Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
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A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
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What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 275 km above the Earth
Answer:
s_400 = 16.5 m , s_700 = 29.4 m
Explanation:
The limit of the human eye's solution is determined by the diffraction limit that is given by the expression
θ = 1.22 λ / D
where you lick the wavelength and D the mediator of the circular aperture.
In our case, the dilated pupil has a diameter of approximately 8 mm = 8 10-3 m and the eye responds to a wavelength between 400 nm and 700 nm.
by introducing these values into the formula
λ = 400 nm θ = 1.22 400 10⁻⁹ / 8 10⁻³ = 6 10⁻⁵ rad
λ = 700 nm θ = 1.22 700 10⁻⁹ / 8 10⁻³-3 = 1.07 10⁻⁴ rad
Now we can use the definition radians
θ= s / R
where s is the supported arc and R is the radius. Let's find the sarcos for each case
λ = 400 nm s_400 = θ R
S_400 = 6 10⁻⁵ 275 10³
s_400 = 16.5 m
λ = 700 nm s_ 700 = 1.07 10⁻⁴ 275 10³
s_700 = 29.4 m
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
soaring birds and glider pilots can remain aloft for hours without expending power. Discuss why this is so.
Answer:
Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection
Explanation:
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?
Answer:
The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
Explanation:
We are told that the particle moves back and forth along a straight line with velocity v(t).
Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.
Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds
The index of refraction of a sugar solution in water is about 1.5, while the index of refraction of air is about 1. What is the critical angle for the total internal reflection of light traveling in a sugar solution surrounded by air
Answer:
The critical angle is [tex]i = 41.84 ^o[/tex]
Explanation:
From the question we are told that
The index of refraction of the sugar solution is [tex]n_s = 1.5[/tex]
The index of refraction of air is [tex]n_a = 1[/tex]
Generally from Snell's law
[tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]
Note that the angle of incidence in this case is equal to the critical angle
Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]
So
[tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]
[tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]
[tex]i = 41.84 ^o[/tex]
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .
A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.
Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.
N/m
Explanation:
Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
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A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago
Answer:
Your answer is( D) - Arago
•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.
[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.
Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]
And final speed is now calculated after clearing it:
[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]
[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Which has more mass electron or ion?
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
A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is
Answer:
0.0081T
Explanation:
The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e
B ∝ I [tex]\frac{N}{L}[/tex]
B = μ₀ I [tex]\frac{N}{L}[/tex] ----------------(i)
Where;
μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²
Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by
L = 2π r
L = 2π (0.042)
L = 0.084π
The number of turns N = 1000
The current in the toroid = 1.7A
Substitute these values into equation (i) to get the magnetic field as follows;
B = 4π x 10⁻⁷ x 1.7 x [tex]\frac{1000}{0.084\pi }[/tex] [cancel out the πs and solve]
B = 0.0081T
The magnetic field along the central radius is 0.0081T
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:
Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus
Answer:
2.55m
Explanation:
Using 1/do+1/di= 1/f
di= (1/f-1/do)^-1
( 1/0.0500-1/0.0510)^-1
= 2.55m
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
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]
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.15 s.
Answer:
206.67NExplanation:
The sum of force along both components x and y is expressed as;
[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]
The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
[tex]a_x = \frac{d^2 x }{dt^2}[/tex]
[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]
Similarly,
[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]
[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]
[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.
Answer:
4.9x10^-6T
Explanation:
See attached file
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz