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.
g Assume you are a farsighted person who has a near point distance of 40 (cm). If you use a converging contact lens with focal length of 10 (cm). What is nearest distance you can vision with you contacts now?
Answer:
object distance p = 13.33 cm
Explanation:
For this problem of finding the image of an object we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image, respectively.
In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)
1 / p = 1 / f - 1 / q
1 / p = 1/10 - 1/40
1 / p = 0.075
p = 13.33 cm
A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as seen from the left)?
Answer:
The induced current is clockwise
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
Take an electric field sensor and move it in a straight line, crossing the equipotential lines. Describe the relationship between the distance between the equipotential lines and the strength of the electric field.
Answer:
E = - dV / dx
Explanation:
The equipotential lines are lines or surfaces that have the same power, therefore we can move in them without carrying out work between equipotential lines, work must be carried out, therefore the electric field changes.
The electric field and the potential are related by
E = - dV / dx
therefore when the change is faster, that is, the equipotential lines are closer, the greater the electric field must be.
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.
a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
A) Applying the energy equation
The positive terms is : ΔUg The negative terms is : ΔEth The zero term are : ΔK and ΔUsB) The energy output by the athlete is ; 800 Joules
C) The metabolic power is : 2000 w
D) When he performs the task in 1.2 s
The metabolic energy he expends is : the same His metabolic power is : moreGiven data :
Weight of barbell = 400 N
Height = 2.0 m
Time = 1.6 secs
efficiency of the human body = 25%
Speed = constant
A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its potential energy increases. ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement
B) Determine the energy output of the athlete
weight of barbell * Height = 400 * 2 = 800 J
C) Determine the metabolic power
Metabolic power = energy output / Time
where ; energy output = 4 * 800 = 3200
∴ Metabolic power = 3200 / 1.6
= 2000 w
D) When performs same task at 1.2 s
The metabolic energy he expends is the same and His metabolic power is more
Hence we can conclude that the answers to your questions are as listed above
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In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?
Answer:
Δy = 1 10⁻⁴ m
Explanation:
In double-slit experiments the constructive interference pattern is described by the equation
d sin θ = m λ
In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle
tan θ = y / L
as the angles are small,
tan θ = sin θ / cos θ = sin θ
substituting
sin θ = y / L
d y / L = m λ
y = m λ / d L
let's apply this formula for each wavelength
λ = 450 nm = 450 10⁻⁹ m
m = 5
d = 5.0 mm = 5.0 10⁻³ m
y₁ = 5 450 10⁻⁹ / (5 10⁻³ 1.4)
y₁ = 3.21 10⁻⁴ m
we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m
y₂ = 5 590 10⁻⁹ / (5 10⁻³ 1.4)
y₂= 4.21 10⁻⁴ m
the separation of these two lines is
Δy = y₂ - y₁
Δy = (4.21 - 3.21) 10⁻⁴ m
Δy = 1 10⁻⁴ m
Large capacitors can hold a potentially dangerous charge long after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 μF capacitor from a camera flash unit retains a voltage of 140 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Answer:
93.3x10^-3s
Explanation:
If
Resistance = 1.8 kΩ
Current = 50 mA
Capacitor = 120 μF
Voltage = 140 V
to calculate the discharge current
Applying the formula of discharge current
io=vo/R
io= 140/ 1.8x 10³
= 0.078A
to calculate the time
Applying the formula of current
io= vo/R e-t/RC
50= 140/1800e-t/RC
0.649= e-t/RC
-t/RC= ln( 0.649)
t = 0.432x 120x10^-6x 1800
t=93.3 x 10^-3seconds
A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t = 0 that the energy stored in the magnetic field of the inductor is a maximum is
Answer:
t = T / 2 all energy is stored in the inductor
Explanation:
The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is
w = √1/LC
2π / T =√1 / LC
T = 2π √LC
The energy is constant and for the initial instant it is completely stored in the capacitor
Uc = Q₀² / 2C
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor
U = L I² / 2
in the intermediate instant the energy is stored in the two elements.
Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor
After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,
[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]
here, T is the period of oscillation.
Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,
[tex]U =\dfrac{Q^{2}}{2C}[/tex]
here, C is the capacitance.
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,
[tex]U'=\dfrac{L I^{2}}{2}[/tex]
here, L is the inductance and I is the current.
Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.
Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
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An intergalactic rock star bangs his drum every 1.30 s. A person on earth measures that the time between beats is 2.50 s. How fast is the rock star moving relative to the earth
Answer:
v = 0.89 c = 2.67 x 10⁸ m/s
Explanation:
The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:
t = t₀/√[1 - v²/c²]
where,
t = time measured by the person on earth = 2.50 s
t₀ = rest time of the intergalactic rock star = 1.30 s
v = relative speed of the rock star = ?
Therefore,
2.5 s = (1.3 s)/√[1 - v²/c²]
√[1 - v²/c²] = 1.3/2.5
√[1 - v²/c²] = 0.52
[1 - v²/c²] = 0.52²
[1 - v²/c²] = 0.2074
v²/c² = 1 - 0.2074
v²/c² = 0.7926
v/c = √0.7926
v = 0.89 c
where,
c = speed of light = 3 x 10⁸ m/s
v = (0.89)(3 x 10⁸ m/s)
v = 0.89 c = 2.67 x 10⁸ m/s
A rock has mass 1.80 kg. When the rock is suspended from the lower end of a string and totally immersed in water, the tension in the string is 10.8 N . What is the smallest density of a liquid in which the rock will float?
