Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.8 AA , how many turns of wire would you need
Answer:
The number of turns of wire needed is 3536 turns.
Explanation:
Given;
length of the wire, L = 8 cm = 0.08 m
magnetic field on the wire, B = 0.1 T
current in the wire, I = 1.8 A
The magnetic field produced by a solenoid is calculated as;
B = μ₀ n I
where;
n is the number of turns per length = N / L
μ₀ is permeability of free space = 4π x 10⁻⁷ N/A²
[tex]B = \frac{\mu_o N I}{L} \\\\N = \frac{BL}{\mu_o I} \\\\N = \frac{0.1 *0.08}{4\pi*10^{-7} *1.8} \\\\N = 3536.32 \ turns[/tex]
Therefore, the number of turns of wire needed is 3536 turns.
Question 8 of 10
On which parts of the heating curve for water does adding thermal energy
mainly cause the particles to move faster?
200
150 -
B
To
100
Temperature ('C)
A
50
С
0
-50
10
40
50
60
70
Time (min)
O A. C and D
B. A and B
O O O O
O C. Band C
OD. B and D
Answer:
The correct answer is A
Explanation:
In this exercise we are given a graph of temperature versus time.
In calorimeter processes there are two types
* one that when giving thermal energy to the system its temperature increases, this fundamentally due to the greater kinetic energy of the molecular ones, this process observes in the graphs as a straight line of constant slope
* A process donates all the thermal energy that is introduced is cracked in breaking the molecular bonds, taking matter from one thermodynamic state to another, for example: liquid to gas.
This process in curves as a horizontal line, that is, there is no temperature change,
When analyzing the graph shown, parts C and D are the one that show a change in temperature with thermal energy. The correct answer is A
Answer:
C and D
Explanation:
Just took the quiz
Gravitational Force: Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one
Answer:
F' = F
Hence, the magnitude of the attractive force remains same.
Explanation:
The force of attraction between two bodies is given by Newton's Gravitational Law:
F = Gm₁m₂/r² --------------- equation 1
where,
F = Force of attraction between balls
G = Universal Gravitational Constant
m₁ = mass of first ball
m₂ = mass of 2nd ball
r = distance between balls
Now, we double the masses of both balls and the separation between them. So, the force of attraction becomes:
F' = Gm₁'m₂'/r'²
here,
m₁' = 2 m₁
m₂' = 2 m₂
r' = 2 r
Therefore,
F' = G(2 m₁)(2 m₂)/(2 r)²
F' = Gm₁m₂/r²
using equation 1:
F' = F
Hence, the magnitude of the attractive force remains same.
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be
Answer:
The current will be 18 A
Explanation:
Given;
potential difference, V = 10 V
current between the resistor, I = 2 A
Apply ohm's law;
V = IR
R = V / I
R = 10 / 2
R = 5Ω
Resistance is given as;
[tex]R = \frac{\rho l}{A}[/tex]
where;
ρ is resistivity
l is length
A is area
[tex]R = \frac{\rho l}{A} \\\\R = \frac{\rho l}{\pi r^2} = \frac{\rho l}{\pi (\frac{d}{2}) ^2} = \frac{\rho l}{\pi (\frac{d^2}{4}) }\\\\R = \frac{4*\rho l}{\pi d^2} \\\\R = (\frac{4*\rho l}{\pi } )\frac{1}{d^2} \\\\R = (k)\frac{1}{d^2} \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = \frac{R_1d_1^2}{d_2^2}[/tex]
When the diameter of the resistor is tripled
d₂ = 3d₁
[tex]R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms[/tex]
The current is now calculated as;
Apply ohms law;
V = IR
I = V / R
I = 10 / 0.556
I = 17.99 A
I = 18 A
Therefore, the current will be 18 A
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]
A positively charged particle has a velocity in the negative z direction at a certain point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data?
a. Bx is positive
b. Bz is positive
c. By is negative
d. By is positive
e. Bx is negative
Answer:
a. Bx is positive
Explanation:
See attached file
A flat slab of material (nm = 2.2) is d = 0.45 m thick. A beam of light in air (na = 1) is incident on the material with an angle θa = 46 degrees with respect to the surface's normal.
Numerically, what is the displacement, D, of the beam when it exits the slab?
Answer:
Explanation:
Formula of lateral displacement
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
t is thickness of slab , i and r are angle of incidence and refraction respectively .
Given t = .45 m
sin i / sin r = 2.2
sin 46 / sin r = 2.2
sin r = .719 / 2.2 = .327
r = 19°
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
[tex]S_{lateral}=\frac{.45}{cos19} \times sin(46-19)[/tex]
= .45 x .454 / .9455
= .216 m
= 21.6 cm .
The displacement, D, of the beam when it exits the slab is; 21.65 cm.
We are given;
Refractive index of slab material; nm = 2.2
Thickness of slab; t = 0.45 m
Refractive index of air; na = 1
Angle of incidence; θa = 46°
From snell's law, we can calculate the angle of refraction from;
na × sin θa = nm × sin θm
Thus;
1 × sin 46 = 2.2 × sin θm
0.7193 = 2.2 × sin θm
sin θm = 0.7193/2.2
θm = sin^(-1) 0.32695
θm = 19.08°
Formula for the displacement of the beam is;
D = (t/cos θm) × sin (θa - θm)
Plugging in the relevant values gives;
D = (0.45/cos 19.08) × sin (46 - 19.08)
D = 0.4783 × 0.4527
D = 0.2165m = 21.65 cm
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If a 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft), how long could she power her 0.4 watt flashlight
Answer: 3217.79 hours.
Explanation:
Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).
Power = 0.4 watt
Mass of climber = 140 lb
= 140 x 0.4535 kg [∵ 1 lb= 0.4535 kg]
⇒ Mass of climber (m) = 63.50 kg
Let [tex]h_1=29,029\ ft= 8848.04\ m\ \ \ \ [ 1 ft=0.3048\ m ][/tex] and [tex]h_2= 4,600 ft = 1402.08\ m[/tex]
Now, Energy saved =[tex]mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J[/tex]
[tex]\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}[/tex]
Hence, she can power her 0.4 watt flashlight for 3217.79 hours.
An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 42 mm and 74 mm. The electric potential of the inner conductor with respect to the outer conductor is -308 V ( = 1/4πε0 = 8.99 × 10^9 N · m^2/C^2).
The maximum energy density of the capacitor is closest to:_______
Correct answer is 2.7 x 10^-3 J/m3
I hope that helps ! <33
The maximum energy density of the capacitor is closest to: 2.7 x 10^-3 J/m3.
What is meant by the energy density of a capacitor?Energy density is defined as the total energy per unit volume of the capacitor. Since, Now, for a parallel plate capacitor, A × d = Volume of space between plates to which electric field E = V / d is confined. Therefore, Energy is stored per unit volume.
How do you calculate energy density?All Answers (14) Energy density is equal to 1/2*C*V2/weight, where C is the capacitance you computed and V should be your nominal voltage (i.e 2.7 V). Power Density is V2/4/ESR/weight, where ESR is the equivalent series resistance.
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A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object placed midway between them.
Answer:
The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Given;
first object with mass, m₁ = 285 kg
second object with mass, m₂ = 585 kg
distance between the two objects, r = 4.3 m
The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m
Gravitational force between the first object and the 42 kg object;
[tex]F = \frac{GMm}{r^2}[/tex]
where;
G = 6.67 x 10⁻¹¹ Nm²kg⁻²
[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]
Gravitational force between the second object and the 42 kg object
[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]
Magnitude of net gravitational force exerted on 42kg object;
F = 3.545x 10⁻⁷ N - 1.727 x 10⁻⁷ N
F = 1.818 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?
a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.
Answer:
the greatest intensity is obtained from c
Explanation:
An electromagnetic wave stagnant by the expression
E = E₀ sin (kx -wt)
when two waves meet their electric fields add up
E_total = E₁ + E₂
the intensity is
I = E_total . E_total
I = E₁² + E₂² + 2E₁ E₂ cos θ
where θ is the phase angle between the two rays
Let's examine the two waves
in this case E₁ = E₂ = E₀
I = Eo2 + Eo2 + 2 E₀ E₀ coasts
I = E₀² (2 + 2 cos θ )
I = 2 I₀ (1 + cos θ )
let's apply this expression to different cases
a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀
b) cos π = -1 this implies that I_total = 0
c) the cosine is 1,
I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ
I = E₀² (5 +4 cos θ)
I = E₀² 9
I = 9 Io
d) in this case the cos pi = -1
I = E₀² (5 -4)
I = I₀
e) we rewrite the equation
I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ
I = Eo2 (10 + 6 cos θ)
cos π = -1
I = E₀² (10-6)
I = 4 I₀
the greatest intensity is obtained from c
The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.
What is an amplitude?An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.
In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.
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An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340V, what current would the motor need to draw from the battery? Neglect any energy losses in getting energy from the battery to the motor.
Answer:
Current = 132.35 A
The motor needs to draw 132.35 Amperes current from the battery.
Explanation:
The formula of electric power is given as follows:
Power = (Voltage)(Current)
Current = Power/Voltage
In this question, we have:
Power = 45 KW = 45000 W
Voltage of Battery Pack = 340 V
Current needed to be drawn = ?
Therefore,
Current = 45000 W/340 V
Current = 132.35 A
The motor needs to draw 132.35 Amperes current from the battery.
Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?
Answer:
The wavelength is [tex]\lambda = 1029 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the left light is [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]
The voltage across A and B is [tex]V_{AB } = 1.2775 \ V[/tex]
Let the stopping potential at A be [tex]V_A[/tex] and the electric potential at B be [tex]V_B[/tex]
The voltage across A and B is mathematically represented as
[tex]V_{AB} = V_A - V_B[/tex]
Now According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side in terms of electron volt is mathematically represented as
[tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]
Where W is the work function of the metal
h is the Planck constant with values [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
And the stopping potential at B for the ejected electron from the right side in terms of electron volt is mathematically represented as
[tex]eV_B = \frac{h * c}{\lambda } - W[/tex]
So
[tex]eV_{AB} = eV_A - eV_B[/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]
=> [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]
=> [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]
Where e is the charge on an electron with the value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]
=> [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]
=> [tex]\lambda = 1.029 *10^{-6} \ m[/tex]
=> [tex]\lambda = 1029 nm[/tex]
A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball The disk starts from rest and after 3 seconds, linear a(3)=1.80m/s2. Find A and then write an expression for the angular acceleration α(t).
Answer:
The value for A is A= 0.6
The angular acceleration is [tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]
Explanation:
From the question we are told that
The radius of the disk is [tex]r = 25.0 \ cm = 0.25 \ m[/tex]
The linear acceleration is [tex]a(t) = At[/tex]
At time [tex]t = 3 \ s[/tex]
[tex]a(3) = 1.80 \ m/s^2[/tex]
Generally angular acceleration is mathematically represented as
[tex]\alpha(t) = \frac{a(t)}{r}[/tex]
Now at t = 3 seconds
a(3) = A * 3
=> 1.80 = A * 3
=.> A = 0.6
So therefore
a(t) = 0.6 t
Now substituting this into formula for angular acceleration
[tex]\alpha (t) = \frac{0.6 t }{R}[/tex]
substituting for r
[tex]\alpha (t) = \frac{0.6 t }{0.25}[/tex]
[tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]
It took a student 30 minutes to drive from his home to campus on
Monday, and it took him 20 minutes on Tuesday driving the same
route. If on Monday he drove 36 mi/hr on average, what was his
average speed on Tuesday?
O 12 mi/hr
O 18 mi/hr
O 48 mi/hr
O 54 mi/hr
O 72 mi/hr
Answer:
48 i believe
Explanation:
An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial, the initial velocity is v0 and in the second it is 4v0. In the second trial, A : the amplitude is twice as great and the maximum acceleration is half as great. B : both the amplitude and the maximum acceleration are four times as great. C : the amplitude is half as great and the maximum acceleration is twice as great. D : both the amplitude and the maximum acceleration are twice as great. E : the amplitude is four times as great and the maximum acceleration is twice as great.
Explanation:
It is given that, in the first trial, the initial velocity is [tex]v_o[/tex] and in the second it is [tex]4v_o[/tex].
The total energy of the system remains constant. So,
[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=\text{constant}[/tex] ....(1)
x is amplitude
It means that the amplitude is directly proportional to velocity. If velcoity increases to four times, then the amplitude also becomes 4 times.
Differentiating equation (1) we get :
[tex]mv\dfrac{dv}{dt}+kx\dfrac{dx}{dt}=0[/tex]
Since,
[tex]\dfrac{dv}{dt}=a,\ \text{acceleration}[/tex] and [tex]\dfrac{dx}{dt}=v,\ \text{velocity}[/tex]
So,
[tex]mva+kxv=0[/tex]
It means that the acceleration is also proportional to the amplitude. So, acceleration also becomes 4 times.
Hence, the correct option is (B) "both the amplitude and the maximum acceleration are four times as great"
Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
Two sound waves W1 and W2, of the same wavelength interfere destructively at point P. The waves originate from two in phase speakers. W1 travels 36m and W2 travels 24m before reaching point P. Which of the following values could be the wave length of the sound waves?
a. 24m
b. 12m
c. 6m
d. 4m
Answer:
a. 24 m
Explanation:
Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.
An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass p=7.95 kg and radius p=0.89 m. The hanging masses are L=32.0 kg and R=17.8 kg. Calculate the magnitude of the masses' acceleration and the tension in the left and right ends of the rope, L and R , respectively.
Answer:
Acceleration(a) = 2.588 m/s²
TL = 230.784 N
TR = 220.5 N
Explanation:
Given:
M = 7.95 kg
mL = 32 kg
mR = 17.8 kg
g = 9.8 m/s²
Find:
Acceleration(a)
TL
TR
Computation:
Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]
Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]
Acceleration(a) = [139.16] / [53.775]
Acceleration(a) = 2.588 m/s²
TL = mL(g-a)
TL = 32(9.8-2.588)
TL = 230.784 N
TR = mR(g+a)
TR = 17.8(9.8+2.588)
TR = 220.5 N
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton box 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda = 622 nm[/tex]
Explanation:
From the question we are told that
The distance of the slit to the screen is [tex]D = 5 \ m[/tex]
The order of the fringe is m = 6
The distance between the slit is [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]
The fringe distance is [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]
Generally the for a dark fringe the fringe distance is mathematically represented as
[tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]
=> [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]
substituting values
=> [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]
=> [tex]\lambda = 6.22 *10^{-7} \ m[/tex]
[tex]\lambda = 622 nm[/tex]
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the electric field between the two plates?
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
ransverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What is the mass of the string
Answer:
0.055 kg
Explanation:
According to the given situation the solution of the mass of the string is shown below:-
Speed of the wave is
[tex]v = \sqrt{\frac{F_T\times Length\ of\ string}{Mass\ of\ string}}[/tex]
[tex]30.0 m/s = \sqrt{\frac{10 kg m/s^2\times 5.00 m}{Mass\ of\ string}[/tex]
Mass of string is
[tex]= \sqrt{\frac{10 kg m^2/s^2\times 5.00 m}{900 m^2 s^2}[/tex]
After solving the above equation we will get the result that is
= 0.055 kg
Therefore for calculating the mass of the string we simply applied the above formula.
If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you predict about the surface elevation for 50 km thick crust with an average density of 2.8 g/cm3
Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
A block of mass m is suspended by a vertically oriented spring. If the mass of a block is increased to 4m, how does the frequency of oscillation change, if at all
Answer:
The frequency will be reduced by a factor of √2/2
Explanation:
Pls see attached file
The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Let the initial mass of block be m.
And new mass is, 4m.
The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,
[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
Here,
k is the spring constant.
If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,
[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]
Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Learn more about the frequency of oscillation here:
https://brainly.com/question/14316711
Suppose a point charge is located at the center of a spherical surface. The electric field at the surface of the sphere and the total flux through the sphere are determined. Now the radius of the sphere is halved. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere
Answer:
The magnitude of flux remains the same, and the field increases.
Explanation:
This is because the number of field lines leaving the sphere remains constant and the electric field increases because the line density increases
The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the electric field strength? (The value is surprisingly large, but correct.) You may assume a uniform E-field.
Answer:
The magnitude of the electric field intensity is [tex]E = 7.89 *10^{6} \ V/m[/tex]
Explanation:
From the question we are told that
The voltage is [tex]\epsilon = 72.7 \ mV = 72.7 *10^{-3} V[/tex]
The thickness of the membrane is [tex]t = 9.22 \ nm = 9.22 *10^{-9} \ m[/tex]
Generally the electric field intensity is mathematically represented as
[tex]E = \frac{\epsilon }{t}[/tex]
substituting values
[tex]E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}[/tex]
[tex]E = 7.89 *10^{6} \ V/m[/tex]
If an object is placed at a distance of 10 cm in front of a concave mirror of focal length 4 cm, find the position and characteristics of the image formed. Also, find the magnification.
Answer:
Explanation:
Focal length f = - 4 cm
Object distance u = - 10 cm
v , image distance = ?
Mirror formula
[tex]\frac{1}{v} +\frac{1}{u} = \frac{1}{f}[/tex]
Putting the given values
[tex]\frac{1}{v} - \frac{1}{10} = - \frac{1}{4}[/tex]
[tex]\frac{1}{v}= - \frac{3}{20}[/tex]
v = - 6.67 cm .
magnification
m = v / u
= - 6.67 / - 10
= .667
so image will be smaller in size in comparison with size of object .
Characteristics will be that ,
1 ) it will be inverted and
2 ) it will be real image .