Answer:
the right answer is true
Answer:
True
Explanation:
differences between
hor
rse
and horse
The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The spring scale [tex]F_2[/tex] reads [tex]F_2 = 2.4225 \ N[/tex]
Explanation:
From the question we are told that
The first force is [tex]F_1 = 10.5 \ N[/tex]
The acceleration by which the cart moves to the right is [tex]a = 2.50 \ m/s^2[/tex]
The mass of the cart is m = 3.231 kg
Generally the net force on the cart is
[tex]F_{net} = F_1 - F_2[/tex]
This net force is mathematically represented as
[tex]F_{net} = m * a[/tex]
So
[tex]m* a = 10 - F_2[/tex]
[tex]F_2 = 10.5 - 2.5 (3.231)[/tex]
[tex]F_2 = 2.4225 \ N[/tex]
A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter of 50 cm. The speed of the top of the wheel is
Answer:
The speed of the top of the wheel is twice the speed of the car.
That is: 72 m/s
Explanation:
To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,
plus the velocity due to the translation of the center of mass (v = 36 m/s).
The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]
Then the addition of these two velocities equals:
[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]
A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)
Answer:
[tex]3.1\times 10^{5}m/s[/tex]
Explanation:
The computation of the speed does the proton gain is shown below:
The potential difference is the difference that reflects the work done as per the unit charged
So, the work done should be
= Potential difference × Charge
Given that
Charge on a proton is
= 1.6 × 10^-19 C
Potential difference = 500 V
[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]
[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]
Simply we applied the above formulas
1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?
Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = [tex]\sqrt{3^{2} +2^{2} }[/tex] = [tex]\sqrt{13}[/tex] = 3.61cm = 0.036m
r₂ = [tex]\sqrt{4^{2} + 3^{2} }[/tex] = [tex]\sqrt{25}[/tex] = 5cm = 0.05m
electric potential V = [tex]\frac{kq}{r}[/tex]
change in potential ΔV = [tex]V_{1} - V_{2}[/tex]
ΔV = [tex]\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }[/tex] , where [tex]q_{1} = q_{2}=[/tex]2.00μC
ΔV = [tex]2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })[/tex]
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × [tex](\frac{1}{0.036} - \frac{1}{0.05} )[/tex]
ΔV= 2.789×10⁵
[tex]\frac{1}{2}mv^{2}[/tex] = ΔV × q₃
[tex]\frac{1}{2}[/tex] ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s
We are given;
Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C
Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C
Distance of charge 1 along x = 3 cm = 3 × 10⁻² m
Distance of charge 2 along x = -3 cm = -3 × 10⁻² m
Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C
mass; m = 0.01 g
distance of charge 3 along y = 4 cm = 4 × 10⁻² m
q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.
Thus;
Distance of charge 1 from the initial position of q₃;
r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)
r₁ = 0.0361 m
Distance of charge 2 from the final position of q₃;
r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)
r₂ = 0.05 m
Now, formula for electric potential is;
V = kq/r
Where k = 9 × 10⁹ N.m²/s²
Thus,change in potential is;
ΔV = V₁ - V₂
Now, Net V₁ = 2kq₁/r₁
Net V₂ = 2kq₂/r₂
Thus;
ΔV = 2kq₁/r₁ - 2kq₂/r₂
ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]
ΔV = 277229.92 V
Now, from conservation of energy;
½mv² = q₃ΔV
Thus;
½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92
v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01
v = √(221.783936)
v = 14.89 m/s
Read more about point charges at;https://brainly.com/question/13914561
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
I wish to use a step up transformer to turn an initial RMS AC voltage of 100 V into a final RMS AC voltage of 200 V. What is the ratio of the number of turns in the primary to the secondary
Answer:
1:2
Explanation:
It is given that,
Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.
We need to find the ratio of the number of turns in the primary to the secondary for step up transformer.
For a transformer, [tex]\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}[/tex]
So,
[tex]\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}[/tex]
So, the ratio of the number of turns in the primary to the secondary is 1:2.
two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister
A centrifuge's angular velocity is initially at 159.0 radians/second to test the stability
of a high speed drill component. It then increases its angular velocity to 999.0
radians/second. If this is achieved in 4,100.0 radians what is the angular acceleration
of the centrifuge?
Answer:
118.6 rad/s²
Explanation:
Δθ = 4100.0 rad
ω₀ = 159.0 rad/s
ω = 999.0 rad/s
Find: α
ω² = ω₀² + 2αΔθ
(999.0 rad/s)² = (159.0 rad/s)² + 2α (4100.0 rad)
α = 118.6 rad/s²
A cube has one corner at the origin and the opposite corner at the point (L,L,L)(L,L,L). The sides of the cube are parallel to the coordinate planes. The electric field in and around the cube is given by
Answer:
Net charge = E• b • L^3.