Answer:
The density is [tex]\rho_z = 2544 \ kg /m^3[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 1.80 \ kg[/tex]
The tension on the string is [tex]T = 10.8 \ N[/tex]
Generally the weight of the rock is
[tex]W = m * g[/tex]
=> [tex]W = 1.80 * 9.8[/tex]
=> [tex]W = 17.64 \ N[/tex]
Now the upward force(buoyant force) acting on the rock is mathematically evaluated as
[tex]F_f = W - T[/tex]
substituting values
[tex]F_f = 17.64 - 10.8[/tex]
[tex]F_f = 6.84 \ N[/tex]
This buoyant force is mathematically represented as
[tex]F_f = \rho * g * V[/tex]
Here [tex]\rho[/tex] is the density of water and it value is [tex]\rho = 1000\ kg/m^3[/tex]
So
[tex]V = \frac{F_f}{ \rho * g }[/tex]
[tex]V = \frac{6.84}{ 1000 * 9.8 }[/tex]
[tex]V = 0.000698 \ m^3[/tex]
Now for this rock to flow the upward force (buoyant force) must be equal to the length
[tex]F_f = W[/tex]
[tex]\rho_z * g * V = W[/tex]
Here z is smallest density of a liquid in which the rock will float
=> [tex]\rho_z = \frac{W}{ g * V}[/tex]
=> [tex]\rho_z = \frac{17.64}{ 0.000698 * 9.8}[/tex]
=> [tex]\rho_z = 2544 \ kg /m^3[/tex]
A thermos bottle works well because:
a. its glass walls are thin
b. silvering reduces convection
c. vacuum reduces heat radiation
d. silver coating is a poor heat conductor
e. none of the above
Answer:
A thermos bottle works well because:
A) Its glass walls are thin
Answer:
A thermos bottle works well because:
C
Vacuum reduces heat radiation
A Van de Graaff generator produces a beam of 2.02-MeV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron.
A) If the beam current is 10.0 μA, how far apart are the deuterons?
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
Answer:
A) The distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) The electrical force of repulsion among them shows a small effect in beam stability.
Explanation:
Given that:
A Van de Graaff generator produces a beam of 2.02-MeV deuterons
If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.
[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]
2 K.E = mv²
[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]
so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.
[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
v = 13911611.49 m/s
v = 1.39 × 10⁷ m/s
So, If the beam current is 10.0 μA.
We all know that:
[tex]I = \dfrac{q}{t}[/tex]
[tex]t = \dfrac{q}{I}[/tex]
[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]
t = 1.6 × 10⁻¹⁴ s
Finally, the distance of the deuterons from one another = v × t
the distance of the deuterons from one another = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)
the distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
The electrical force of repulsion among them shows a small effect in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
In an adiabatic process:
a. the energy absorbed as heat equals the work done by the systemon its environment
b. the energy absorbed as heat equals the work done by theenvironment on the system
c. the work done by the environment on the system equals the changein internal energy
Answer:
c. the work done by the environment on the system equals the changein internal energy.
Explanation:
Adiabatic process:
When the boundary of a system is perfectly insulated, it means that the energy can not flow from the system and into the system ,these system is known as adiabatic system.
When the energy transfer in the system is zero ,then these type of process is known as adiabatic process.
From the first law of thermodynamics
Q= ΔU + W
Q=Heat transfer
ΔU=Change in internal energy
W=Work transfer
In adiabatic process , Q= 0
Therefore
0=ΔU +W
W=- ΔU
Negative sign indicates that ,the work done by the environment.
Therefore the correct option will be (c).
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
1.24Mev
Explanation:
Using
E= hc/lambda
= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9
= 1.24Mev
A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V
Answer:
C. 450v
Explanation:
Using
Voltage= B*distance of separation*velocity
3mm x 0.3T x 5E5m/s
= 450v
g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
What is an understood decimal
an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate
Answer:
The electric field can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.
Explanation:
Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.
A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A What is the focal length of the lens
Answer:
The focal length of the lens is 2.5 cm
Explanation:
Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;
[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]
Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:
[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]
then, solving for the focal distance knowing that the object distance is 1.5 cm:
[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]
1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 6.76 s. How much time does the driver of the car measure for his trip between the poles
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
Answer:
The wave is travelling in the ±z-axis direction.
Explanation:
An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the electric and magnetic field.
In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.
Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?
Answer:
L = 0.8 m
Explanation:
Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:
Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m
but,
Δx = λL/d
λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m
d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m
L = Distance between slits and screen = ?
Therefore, using the values, we get:
2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)
L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)
L = 0.8 m
wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
A single-slit diffraction pattern is formed on a distant screen. Assume the angles involved are small. Part A By what factor will the width of the central bright spot on the screen change if the wavelength is doubled
Answer:
If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Explanation:
For a single-slit diffraction, diffraction patterns are found at angles θ for which
w sinθ = mλ
where w is the width
λ is wavelength
m is an integer, m = 1,2,3, ....
From the equation, w sinθ = mλ
For the first case, where nothing was changed
w₁ = mλ₁ / sinθ
Now, If the wavelength is doubled, that is, λ₂ = 2λ₁
The equation becomes
w₂ = mλ₂ / sinθ
Then, w₂ = m(2λ₁) / sinθ
w₂ = 2(mλ₁) / sinθ
Recall that, w₁ = mλ₁ / sinθ
Therefore, w₂ = 2w₁
Hence, If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position, the motion of the object is simple harmonic motion. Mary says this is true only if the acceleration is opposite in direction to the displacement. Which one, if either, is correct
Answer:
Both Ellen and Mary are correct.
Explanation:
Both are correct, it's just different ways of saying the same thing.
When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position