Answer:
λ = 5.773 x 10⁻⁷ m = 577.3 nm
Explanation:
In order to solve this problem we will use the grating equation:
mλ = d Sin θ
where,
m = order = 3
λ = wavelength of light = ?
d = slit separation = 2 μm = 2 x 10⁻⁶ m
θ = angle = 60°
Therefore,
(3)λ = (2 x 10⁻⁶ m)Sin 60°
λ = 1.732 x 10⁻⁶ m/3
λ = 5.773 x 10⁻⁷ m = 577.3 nm
1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the direction of the acceleration of the car? *
A- outside track, and normal to track
B- towards the center and normal to the track
C- up
D- down
Answer:
B.
Explanation:
What resistance should be added in series with a 3.0-H inductor to complete an LR circuit with a time constant of 4.0 ms? A)0.75 k Ω Β) 12 Ω C) 0.75 Ω D) 2.5 Ω
Answer:
O.75KΩ
Explanation:
We measure the time constant τ, using the formula τ = L/R,
t is in seconds, then we have R to be the value of the resistor which is measured in ohms and also L is the value of the inductor which is measured in Henries.
Since t = L/R
We make R subject of the formula
R = L/τ
= 3/4x10-3
= 0.00075
= 0.75 KΩ
So we have it that the first Option (A) is the correct answer to the question
The resistance to be added is required.
The resistance added should be A. [tex]0.75\ \text{k}\Omega[/tex]
L = Inductance = 3 H
[tex]\tau[/tex] = Time constant = 4 ms
R = Resistance
Time constant is given by
[tex]\tau=\dfrac{L}{R}\\\Rightarrow R=\dfrac{L}{\tau}\\\Rightarrow R=\dfrac{3}{4\times 10^{-3}}=750\ \Omega=0.75\ \text{k}\Omega[/tex]
The resistance added should be [tex]0.75\ \text{k}\Omega[/tex]
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What ate the two safety precautions that should be taken before driving your car?
Answer:
When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.
Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.
A sample of an ideal gas has a volume of 0.0100 m^3, a pressure of 100 x 10^3 Pa, and a temperature of 300K. What is the number of moles in the sample of gas?
Answer:
Explanation:
pV = nrT
n = PV/RT
n = (100*10^3)(.01)/(300*0.082057)
n = 40.62 moles
A cylindrical wire of radius 2 mm carries a current of 3.0 A. The potential difference between points on the wire that are 44 m apart is 3.8 V.
Required:
a. What is the electric field in the wire?
b. What is the resistivity of the material of which the wire is made?
Answer:
a. E = 86.36 x 10⁻³ V/m = 86.36 mV/m
b. ρ = 3.6 x 10⁻⁷ Ωm
Explanation:
a.
The electric field in terms of the voltage is given by the following formula:
E = V/d
where,
E = Electric Field in the Wire = ?
V = Potential Difference = 3.8 V
d = distance between the points = 44 m
Therefore,
E = 3.8 V/44 m
E = 86.36 x 10⁻³ V/m = 86.36 mV/m
b.
Now, from Ohm's Law:
V = IR
R = V/I
where,
R = Resistance of wire = ?
I = Current = 3 A
Therefore,
R = 3.8 V/3 A
R = 1.27 Ω
Now, the resistance of a wire can be given as:
R = ρL/A
where,
ρ = resistivity of material = ?
L = Length = 44 m
A = Cross-sectional area = πr² = π(0.002 m)² = 1.25 x 10⁻⁵ m²
Therefore,
1.27 Ω = ρ*44 m/1.25 x 10⁻⁵ m²
(1.27 Ω)(1.25 x 10⁻⁵ m²)/44 m = ρ
ρ = 3.6 x 10⁻⁷ Ωm
An inductor is connected to a 26.5 Hz power supply that produces a 41.2 V rms voltage. What minimum inductance is needed to keep the maximum current in the circuit below 126 mA?
