
Temperature [tex]T_w=300K[/tex]
Pressure [tex]P_w=100kPa[/tex]
Since we have The same Reynolds number
a)
Generally the equation for equal Reynolds number is mathematically given by
Re_{air}=Re_{water}
Therefore
[tex]\frac{\rho V D}{\mu_{air}}=\frac{p_{water}*V*D}{\mu_{water]}}[/tex]
[tex]\frac{100*12}{300*0.28*1.81*10^{-3}}}=\frac{998*D}{0.000890}[/tex]
[tex]D=0.7[/tex]
b)
Generally the equation for Weight of scale is mathematically given by
[tex]W=\rho*V*g[/tex]
[tex]W=998*\frac{4}{3}*\pi*0.35^3*9.81[/tex]
[tex]W=1787.5N[/tex]
c)
Generally the equation for Density of buoyant material is mathematically given by
[tex]\rho'=\frac{w}{g*V}[/tex]
[tex]\rho'=\frac{1781.5}{\frac{4}{3}*\pi*0.35^3*9.81}[/tex]
[tex]\rho'=998.19kg/m^3[/tex]
Una persona de 76 kg está siendo retirada de un edificio en llamas mientras se muestra en la figura. Calcule la tensión
en las dos cuerdas si la persona está momentáneamente inmovil.
Ayuda por favor.
Answer:
T1 = 736.6 N, T2 = 193.5 N
Explanation:
W = 76 N
The tension is T1 and T2.
By use of Lami's theorem
[tex]\frac{T_1}{Sin100}=\frac{T_2}{Sin165}=\frac{W}{Sin 95}\\\\So, \\\\T_1 = \frac{76\times 9.8\times Sin 100}{Sin 95} = 736.6 N \\And\\T_2 = \frac{76\times 9.8\times Sin 165}{Sin 95} = 193.5 N \\[/tex]
Which of the following statements about magnetism is TRUE?
a) The direction of the magnetic force on a current-carrying wire is parallel to the wire.
b) Magnetic poles always occur in pairs (N and S).
c) Magnetic field lines begin at south poles and end on north poles.
d) Moving charges do not experience a force in magnetic fields.
If ∆H = + VE , THEN WHAT REACTION IT IS
1) exothermic
2) endothermic
Answer:
endothermic
Explanation:
An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.
It is easy to produce a potential difference of several thousand volts between your body and the floor by scuffing your shoes across a nylon carpet. When you touch a metal doorknob, you get a mild shock. Yet contact with a power line of comparable voltage would probably be fatal. Why is there a difference?
Answer:
In sof the friction with the nylon is very small and the current with the line e is largeummary
Explanation:
When we rub the shoes against the carpet, static electricity is produced, when you touch the metal knob you close the circuit and the current can circulate to three of the body, the value of this current is of the order of micro volts, for which a small discharge, the power that circulates through the body is very small of the order of 0.005 A
When you touch the power line, the voltage may be small, but the amount of current that can generate them is of the order of tens of amps, the electric shock is much greater per location.
In general there is a rule that if the body resumes more than P = 4000W the discharge could be fatal.
In sof the friction with the nylon is very small and the current with the line e is largeummary, the difference is that the current at the stop , so the paper that passes through the body is large and can be dangerous.
In contact with metal doorknob, get a mild shock while with power line of same voltage, fatal the body as the amount of current is more.
What is charging by friction?When the two materials are rubbed each other, then the electric charged generated between them.
This charging of materials, due to the rubbing of two materials against each other, is called the charging by friction.
It is easy to produce a potential difference of several thousand volts between the body and the floor by scuffing your shoes across a nylon carpet.
In this case, the potential difference may be higher, but the value of current is very low. Thus, when the body touches a metal doorknob, it will get a mild shock.
Now, in another case, the contact with a power line of comparable voltage would probably be fatal. This is because in the power line the amount of current is much higher.