Explanation:
NB: here, the symbol representation of the flux is "p" = electric Field • Area(dot Product).
So, we will take a look at the flux through -x face, through x face and through -y face, through y face and through - z face and through z face.
(1). Starting from -z and z faces which are the back and front faces of the cube:
Thus, We have that the flux,p = 0 for -z and z.
(2). Recall that we are given that E = =(a+bx)i^+cj^.
Thus, p_-y = (a + bx)i + cj (-j) (L^2)
Where y = 0
p_-y = -cL^2.
Obviously for p_j, we will have cL^2 and y = L
(3). For p_-x = =(a + bx)i + cj (-i) (L^2).
p_-x = -aL^2
Where x = 0.
When x = L and p_x = (a + bL)L^2.
This, adding all together gives Net charge = E • b • L^3.
The right-hand rule, which is a convention for identifying the direction of the force on a current or a moving charged positively charged particle, has several correct versions. Which one of the descriptions below is the right-hand rule for the magnetic force exerted on a current or a moving charged particle recommended in this textbook?
A. Thumb of the right hand points in the direction of current or the velocity of the charged particle, the fingers in the direction of B, and the force (F) is directed perpendicular to the right hand palm.
B. Keeping your right hand flat, point your thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.
C. Using your right hand, point your thumb in the direction of the current or the velocity of the charged particle, your fingers in the direction of magnetic field, and your palm points in the direction of the cross-product.
D. Using your right-hand, point your index finger in the direction of the current or the velocity of the charged particle. Point your middle finger in the direction of the magnetic field. Your thumb now points in the direction of the magnetic force.
E. Using the right hand, the direction of the thumb is the direction of the force, the direction of the index finger indicates the direction of the magnetic field, and the direction of the middle finger is the direction of the electric current. Submit
Answer:
B. Keeping your right hand flat, point your thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.
Explanation:
This is the Fleming's right hand rule, which was stated to explain the relationship or induction ability of the magnetic field, current or velocity of charged particles and magnetic force. These three variables are held mutually perpendicularly to one another.
The most suitable description of the right-hand rule is option B which clarifies the perpendicular mutual relationship of the thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.
A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed
Answer:
The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
Explanation:
This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery
change in entropy of universe during 900g of ice at 0 degree celcus to water at 10 degree celcius at room temp=30 degree celcius
Answer:
4519.60 J/KExplanation:
Change in entropy is expressed as ΔS = ΔQ/T where;
ΔQ is the total heat change during conversion of ice to water.
T is the room temperature
First we need to calculate the total change in heat using the conversion formulae;
ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)
m is the mass of ice/water = 900g = 0.9kg
L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg
c is the specific heat capacity of water = 4200J/kgK
Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K
Substituting the given values into ΔQ;
ΔQ = 0.9(333000)+0.9(4200)(283)
ΔQ = 299700 + 1069740
ΔQ = 1,369,440 Joules
Since Change in entropy ΔS = ΔQ/T
ΔS = 1,369,440/30+273
ΔS = 1,369,440/303
ΔS = 4519.60 J/K
Hence, the change in entropy of the universe is 4519.60 J/K
An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)
Answer:
The induced emf is [tex]\epsilon =7.68 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]
The initial magnetic field is [tex]B_i = 0.14 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2 }[/tex]
substituting values
[tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]
[tex]r = 1.1*10^{-2} \ m[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]
Where [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as
[tex]d\phi = dB * A * cos\theta[/tex]
=> [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]
Here [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field
Also A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]
[tex]A = 3.8 *10^{-4] \ m^2[/tex]
So
[tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]
[tex]d\phi = -5.32*10^{-5} \ weber[/tex]
So
[tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]
[tex]\epsilon =7.68 \ V[/tex]
g The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit
Answer:
Explanation:
Let the original resistance be R and voltage be V
Applying ohm's law
V / R = 15
V = 15 R
In second case
V / (R+8 ) = 12
V = 12 R + 96
15 R = 12 R + 96
3R = 96
R = 32 ohm .
A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s, the voltmeter reads 3.0 V.
A) What are the capacitance?
B) What is the time constant of the circuit?