Answer:
The minimum inductance needed is 2.78 H
Explanation:
Given;
frequency of the AC, f = 26.5 Hz
the root mean square voltage in the circuit, [tex]V_{rms}[/tex] = 41.2 V
the maximum current in the circuit, I₀ = 126 mA
The root mean square current is given by;
[tex]I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_{rms} = \frac{126*10^{-3}}{\sqrt{2} }\\\\I_{rms} =0.0891 \ A[/tex]
The inductive reactance is given by;
[tex]X_l = \frac{V_{rms}}{I_{rms}} \\\\X_l= \frac{41.2}{0.0891}\\\\X_l = 462.4 \ ohms[/tex]
The minimum inductance needed is given by;
[tex]X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2\pi f}\\\\L = \frac{462.4}{2\pi *26.5}\\\\L = 2.78 \ H[/tex]
Therefore, the minimum inductance needed is 2.78 H
HELP ASAP!
What happens to a circuit's resistance (R), voltage (V), and current (I) when you increase the length of the wire in the circuit?
Answer:
B R is constant V increases /increases
If one increase the length of the wire in the circuit, then, R increases, V decreases, I decreases. The correct option is D.
The resistance (R) of a wire increases as its length is increased in a circuit. This is because longer wires have more resistance because resistance increases with length.
Ohm's Law states that the current (I) in the circuit will drop if the resistance (R) rises while the voltage (V) stays constant. This is due to the inverse proportionality between current and resistance.
Thus, the correct option is D.
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Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat of 1,700 J/(kg°C).
Answer:
Q = 425 kJ
Explanation:
Given that,
Mass, m = 25 kg
The clown doll head that heats up from 25°C to 35°C
The specific heat is 1700 J/kg°C
We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :
[tex]Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ[/tex]
So, 425 kJ of thermal energy is severed.
a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the post. determine the stress in the post
Answer:
The stress is [tex]\sigma = 1.218*10^{6} \ N/m^2[/tex]
Explanation:
From the question we are told that
The diameter of the post is [tex]d = 29 \ cm = 0.29 \ m[/tex]
The length is [tex]L = 2.0 \ m[/tex]
The weight of the loading mass
Generally the radius of the post is mathematically represented as
[tex]r = \frac{0.29}{2}[/tex]
=> [tex]r = 0.145 \ m[/tex]
Generally the area of the post is
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.14 * 0.145 ^2[/tex]
=> [tex]A = 0.066 \ m^2[/tex]
Generally the weight exerted by the load is mathematically represented as
[tex]F = m * g[/tex]
=> [tex]F = 8200 * 9.8[/tex]
=> [tex]F = 80360 \ N[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]\sigma = \frac{80360 }{0.066}[/tex]
=> [tex]\sigma = 1.218*10^{6} \ N/m^2[/tex]
Use the information below for the next five questions:
An open organ pipe emits B (494 Hz) when the temperature is 14°C. The speed of sound in air is v≈(331 + 0.60T)m/s, where T is the temperature in °C
Determine the length of the pipe.
What is the wavelength of the fundamental standing wave in the pipe?
What is frequency of the fundamental standing wave in the pipe?
What is the frequency in the traveling sound wave produced in the outside air?
What is the wavelength in the traveling sound wave produced in the outside air?
How far from the mouthpiece of the flute should the hole be that must be uncovered to play D above middle C at 294 Hz? The speed of sound in air is 343 m/s.
Answer: Please see answer in explanation column.
Explanation:
Given that
v≈(331 + 0.60T)m/s
where Temperature, T = 14°C
v≈(331 + 0.60 x 14)m/s
v =331+ 8.4 = 339.4m/s
In our solvings, note that
f= frequency
λ=wavelength
L = length
v= speed of sound
a) Length of the pipe is calculated using the fundamental frequency formulae that
f=v/2L
Length = v/ 2f
= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m
b) wavelength of the fundamental standing wave in the pipe
L = nλ/2,
λ = 2L/ n
λ( wavelength )= 2 x 0.3435/ 1
= 0.687m
c) frequency of the fundamental standing wave in the pipe
F = v/ λ
= 339.4m/s/0.687m=
494.03s^-1 = 494 Hz
d) the frequency in the traveling sound wave produced in the outside air.