Hence, in contact with metal doorknob, get a mild shock while with power line of same voltage, fatal the body as the amount of current is more.
Learn more about the charging by friction here;
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Imagine that you and your lab partner are standing on smooth ice holding onto opposite ends of a long rope. There is no friction between your feet and the ice. You tug gently on the rope. (a)Describe the motion of you and your lab partner before, during, and after the tug. Be as specific as you can. Expla
Answer:
Following are the complete solution to the given question:
Explanation:
In this question, before tug they are at rest, thus their initial momentum is 0 [tex]\pi=0[/tex], according to the law of conservation [tex]p_f=0[/tex] thus both should move in the opposite direction whose sum of the momentums is equaled to zero as there is no friction they will continue to move with their velocities until their collision. Its collision doesn't happen it will continue up to an infinite distance.
The motion of both you and your partner depends on the tension in the rope.
We must note that frictional force is the force that acts between two surfaces in contact with each other. The frictional force depends on the nature of the surfaces in contact.
Now, since there is no friction between your leg and the ice. The motion of both you and your partner depends on the tension in the rope.
Learn more about tension force: https://brainly.com/question/1195122
1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?
Explanation:
the car will be brought back
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?
Answer:
opposite direction
Explanation:
An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.
Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.
Answer:
Explanation:
negatively charged particle is placed inside uniform electric field
The force on the charge due to the electric field is
F = q E
when the charge is negative so the force on the charge is opposite to the direction of electric field.
The electric field is opposite to the force.
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of air is 1.29 kg/m^3. The skin and structure of the balloon has a mass of 910 kg. Neglect the buoyant force on the cargo volume itself.
Determine the largest mass of cargo the balloon can lift.
The largest mass of cargo the balloon can lift is 791.06 kg
First, we need to calculate the mass of helium.
Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.
The volume of the balloon also equals the volume of helium present.
Now, the mass of helium m = density of helium, ρ × volume of helium, V
m = ρV
Since ρ = 0.179 kg/m³
m = ρV
m = ρ4πr³/3.
m = 0.179 kg/m³ × 4π(7.15 m)³/3
m = 0.179 kg/m³ × 4π(365.525875 m³)/3
m = 0.179 kg/m³ × 1462.1035π m³/3
m = 261.7165265π/3 kg
m = 822.207/3 kg
m = 274.07 kg
Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.
The weight of this mass W = Mg where g = acceleration due to gravity.
The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.
W' = m'g
Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'
We know that the volume of air displaced, V' = volume of balloon, V
So, V' = V = 4πr³/3.
Since the density of air, ρ' = 1.29 kg/m³,
m' = ρ'V
m = 1.29 kg/m³ × 4π(7.15 m)³/3
m = 1.29 kg/m³ × 4π(365.525875 m³)/3
m = 1.29 kg/m³ × 1462.1035π m³/3
m = 1886.113515π/3 kg
m = 5925.4/3 kg
m = 1975.13 kg
So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.
So, the net mass m" = W"/g = 791.06g/g = 791.06 kg
This net mass is the largest mass of cargo that the balloon can lift.
Thus, the largest mass of cargo the balloon can lift is 791.06 kg
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In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage of 0.0320 V is applied across the ends of the wire. From this information calculate the resistance of the wire.
Answer:
0.023 Ohms
Explanation:
Given data
Length= 2.8m
radius= 1.03mm
current I= 1.35 A
voltage V= 0.032V
We know that from Ohm's law
V= IR
Now R= V/I
Substitute
R= 0.032/1.35
R= 0.023 Ohms
Hence the resistance is 0.023 Ohms
Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?
Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?
Answer:
The charges are + 74.3 μC and - 74.3 μC
Explanation:
Let the charges be q and q'.
Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by
F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m
When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.
They now repel each other.
So, the magnitude of the force of repulsion is given by
F' = k[(q + q')/2][(q + q')/2]/r²
F' = k[(q + q')²/4r²
Since the magnitude of the force of attraction and repulsion are the same, we have that
F = F'
kqq'/r² = k[(q + q')²/4r²
qq' = (q + q')²/4
(q + q')² = 4qq'
q² + 2qq' + q'² = 4qq'
q² + 2qq' - 4qq' + q'² = 0
q² - 2qq' + q'² = 0
(q - q')² = 0
q - q' = 0
q = q'
Substituting q = q' into F, we have
F = kqq'/r²
F = kq²/r²
making q subject of the formula, we have
q² = Fr²/k
q = √(Fr²/k)
q = r√(F/k)
Substituting the values of the variables into the equation, we have
q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)
q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)
q = 0.189 m(0.3923 × 10⁻³ C/m)
q = 0.0743 × 10⁻³ C
q = 74.3 × 10⁻³ × 10⁻³ C
q = 74.3 × 10⁻⁶ C
q = 74.3 μC
Since q and q' initially attract, it implies that they initially had opposite charges.
So, q = 74.3 μC and q' = -74.3 μC
So, the charges are + 74.3 μC and - 74.3 μC
True or false : conservation of energy gives a relationship between the speed of a falling object and the height from which it was dropped
Answer:
truee
Explanation:
A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?
Answer:
Explanation:
The formula for angular momentum is
L = mvr where L is the angular momentum, m is the mass of the object, v is the velocity of the object, and r is the radius of the object. The problem we have that prevents us from just throwing those numbers in there is that mass has to be in kg and it's not, and radius has to be in meters and it's not.
Changing the mass to kg:
750 g = .750 kg
Changing the radius to m:
35 cm = .35 m
Now we can fill in the variables with their respective values:
L = .750(10.0)(.35) gives us
[tex]L=2.625\frac{kg*m^2}{s}[/tex]
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 21 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 61 cm. a. What is the wavelength of the sound
Answer:
The answer is "80 cm".
Explanation:
The distance of 21 cm between the speaker's effect of high strength but a spacing of 61 cm corresponds to a zero to zero intensity, that also is, the waves are all in phase with others [tex]\Delta \ x_1 = 21 \ cm[/tex] this is out of phase [tex]\Delta\ x_2 = 61\ cm[/tex]
[tex]\therefore\\\\\Delta\ x_2 -\Delta\ x_1 = \frac{\lambda}{2}\\\\\lambda= 2( \Delta\ x_2 -\Delta\ x_1)[/tex]
[tex]= 2 ( 61\ cm - 21\ cm)\\\\ = 2(40\ cm)\\\\= 80\ cm[/tex]
A cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel. The bucket starts from rest and falls for 4.00 s.
Required:
a. What is the linear acceleration of the falling bucket?
b. How far does it drop?
c. What is the angular acceleration of the reel?
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
Learn more about object kinetic energy of objects in free fall here;
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A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)
Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance:___________.
A) must be directed inward.
B) acts like it originated in a point charge Q at the center of the distribution.
C) must be directed outward.
D) must be greater than zero.
E) must be zero.
Answer:
Q at the center of the distribution.
Explanation:
The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels
Answer: [tex]53\ rev/min[/tex]
Explanation:
Given
angular speed of wheel is [tex]\omega_1 =530\ rev/min[/tex]
Another wheel of 9 times the rotational inertia is coupled with initial wheel
Suppose the initial wheel has moment of inertia as I
Coupled disc has [tex]9I[/tex] as rotational inertia
Conserving angular momentum,
[tex]\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min[/tex]
If Light is travels from air into pure water with an incident angle of 30°. What is the angle of refraction?
Explanation:
For air, n1 = 1.00003; for water, n2 = 1.3330
Given: θ2 = 30 degrees, then
θ1 = arcsin [(n2/n1) sin θ2]
= arcsin [(1.3330/1.0003) sin (40)]
= 58.93 degrees
Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.