Answer:
a. 0.849 micro farad
b. 2.89 s
Explanation:
a) V=V0 e^-t/RC
3=12*e^-4/3.4*10^6*C
3/12=e^-4/3.4*10^6*C
-1.3863 =-4/3.4*10^6*C
C=8.49*10^-7 F
=0.849 micro farad
B) time constant= R*C
=3.4*10^6*8.49*10^-7
=2.89 S
a. The capacitance is 0.849 micro farad
b. The time constant of the circuit is 2.89 s
Calculation of capacitance & time constant:a)
We know that
V=V0 e^-t/RC
3=12*e^-4/3.4*10^6*C
3/12=e^-4/3.4*10^6*C
-1.3863 =-4/3.4*10^6*C
C=8.49*10^-7 F
=0.849 micro farad
B)
Now
time constant= R*C
=3.4*10^6*8.49*10^-7
=2.89 S
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Asteroid A has 3.5 times the mass and 2.0 times the velocity of Asteroid B. If
Asteroid B has a kinetic energy of 2,300,000 J then what is the kinetic energy of
Asteroid A?
Answer:
K_A = 32.2 10⁶ J
Explanation:
In this exercise we must relate the quantities given to find the kinetic energy
Asteroid A data
m_A = 3.5 m_B
v_A = 2.0 v
they also give the value of the kinetic energy of asteroid A
K_B = 2.3 10⁶ J
the expression for scientific energy is
K = ½ m v²
let's replace
K_A = ½ m_a V_a2
K_A = ½ 3.5 m_B (2.0 v_B)^2
K_A = 3.5 2² (½ m_B v_B²)
K_A = 14 K_B
K_A = 32.2 10⁶ J
A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
The Goliath six flags Magic Mountain roller coaster ride starts at 71.6 m (235 feet) above the ground. Assuming the coaster starts from rest and ignoring any friction, what is the speed of the coaster when it reaches the ground level
Answer:
The velocity is [tex]v = 37 .46 \ m/s[/tex]
Explanation:
From the question we are told that
The start distance above the ground is [tex]h = 71.6 \ m[/tex]
Generally according to the law of energy conservation we have that
[tex]PE_{top} = KE_{bottom }[/tex]
Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as
[tex]PE_{top} = m * g * h[/tex]
And [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as
[tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]
Therefore
[tex]m * g * h = \frac{1}{2} * m * v^ 2[/tex]
=> [tex]v = \sqrt{2 * g * h }[/tex]
substituting value
[tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]
[tex]v = 37 .46 \ m/s[/tex]
What is the major cause of the muffled noise from a radio station?
Answer:
The major cause is "lack of high frequencies in a sound wave".
Explanation:
Muffling derives from either the absence of such a radio signal of a higher or specific frequency. This very same phenomenon has been observed whenever you overhear conversations through some kind of wall and perhaps door. The approach is equalization. This method helps them to raise those frequencies although these overprotective wavelengths decrease.So that the above would be the correct solution.
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
Find the distance to a Sun-like star (L=3.8x1026 watts) whose apparent brightness at Earth is 1.0 x10-10 watt/m2.
Answer:
5.49 x 10^17 m is the distance between the sun-like star to the earth
Explanation:
Radiation intensity on Earth = 1.0 x 10^-10 W/m^2
Power of radiation of the star = 3.8 x 10^26 W
Recall that the intensity of radiation is given as
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where
[tex]I[/tex] = intensity of radiation
P = power of radiation
A is the area through which the radiation spreads out in all three dimensional direction.
A = [tex]\frac{P}{I}[/tex] = [tex]\frac{3.8*10^{26} }{1.0*10^{-10} }[/tex] = 3.8 x 10^36 m^2
This area is spread out in the form of a sphere of area
A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]r^{2}[/tex]
3.8 x 10^36 = 12.568[tex]r^{2}[/tex]
[tex]r^{2}[/tex] = (3.8 x 10^36)/12.568 = 3.02 x 10^35
r = [tex]\sqrt{3.02*10^{35} }[/tex] = 5.49 x 10^17 m this is the distance of the star to the Earth
15. The blank
of a sine wave is the time it takes to complete one cycle of the wave.
O A. maximum amplitude
O B. minimum amplitude
O C. average value
O D. wavelength
That time is the "period" of the wave.
(It's not one of the choices.)
The blank of a sine wave is the time it takes to complete one cycle of the wavelength, the correct answer is D.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
It is the total length of the wave for which it completes one cycle.
The wavelength is inversely proportional to the frequency of the wave as from the following relation.
C = νλ
where c is the speed of light
ν is the frequency of the wave
λ is the wavelength of the wave
The time taken by the sine wave to complete one cycle of the wavelength is called blank the correct answer is D.
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A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?