This is the same as the frequency in the open organ pipe = 494Hz
e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m
f) To play D above middle c . the distance is given by
L =v/ 2 f
= 343/ 2 x 294
=0.583m
A wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of magnitude 11.4 à 10-3 T. The magnitude of the force on 11.4 cm of this wire is:____.a) 11.4 * 10^-3 N
b) 0.130 N
c) 1.48 * 10^-2 N
d) 2.93 * 10^-3 N
Answer:
(d) 2.93 x 10⁻³ N
Explanation:
Given;
current in the wire, I = 11.4 A
angle of inclination, θ = 11.4⁰
magnetic field on the wire, B = 11. 4 x 10⁻³
length of the wire, L = 11.4 cm = 0.114 m
The magnitude of magnetic force on the wire is given by;
F = BILsinθ
F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)
F = 0.00293 N
F = 2.93 x 10⁻³ N
Therefore, the correct option is "D"
A seated musician plays a C4 note at 262 Hz . How much time Δ does it take for 346 air pressure maxima to pass a stationary listener?
Answer:
t = 1.32 s
Explanation:
We are given;. Frequency of C4 note; F_c = 262 Hz
In conversions, we know that 1 Hz = 1 cycle/s
Thus, F_c = 262 cycles/s
Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.
346 air pressure maxima denotes that the air pressure maxima is 346 cycles.
Thus, time will be;
t = 346 cycles/262 cycles/s
t = 1.32 s
The time taken for the musical note to pass the stationary listener is 1.32 s.
The given parameters:
frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346The frequency of a sound wave is defined as the number of cycles completed per second by the wave.
[tex]F = \frac{n}{t}[/tex]
where;
t is the time to compete the maximum cycleThe time taken for the musical note to pass the stationary listener is calculated as follows;
[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]
Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.
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How do lenses and mirrors compare in their interactions with light?
A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.
Answer:
C. lenses refract light; mirrors do not
This question involves the concepts of reflection and refraction.
The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".
LENSES AND MIRRORSWhen it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the bending of light rays, while passing through a medium, without any rebound or absorption.
Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.
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carrier concentration for n type
Answer:
Consider an n-type silicon semiconductor at T = 300°K in which Nd = 1016 cm-3 and Na = 0. The intrinsic carrier concentration is assumed to be ni = 1.5 x 1010 cm-3. - Comment Nd >> ni, so that the thermal-equilibrium majority carrier electron concentration is essentially equal to the donor impurity concentration.
Explanation:
During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young's modulus of 1.5×109 N/m2?
Answer:
[tex]3978873.58\ \text{Pa}[/tex]
[tex]0.00002122\ \text{m}[/tex]
Explanation:
F = Force = 5000 N
r = Radius of circular cross section = 2 cm
l = Length of disk = 0.8 cm
A = Area = [tex]\pi r^2[/tex]
Y = Young's modulus = [tex]1.5\times 10^9\ \text{N/m}^2[/tex]
Pressure is given by
[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{5000}{\pi (2\times 10^{-2})^2}\\\Rightarrow P=3978873.58\ \text{Pa}[/tex]
The pressure on the cross section is [tex]3978873.58\ \text{Pa}[/tex]
The change in length of the cross section is given by
[tex]\Delta L=\dfrac{PL}{Y}\\\Rightarrow \Delta L=\dfrac{3978873.58\times 0.8\times 10^{-2}}{1.5\times 10^9}\\\Rightarrow \Delta L=0.00002122\ \text{m}[/tex]
The deformation produced is [tex]0.00002122\ \text{m}[/tex]
A nonmechanical water meter could utilize the Hall effect by applying a magnetic field across a metal pipe and measuring the Hall voltage produced.What is the average fluid velocity in m/s for a 4.25 cm diameter pipe, if a 0.575 T field across it creates a 60.0 mV Hall voltage?
Answer:
The velocity is [tex]v =2.455 \ m/s[/tex]
Explanation:
From the question we are told that
The diameter of the pipe is [tex]d = 4.25 \ cm = 0.0425 \ m[/tex]
The magnetic field is [tex]B = 0.575 \ T[/tex]
The hall voltage is [tex]V_H = 60.0 mV = 60 *10^{-3} \ V[/tex]
Generally the average fluid velocity is mathematically represented as
[tex]v = \frac{V}{ B * d }[/tex]
=> [tex]v = \frac{60*10^{-3}}{ 0.575 * 0.0425 }[/tex]
=> [tex]v =2.455 \ m/s[/tex]
Suppose two parallel-plate capacitors have the same charge Q, but the area of capacitor 1 is A and the area of capacitor 2 is 2 A.If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2?Express your answer in terms of V but do not type in the symbol "V"
Answer:
V' = V/2
Explanation:
The voltage across a parallel plate capacitor is given as follows:
V = Q/C
where,
V = Voltage across capacitor
Q = Charge on Capacitor
C = Capacitance of Capacitor = A∈₀/d
Therefore,
V = Qd/A∈₀
where,
A = Area of plate
d = distance between plates
∈₀ = permittivity of free space
FOR CAPACITOR 1:
Q = Q
d = d
A = A
V = V
Therefore,
V = Qd/A∈₀ --------------- equation (1)
FOR CAPACITOR 2:
V' = ?
Q' = Q
d' = d
A' = 2A
Therefore,
V' = Q'd'/A'∈₀
V' = Qd/2A∈₀
V' = (1/2)(Qd/A∈₀)
using equation (1):
V' = V/2
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is launched to a maximum height 50.2 cm. How much should the spring be compressed to send the ball twice as high?
We know, by conservation of energy :
[tex]\dfrac{kx^2}{2}=mgh[/tex]
Therefore,
[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}[/tex]
Putting given values, we get :
[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm[/tex]
Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.
Hence, this is the required solution.
A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the horizontal. The ball rolls without slipping down the incline and at the bottom has a speed of 4.9 m/s. How many revolutions does the ball rotate through as it rolls down the incline
Answer:
Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!
The ball rotates 6.78 revolutions.
Explanation:
Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!
At the bottom the ball has the following angular speed:
[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]
Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:
[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]
To find the revolutions we need the time, which can be found using the following equation:
[tex] v_{f} = v_{0} + at [/tex]
[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)
So first, we need to find the acceleration:
[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]
Since it starts from rest (v₀ = 0):
[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]
Finally, we can find the revolutions:
[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]
Therefore, the ball rotates 6.78 revolutions.
I hope it helps you!
3. Sarah drops a 10.0 kg objed from a 200 m high bridge over a river. a) What work is done by the objec as a result of its fall?
Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled eastward at 8 m/s while billiard ball B originally traveled westward at 2 m/s. Calculate the speed and direction of each ball after the collision.
Answer:
lucky mauld mauldgomary was an british poet...
It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s).
Let B = 0.86 T , I = 2300 A , m = 20 kg , and L = 55 cm . For simplicity, assume the net force on the object is equal to the magnetic force, as in parts A and B, even though gravity plays an important role in an actual launch into space.
Express your answer using two significant figures.
Answer:
The distance of the bar D = 1153 km
Explanation:
The electric force is the one that takes place between electric charges.
The electric force with which two point charges are attracted or repelled at rest is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them.
Recall that:
Electrical force(F) = I*B*L
where;
I = the current,
B = the magnetic field strength,
L = the length of the bar
However;
From the second equation of motion,
F = Ma
Since; (F) = I*B*L
Then,
Ma = IBL,
where;
M is the mass;
a is the acceleration
Making the acceleration (a) the subject of the formula, we have
a = IBL/M
Similarly;
From the third equation of motion;
v^2= u^2+2as,
where v and u are the final velocity and the initial velocity respectively
Here u = 0
Also; let distance s = D
Then
v^2 = 2aD
where;
a = IBL/M
Making the distance D the subject of the formula, we get:
D = v^2/2a = v2*M/(2IBL)
D = 11200² × 20/(2×2300×0.86×0.55)
D = 1153047.155 m
D = 1153 km
A mass m at the end of a spring of spring constant k is undergoing simple harmonic oscillations with amplitude A.
Part (a) At what positive value of displacement x in terms of A is the potential energy 1/9 of the total mechanical energy?
Part (b) What fraction of the total mechanical energy is kinetic if the displacement is 1/2 the amplitude?
Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3?
Answer:
a) The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].
b) The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) The maximum kinetic energy is increased by a factor of 9.
Explanation:
a) From Mechanical Physics, we remember that the mechanical energy of mass-spring system ([tex]E[/tex]), measured in joules, is the sum of the translational kinetic energy ([tex]K[/tex]), measured in joules, and elastic potential energy ([tex]U[/tex]), measured in joules. That is:
[tex]E = K + U[/tex] (1)
By definitions of translational kinetic energy and elastic potential energy, we have the following expressions:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
[tex]U = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (3)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity of the mass, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Elongation of the spring, measured in meters.
If we know that [tex]U = \frac{1}{9}\cdot E[/tex], [tex]k = k[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then:
[tex]\frac{1}{18}\cdot k\cdot A^{2} = \frac{1}{2}\cdot k\cdot x^{2}[/tex]
[tex]\frac{1}{9}\cdot A^{2} = x^{2}[/tex]
[tex]x= \frac{1}{3}\cdot A[/tex]
The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].
b) If we know that [tex]k = k[/tex], [tex]x = \frac{1}{2}\cdot A[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then the equation of energy conservation associated with the system is:
[tex]\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{4}\cdot k\cdot A^{2}+K[/tex]
[tex]K = \frac{1}{4}\cdot k\cdot A^{2}[/tex]
The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.
c) From the Energy Conservation equation associated with the system, we know that increasing the amplitude by a factor of 3 represents an increase in the elastic potential energy by a factor of 9. Then, the maximum kinetic energy is increased by a factor of 9.
Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
Which environment is least likely to support protists
A soil
B open ocean
C shallow pod
D organisms blood
Answer:
A: Soil
Explanation:
Protists need a moist environment to survive, and shallow ponds, oceans, and blood is all moist. So, the answer would be the soil, because that is the least moist environment out of these options.
A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?
Answer:
The number is [tex]N = 300[/tex]
Explanation:
From the question we are told that
The net charge is [tex]Q = -4.8 *10^{-17 } \ C[/tex]
Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]
Generally the number of excess electrons is mathematically represented as
[tex]N = \frac{Q}{e}[/tex]
=> [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]
=> [tex]N = 300[/tex]
define these terms about speed and state their units
speed
distance covered
time taken
Explanation:
Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.
Speed = [tex]\frac{distance}{time}[/tex]
The unit is m/s or km/hr or mile/hr
Distance covered is simply the length of the path traveled.
The unit is m or km or miles
Time taken is the duration of an event.
The unit is seconds or minutes or hour.
What is the electric force between two point charges when 91 = -4e, 92 = +3 e,
and r= 0.05 m?
e = 1.6 x 10-1C, k = 9.00 x 10°Nom/C2)
Kouch
A. -1.1 * 10-24 N
B. 1.1 * 10-24 N
C. 5,5 x 10-25 N
D. -5.5 x 10-25 N
Answer:option B
Explanation:
I have a pen
I have a apple
what do I have now?
Answer:
You have an apple pen. :)
Answer:
I have a pen
I have a apple
apple pen
Explanation:
If the peak wavelength of a star at rest is 615 nm, then what peak wavelength is observed when the star is traveling 2,500,000 m/s toward the Earth.
Answer:
1612903nm
Explanation:
Doppler effect can be regarded as the change in frequency of a wave with respect toobserver that move relative to the wave source
We can expressed as
(λo - λs)/λs = v/s
λo= peak wavelength
λs= peak wavelength observed
C= speed of light
(λo -615×10^9 )/615×10^9 = 2,500,000/(3×10^8)
1.5375=3×10^8λo -184.5
1.5375+184.5=3×10^8λo
186=3×10^8λo
λo=1612903nm
Therefore the peak wavelength that is observed when the star is traveling away from the eart to the velocity given is 1612903nm