Answer:
The wavelength is [tex]\lambda = 1805 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]
The distance of the screen is D = 3.0 m
The fringe width is [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]
Generally the fringe width for a bright fringe is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
=> [tex]d = \frac{ \lambda * D }{ y }[/tex]
=> [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]
=> [tex]d = 0.000373 \ m[/tex]
Generally the fringe width for a dark fringe is mathematically represented as
[tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
Here m = 0 for first order dark fringe
So
[tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
looking at which we see that [tex]y_d = y[/tex]
[tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]
=> [tex]\lambda = 1805 *10^{-9} \ m[/tex]
=> [tex]\lambda = 1805 nm[/tex]
If the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are
Answer:
far away
Explanation:
There are different types of eye defect ranging from short sightedness, longsighted, astigmatism, presbyopia etc.
If someone is only able to see close ranged object clearly but not far distant object, then such person is suffering from short sightedness or myopia. This occurs when the light rays entering the eye does not converge on the retina. Instead of converging on the retina, the light ray is formed on a point in front of the retina. This causes the distance from the eye's lens to the retina shorter compared to that of a normal eye. This eye defect is usually corrected using concave lens in order to diverge the rays thereby allowing it to focus on the retina.
Hence, if the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are far away (at a far distant).
How many heartbeats in a typical human lifetime? Enter your answer as a number (NOT as a power of ten) and in one significant figure.
Answer:
20,000,000,000Explanation:
As we've seen, humans have on average a heart rate of around 60 to 70 beats per minute, give or take. We live roughly 70 or so years, giving us just over 2 billion beats all up.Apr
A wave travelling along the positive x-axis side with a
frequency of 8 Hz. Find its period, velocity and the distance covered
along this axis when its wavelength and amplitude are 40 and 15 cm
respectively.
Explanation:
The frequency is given to be f = 8 Hz.
Period is the inverse of frequency.
T = 1/f = 0.125 s
Velocity is wavelength times frequency.
v = λf = (0.40 m) (8 Hz) = 3.2 m/s
The wave travels 3.2 meters every second.
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.
Hope this helps!
Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
Please help!
Much appreciated!
Answer:
F = 2.7×10¯⁶ N.
Explanation:
From the question given:
F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²
Thus we can obtain the value value of F by carrying the operation as follow:
F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²
F = 2.7648×10¯⁷ / 0.1024
F = 2.7×10¯⁶ N.
Therefore, the value of F is 2.7×10¯⁶ N.
what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor
Answer
The effect is that it Decreases the field current IF and increases slope K1
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
We have that for the Question, it can be said that the average induced emf in the coil is
E=0.028565V
From the question we are told
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Generally the equation for the Average emf induced is mathematically given as
[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]
Hence
[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]
Therefore
[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]
E=0.028565V
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A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The distance it moves in the third second.
Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= - 4 m/s
A cart rolls 2 m to the right then rolls back 1 m to the left.
a. What is the total distance rolled by the cart?
Explanation:
It is given that,
Distance covered by the cart to the right is 2 m
Distance covered by the cart to the left is 1 m
We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.
Total distance = 2 m + 1 m
D = 3 m
The cart rolled to a total distance of 3 m.
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave travels at 5 000 m/s. If P-waves are received at a seismic station 1.00 minute before an S-wave arrives, how far away is the earthquake center?
Assuming constant speeds, the P-wave covers a distance d in time t such that
9000 m/s = d/(60 t)
while the S-wave covers the same distance after 1 more minute so that
5000 m/s = d/(60(t + 1))
Now,
d = 540,000 t
d = 300,000(t + 1) = 300,000 t + 300,000
Solve for t in the first equation and substitute it into the second equation, then solve for d :
t = d/540,000
d = 300,000/540,000 d + 300,000
4/9 d = 300,000
d = 675,000
So the earthquake center is 675,000 m away from the seismic station.
A parallel plate air-filled capacitor is being charged.The circular plates have a radius of 4.00cm, and at a particular instant the conduction current in the wiresis 0.280A.
a. What is the displacement current density (jD) in theair space between the plates?
b. What is the rate at which the electric field between the platesis changing?
c. What is the induced magnetic field between the plates at adistance of 2.00 cm from the axis?
d. What is the induced magnetic field between the plates at adistance of 1.00 cm form the axis?
Answer:
Given that
capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴
displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m
( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2
( B ) . displacement current density j
D = ε( dE / dt )the rate at which the electric field between the plates is changing(
dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer
What is the force that attracts objects with mass toward each other?
Explanation:
gravitional force attracts objects with mass toward each other.
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
The direction for each field vector is perpendicular to equipotential lines.
Take a snapshot of the simulation showing equipotential lines and paste to a word document.
....................
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 10-7 N/A2 .) A. 0.02219 m B. 327 m C. 52 m D. 0.00199 m
Answer:
A. 2.2*10^-2m
Explanation:
Using
Area = length x L/ uo xN²
So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*
3000²)
A = 17.5*10^-3/ 1.13*10^-5
= 15.5*10^-2m²
Area= π r ²
15.5E-2/3.142 = r²
2.2*10^2m
Explanation:
UV radiaGon having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kineGc energy of the ejected photoelectrons
Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".
Kinetic energyAccording to the question,
Speed of light, c = 3 × 10⁸ m/s
Wavelength, λ = 120 nm or,
= 1.2 × 10⁻⁷ m
Plank's Constant, h = 6.626 × 10⁻³⁴ J.s
Now,
The energy of photon will be:
→ E = [tex]\frac{hc}{\lambda}[/tex]
By substituting the values,
= [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]
= [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]
= 10.35 eV
By using Einstein's Photoelectric equation,
Energy of Photon = Work function + K.E
10.35 = 4.82 + K.E
K.E = 10.35 - 4.82
= 5.53 eV or,
= 8.85 × 10⁻¹⁹ J
Thus the response above is correct.
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A scientist is carrying out an experiment to determine the index of refraction for a partially reflective material. To do this, he aims a narrow beam of light at a sample of this material, which has a smooth surface. He then varies the angle of incidence. (The incident beam is traveling through air.)
The light that gets reflected by the sample is completely polarized when the angle of incidence is 46.5°.
(a) What index of refraction describes the material?
n =
(b) If some of the incident light (at θi = 46.5°) enters the material and travels below the surface, what is the angle of refraction (in degrees)?
Answer:
a) 1.05
b) 43.6°
Explanation:
a) The index refraction that describes the material can be found using Brewster's law:
[tex] \theta_{1} = arctan(\frac{n_{2}}{n_{1}}) [/tex]
where:
n₁ is the refractive index of the initial medium through which the light propagates (air) = 1
n₂ is the index of the material=?
θ₁ = 46.5°
[tex] n_{2} = n_{1}tan(\theta_{1}) = tan(46.5) = 1.05 [/tex]
Hence, the material's index refraction is 1.05.
b) The angle of refraction can be found as follows:
[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]
[tex]sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{sin(46.5)}{1.05} = 0.69[/tex]
[tex] \theta_{2} = arcsin(0.69) = 43.6^{\circ} [/tex]
Therefore, the angle of refraction is 43.6°.
I hope it helps you!
An object is inside a room that has a constant temperature of 289 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object
Answer:
T_object = 380.35 K
Explanation:
From Stefan–Boltzmann law, the power output is given by the formula:
P = σAT⁴
where;
σ is Stefan-Boltzmann constant
A is area of the radiating surface.
T is temperature of the body
Now, we are told that the power the object emitted is 3 times the power absorbed from the room.
Thus, we have;
P_e = 3P_a
Where P_e is power emitted and P_a is power absorbed.
So, we have;
σA(T_object)⁴ = 3σA (T_room)⁴
σA will cancel out to give;
(T_object)⁴ = 3(T_room)⁴
We are given T_room = 289 K
Thus;
(T_object)⁴ = 3 × 289⁴
(T_object) = ∜(3 × 289⁴)
T_object = 380.35 K
A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?
Answer:
3.067MHzExplanation:
The formula for calculating the voltage across an inductor is expressed as
[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]
Given parameters
current amplitude I = 1.50mA = 1.5*10⁻³A
inductance L = 0.450mH = 0.450*10⁻³H
Voltage across the inductor [tex]V_l[/tex] = 13.0V
Required
frequency f
Substituting the given parametres into the formula, we have;
[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]
Hence, the frequency required is 3.067MHz
Your favorite radio station broadcasts at a frequency of 91.5 MHz with a power of 11.5 kW. How many photons does the antenna of the station emit in each second?
Answer:
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
Explanation:
Given:
Frequency = 91.5 MHz
Power = 11.5 Kw = 11,500 J/s
Find:
Number of photons emit per second
Computation:
Total energy with frequency (E) = hf
Total energy with frequency (E) = 6.626×10⁻³⁴ × 91.5×10⁶
Total energy with frequency (E) = 6.06×10⁻²⁶ J
Number of photons emit per second = 11,500 / 6.06×10⁻²⁶
Number of photons emit per second = 1897.689 × 10²⁶
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Answer:
The value is [tex]\mu = 0.76[/tex]
Explanation:
From the question we are told that
The acceleration is [tex]a = 7.4 \ m /s^2[/tex]
Generally the force by which the truck bed (truck) is moving with is mathematically represented as
[tex]F = ma[/tex]
Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet is equal the force by which the the truck bed is moving with that is
[tex]F_f = F[/tex]
Here [tex]F_f[/tex] is the frictional force which is mathematically represented as
[tex]F_f = \mu * m * g[/tex]
substituting into above equation
[tex]\mu * m * g = ma[/tex]
=> [tex]\mu = \frac{a}{g}[/tex]
substituting values
[tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]
[tex]\mu = 0.76[/tex]
A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s
(a) The car is undergoing an acceleration of
[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]
so that in 9.02 s, it will have covered a distance of
[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]
The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:
[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]
(b) The wheels have average angular velocity
[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]
where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.
The wheels start at rest, so
[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]
Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Answer:
λ = 6.61 x 10⁻⁷ m = 661 nm
Explanation:
From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:
Δx = λL/d
where,
Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m
L = Distance between slits and screen = 2.12 m
d = slit separation = 0.49 mm = 0.49 x 10⁻³ m
λ = wavelength of light = ?
Therefore,
2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)
(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ
λ = 6.61 x 10⁻⁷ m = 661 nm
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?
Answer:
a) the distance between the interference fringes is reduced by half
b) the distance between stripes is doubled
Explanation:
Interference experiments constructive interference is described by the expression
d sin θ = m λ
let's use trigonometry to find the distance between the interference fringes
tan θ= y / L
dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ / d L
a) it asks us when the screen doubles its distance
L ’= 2 L
subtitute in the equation
y ’= m λ / (d 2L)
y ’=( m λ / d L) /2
y ’= y / 2
the distance between the interference fringes is reduced by half
b) the density of the network doubles
if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half
d ’= d / 2
we substitute
y ’= m λ (L d / 2)
y ’= m λ / (L d) 2
y ’= y 2
the distance between stripes is doubled
Calculate the focal length (in m) of the mirror formed by the shiny bottom of a spoon that has a 3.40 cm radius of curvature. m (b) What is its power in diopters? D
Answer:
The power of the mirror in diopters is 58.8 D
Explanation:
Given;
radius of curvature of the spoon, R = 3.4 cm = 0.034 m
The focal length of a mirror is given by;
[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]
The focal length of the mirror is 0.017 m
(b) The power of the mirror is given by;
[tex]P = \frac{1}{f}[/tex]
where;
P is the power of the mirror
f is the focal length
[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]
Thus, the power of the mirror in diopters is 58.8 D
The two metallic strips that constitute some thermostats must differ in:_______
A. length
B. thickness
C. mass
D. rate at which they conduct heat
E. coefficient of linear expansion
Answer:
E. Coefficient of linear expansion
Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire
Answer:
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Explanation:
This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.
Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.
Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Suppose you want a telescope that would allow you to see distinguishing features as small as 3.5 km on the Moon some 384,000 km away. Assume an average wavelength of 550 nm for the light received.Required:What is the minimum diameter mirror on a telescope?
Explanation:
[tex]\theta=1.22 \frac{\lambda}{D}[/tex]
And, from equation ( 2 ), we get
[tex]\theta=\frac{x}{d}[/tex]
Thus,
[tex]\frac{x}{d} &=1.22 \frac{\lambda}{D}[/tex]
[tex]D &=1.22 \frac{\lambda d}{x}[/tex]
[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]
[tex]=0.0515 \mathrm{m}[/tex]
Thus, the diameter of the telescope's mirror that would allow us to see details as small as is