Answer:
a
[tex]F =326.7 \ N[/tex]
b
[tex]M = 6[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 200 \ kg[/tex]
The length of the small object from the rock is [tex]d = 2 \ m[/tex]
The length of the small object from the branch [tex]l = 12 \ m[/tex]
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
[tex]W = m_r * g[/tex]
substituting values
[tex]W = 200 * 9.8[/tex]
[tex]W = 1960 \ N[/tex]
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
[tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]
Here [tex]\theta[/tex] is very small so [tex]cos\theta * l = l[/tex]
and [tex]cos\theta * d = d[/tex]
Hence
[tex]F * l - W * d = 0[/tex]
=> [tex]F = \frac{W * d}{l}[/tex]
substituting values
[tex]F = \frac{1960 * 2}{12}[/tex]
[tex]F =326.7 \ N[/tex]
The mechanical advantage is mathematically evaluated as
[tex]M = \frac{W}{F}[/tex]
substituting values
[tex]M = \frac{1960}{326.7}[/tex]
[tex]M = 6[/tex]
A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .
Answer:
Option 3 = both spheres are at the same potential.
Explanation:
So, let us complete or fill the missing gap in the question above;
" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"
The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:
=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.
11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length
Answer:
The frequency is [tex]f_n = 257.1 \ Hz[/tex]
Explanation:
From the question we are told that
The third harmonic frequency of the tight guitar string is [tex]f_3 = 540 \ Hz[/tex]
Let the original length be L
Then the length at which it is fingered is 0.7 L
Generally the fundamental is mathematically represented as
[tex]f = \frac{v_s}{ 2L}[/tex]
Now when it finger at 70% it original length is
[tex]f_n = \frac{v}{2 * (0.7 L)}[/tex]
[tex]f_n = \frac{v}{1.4 L}[/tex]
Here v the velocity of sound
So
[tex]\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]
Also the fundamental frequency for the original length can also be represented as
[tex]f = \frac{f_3}{3}[/tex]
substituting values
[tex]f = \frac{540}{3}[/tex]
[tex]f = 180 \ Hz[/tex]
So
[tex]\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]
=> [tex]f_n =\frac{180}{0.7}[/tex]
=> [tex]f_n = 257.1 \ Hz[/tex]
The fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1 Hz.
What is the frequency?Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.
f is the frequency of tight guitar string = 540 Hz
Let's call the original length L.
The amount of time it is fingered is then 0.7 L.
In general, the fundamental frequency is expressed mathematically as;
[tex]\rm f = \frac{v_0}{2L} \\\\[/tex]
For the given conditions;
[tex]\rm f_n=\frac{v}{2 \times 0.7L} \\\\ \rm f_n=\frac{v}{1.4L}[/tex]
The ratio of the frequency is;
[tex]\rm \frac{f_n}{f} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }[/tex]
Also, the fundamental frequency for the original length can also be represented as;
[tex]\rm f= \frac{f'}{3} \\\\ f=\frac{540}{3} \\\\ \rm f=180\ Hz[/tex]
On putting the given data;
[tex]\rm \frac{f_n}{180} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }\\\\ \rm f_n=\frac{180}{0.7}\\\\\ \rm f_n=257.1\ Hz[/tex]
Hence the fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1 Hz.
To learn more about the frequency reference the link;
https://brainly.com/question/14926605
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).
Answer:
The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m
The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m
Explanation:
Thermal kinetic energy of electron or proton = KE
∴ KE = 3kbT/2
given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K
so we substitute
KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2
kE = 4.0365 × 10⁻²⁰ ( is the kinetic energy for both electron and proton at temperature T )
Now we know that
mass of electron M'e = 9.109 × 10⁻³¹
mass of proton M'p = 1.6726 × 10⁻²⁷
We also know that
KE = p₂ / 2m
from the equation, p = √ (2mKE)
{ p is momentum, m is mass }
de Broglie wavelength = β
so β = h / p = h / √ (2mKE)
h = Planck's constant = 6.626 × 10⁻³⁴
∴ βe = h / √ (2m'e × KE)
βe = 6.626 × 10⁻³⁴ / √ (2 × 9.109 × 10⁻³¹ × 4.0365 × 10⁻²⁰ )
βe = 6.626 × 10⁻³⁴ / √ 7.3536957 × 10⁻⁵⁰
βe = 6.626 × 10⁻³⁴ / 2.71176984642871 × 10⁻²⁵
βe = 2.443422 × 10⁻⁹ m
βp = h / √ (2m'p ×KE)
βp = 6.626 × 10⁻³⁴ / √ (2 × 1.6726 × 10⁻²⁷ × 4.0365 × 10⁻²⁰ )
βp = 6.626 × 10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶
βp = 6.626 × 10⁻³⁴ / 1.16201978468527 × 10⁻²³
βp = 5.702140 × 10⁻¹¹ m
A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe
Answer:
0.0127m
Explanation:
Using
Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m
The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm