An emf E is connected to six identical resistors, each with resistance R.
(A) When identical resistors are connected in parallel, the current through each resistor is the same and is given by [tex]\begin{equation}I = \frac{E}{R}[/tex].
(B) When identical resistors are connected in series, the total resistance is 6R, and the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
Part A: When the identical resistors are connected in parallel, the current (I) through each resistor is the same. To calculate the current, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R):
[tex]\begin{equation}I = \frac{V}{R}[/tex]
In this case, the voltage across each resistor is the same, and it is equal to the emf (E). Therefore, the current through each resistor connected in parallel is:
[tex]\begin{equation}I = \frac{E}{R}[/tex]
Part B: When the identical resistors are connected in series, the total resistance ([tex]R_total[/tex]) is the sum of the individual resistances. Therefore, the current (I) flowing through the resistors in series is given by Ohm's Law:
[tex]\begin{equation}\I = \frac{E}{R_\text{total}}[/tex]
Since the resistors are identical, the total resistance can be calculated as:
[tex]R_total[/tex] = R + R + R + R + R + R = 6R
Substituting this value into the equation for the current, we get:
[tex]\begin{equation}I = \frac{E}{6R}[/tex]
So, when the resistors are connected in series, the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
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Who sponsored Felix Baumgartner in the second space jump that took placed in
2008?
Alban Geissler, who developed the SKYRAY carbon fiber wing with Christoph Aarns, suggested after Baumgartner's jump that the wing he used was a copy of two prototype SKYRAY wings sold to Red Bull (Baumgartner's sponsor) two years earlier. - wiki
a container of mass 200kg contains 160cm
of liquid. The total mass of the container and liquid is 520g. what is the density of the liquid?
Answer:
mass of liquid = total mass - mass of container
m = 52000kg - 200kg
m = 519800
D = m/v
D = 519800/160
D = 3,248.75 kgm-³
If it is known that a motor battery has an input voltage of 12V and a capacity of 6 Ah, how much power and resistor value is required to turn on 8 lamps with a parallel circuit, with the specifications of each lamp having a maximum voltage of 3V and an electric current of 140 mA? How long did all the lights go on until they off?
Answer:
Part A
The power to turn on the lamp, ∑P = 3.36 W
Part B
The Resistor required is approximately 8.04 Ohms
Part C
The time for all the lights to go out is approximately 21.43 hours
Explanation:
The input voltage of the motor battery , V = 12 V
The capacity of the battery, Q = 6 Ah
The number of lamps in parallel = 8 lamps
The maximum voltage of each lamp, = 3 V
The electric current in each lamp = 140 mA
The energy available in a battery, E = Q × V
For the battery, we have;
E = 6 Ah × 12 V = 72 Wh
The energy available in a battery, E = 72 Wh
Part A
The power used by the lamps, [tex]P_i[/tex] = [tex]I_i[/tex] × [tex]V_i[/tex]
∴ The total power used by the lamp, ∑P = 8 × 0.14 A × 3 V = 3.36 W
The power to turn on the lamp, ∑P = 3.36 W
Part B
The resistance required, is given as follows;
Resistor required = (Battery voltage - Lamp voltage)/(The sum of bulb current)
∴ Resistor required = (12 V - 3 V)/(8 × 0.14 A)
The Resistor required = 8.03571429 Ohms
The Resistor required ≈ 8.04 Ohms
Part C
The time for all the lights to go out = The time for the lamps to use all the power available in the battery
The time for all the lights to go out, t = E/∑P
∴ t = 72 Wh/(3.36 W) = 21.4285714 h
∴ The time for all the lights to go out, t ≈ 21.43 h
The time for all the lights to go out = The time for the lamps to use all the power available in the battery = t ≈ 21.43 h
∴ The time for all the lights to go out ≈ 21.43 hours.
Two objects collide elastically. The first has a mass of 5.00kg and a
velocity of 8.00 m/s. The second has a mass of 2.50 kg and a velocity
of -4.00m/s. If the velocity of object 1 after the collision is -4.00m/s,
what is the velocity of object 2?
Answer:
Explanation:
General formula
m1 * vi + m2*v2 = m1*v3 + m2*v4
Givens
m1 = 5
m2 = 2.5
v1 = 8 m/s
v2 = - 4 m/s
v3 = -4 m/s
v4 = x
Solution
5 * 8 - 2.5 * 4 = 5 * -4 + 2.5*x
40 - 10 = -20 + 2.5x
30 = - 20 + 2.5x
50 = 2.5x
x = 50/2.5
x = 20 m/s in the positive direction
Remark
Does this answer make sense? It should. You have 5 kg moving 8m/s in the plus direction. That's a lot of momentum. In addition after the collision, it turns around which is more momentum needed.
It has to give up that extra momentum to the 2.5 kg mass.
Purpose:You will use the radioactive decay rate and original-daughter element ratios of carbon-14and uranium-238 to determine the ages of different objects.
Procedure:1.Load PhET Radioactive Dating Game
2.Click on the tab for Decay Rates
3.Select Carbon-14.Using the graph, the estimated half-life for C-14 is5,200years.
4.Move the bucket slider all the way to the right.This will place 1000 C-14 atoms onto the screen.a.Click on the Start/Stop to stop the C-14 decay.Click on Reset All Nucleib.Click on theStart/Stop to start the C-14 decay.Stop the decay as you get close to one half-life.c.Use the Step button to stop decay at one half-life.After 1 half-life, how many C-14 atoms of the 1000 original remain?
After one half-life, the number of remaining C-14 atoms can be calculated by multiplying the initial number of atoms (1000) by 0.5 (since half of the atoms decayed).
Based on the given procedure, after one half-life of carbon-14 (C-14), the number of C-14 atoms remaining can be determined. Since the estimated half-life of C-14 is 5,200 years, we can use this information to answer the question. After one half-life, the number of remaining C-14 atoms can be calculated as half of the original number of C-14 atoms. Given that the initial number of C-14 atoms is 1000, after one half-life: Remaining C-14 atoms = (1/2) * 1000. Remaining C-14 atoms = 500
after one half-life of carbon-14 (C-14), the number of C-14 atoms remaining can be determined.
Therefore, after one half-life, 500 C-14 atoms of the 1000 original atoms remain.
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n a certain series rlc circuit, irms = 9.00 a, δvrms = 190 v, and the current leads the voltage by 38.0°. (a) what is the total resistance (in ω) of the circuit?
The total resistance of the circuit is approximately 21.11 Ω. To determine the total resistance (R) of the series RLC circuit, we can use the relationship between the current (I), voltage (V), and resistance.
To determine the total resistance (R) of the series RLC circuit, we can use the relationship between the current (I), voltage (V), and resistance:
V = I * R
Given:
Irms = 9.00 A (root mean square value of current)
δvrms = 190 V (root mean square value of voltage)
The current leads the voltage by 38.0°
First, we need to find the peak values of current (I_peak) and voltage (V_peak). Since the root mean square (rms) values are given, we can use the following relationship:
I_peak = Irms
V_peak = δvrms
Now, we can calculate the total resistance (R) using the peak values:
R = V_peak / I_peak
R = 190 V / 9.00 A
R ≈ 21.11 Ω
Therefore, the total resistance of the circuit is approximately 21.11 Ω.
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What is happening to the ATOMS INSIDE of a magnet that gives the
material its magnetic properties?
Answer:
The atoms are aligned in a particular direction
Explanation:
The atoms become aligned in a particular direction in regions called domains, thus resulting in an overall resultant magnetism due to the spin of the electrons.
Jenise is buying a car for $7,020. The TAVT rate is 9.1%.
What is the amount of tax that Jenise will have to pay on her car?
Answer:
$7,658.82
Explanation:
Sales Tax Calculations:
Sales Tax Amount = Net Price x (Sales Tax Percentage / 100)
Total Price = Net Price + Sales Tax Amount
Net Price: $ 7,020.00
+Sales Tax (9.1%): $ 638.82
Total Price: $ 7,658.82
Therefore, the amount of tax that Jenise has to pay on her car is $7,658.82
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sciencephysicsphysics questions and answersa 6.50 μf capacitor that is initially uncharged isconnected in series with a 4500 ω resistor and a503 v emf source with negligible internal resistance.a)just after the circuit is completed, what is the voltagedrop across the capacitor? vc= vb)just after the circuit is completed, what is the voltagedrop across the resistor?
Question: A 6.50 ΜF Capacitor That Is Initially Uncharged Isconnected In Series With A 4500 Ω Resistor And A503 V Emf Source With Negligible Internal Resistance.A)Just After The Circuit Is Completed, What Is The Voltagedrop Across The Capacitor? Vc= Vb)Just After The Circuit Is Completed, What Is The Voltagedrop Across The Resistor?
A 6.50 μF capacitor that is initially uncharged isconnected in series with a 4500 Ω resistor and a503 V emf source with negligible internal resistance.
a)Just after the circuit is completed, what is the voltagedrop across the capacitor?
Vc= V
b)Just after the circuit is completed, what is the voltagedrop across the resistor?
VR = V
c)Just after the circuit is completed, whatis the charge on the capacitor?
Qo= C
d)Just after the circuit is completed, whatis the current through the resistor?
IR= A
e)A long time after the circuit is completed(after many time constants), what are the values of the precedingfour quantities?
Vc, V R, I,Q= V, V, A, C
The voltage across the capacitor (Vc) will be equal to the emf of the source (V) which is 503 V, the voltage drop across the resistor ([tex]V_R[/tex]) will be zero, the current through the resistor ([tex]I_R[/tex]) will be 0.112 A, and the charge on the capacitor (Q) will be at its maximum value (Q = CV).
a) Just after the circuit is completed, the voltage drop across the capacitor (Vc) is equal to the emf of the source (V). Therefore, Vc = V = 503 V.
b) Just after the circuit is completed, the voltage drop across the resistor ([tex]V_R[/tex]) is zero since the capacitor is initially uncharged and behaves like a open circuit. Therefore, [tex]V_R = 0 V[/tex].
c) Just after the circuit is completed, the charge on the capacitor (Q₀) is zero since the capacitor is initially uncharged. Therefore, Q₀ = 0 C.
d) Just after the circuit is completed, the current through the resistor ([tex]I_R[/tex]) can be calculated using Ohm's Law:
[tex]I_R[/tex] = V / R = 503 V / 4500 Ω ≈ 0.112 A
e) A long time after the circuit is completed (after many time constants), the capacitor will be fully charged and behave like an open circuit. The charge on the capacitor (Q) will be at its maximum value (Q = CV).
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an iron anchor of density 7810.00 kg/m3 appears 152 n lighter in water than in air. (a) what is the volume of the anchor? (b) how much does it weigh in air
The volume of the anchor is approximately 0.0195 m^3. The weight of the anchor in air is approximately 1492 N.
To find the volume of the anchor, we can use Archimedes' principle, which states that the buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Given that the anchor appears 152 N lighter in water than in air, we can equate this weight difference to the buoyant force experienced by the anchor in water.
Weight difference = Buoyant force
= Weight in air - Weight in water
Let's assume the weight of the anchor in air is W_air and the weight of the anchor in water is W_water.
W_air - W_water = 152 N
We know that weight = mass × acceleration due to gravity (W = m × g), and density is defined as mass divided by volume (ρ = m/V), where ρ is the density, m is the mass, and V is the volume.
Therefore, W_air = ρ_anchor × g × V × (1), and
W_water = ρ_water × g × V × (2).
Given that the density of water, ρ_water, is 1000 kg/m^3, and the density of the anchor, ρ_anchor, is 7810.00 kg/m^3, we can substitute these values into equations (1) and (2):
7810.00 × g × V - 1000 × g × V = 152
Simplifying the equation:
6810.00 × g × V = 152
V = 152 / (6810.00 × g)
Using the standard acceleration due to gravity, g = 9.8 m/s^2:
V = 152 / (6810.00 × 9.8)
≈ 0.0195 m^3
Therefore, the volume of the anchor is approximately 0.0195 m^3.
To calculate the weight of the anchor in air, we can use the formula:
Weight in air = ρ_anchor × g × V
Substituting the values:
Weight in air = 7810.00 × 9.8 × 0.0195
≈ 1492 N
Therefore, the weight of the anchor in air is approximately 1492 N.
The volume of the anchor is approximately 0.0195 m^3, and its weight in air is approximately 1492 N.
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An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only force on each of the two particles is the force from the uniform electric field due to the capacitor. Each particle accelerates until striking one of the plates of the capacitor. (There is no gravity in this problem and we ignore the small force between the electron and the proton.) How do the final kinetic energies and final speeds (just before striking a plate) compare
Answer:
Explanation:
Let the potential difference between the middle point and one of the plate be ΔV .
electric potential energy will be lost and it will be converted into kinetic energy .
Electrical potential energy lost = Vq , where q is charge on charge particle .
For proton
ΔV× q = 1/2 M V² ( kinetic energy of proton )
where M is mass and V be final velocity of proton .
For electron
ΔV× q = 1/2 m v² ( kinetic energy of electron )
where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .
As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same
1/2 M V² = 1/2 m v²
(V / v )² = ( m / M )
(V / v ) = √ ( m / M )
In other words , their velocities are inversely proportional to square root of their masses .
It is observed that 55.00 mLmL of water at 20∘C∘C completely fills a container to the brim. When the container and the water are heated to 60∘C∘C, 0.35 gg of water is lost.
What is the coefficient of volume expansion of the container? Density of water at 60 ∘C∘C is 0.98324 g/mLg/mL.
The coefficient of volume expansion of the container is 1.64 x 10⁻⁴ °C⁻¹.
Initial volume of water, V₁ = 55 mL
Initial temperature of the water, T₁ = 20°C
Final temperature of the water, T₂ = 60°C
Density of water at 60°C, d = 0.983 g/mL
Mass of water lost during heating, m = 0.355 g
The change in volume of water is,
ΔV = m/d
ΔV = 0.355/0.983
ΔV = 0.361 mL
Volume expansion occurs when a solid, whether it be in the form of a cube, cuboid, sphere, or another shape, rises in volume as a result of heating.
The expression for the coefficient of volume expansion of the container is given by,
α = ΔV/VΔT
α = 0.361/[55 x (60 - 20)]
α = 0.361/(55 x 40)
α = 0.361/2200
α = 1.64 x 10⁻⁴ °C⁻¹
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what tangential speed v must the bob have so that it moves in a horizontal circle with the string making an angle 21.0 ∘ with the vertical?
To find the tangential speed required for the bob to move in a horizontal circle with the string making an angle of 21.0 degrees with the vertical, we can use the concept of centripetal force.
The centripetal force required to keep the bob moving in a circular path is provided by the tension in the string. The tension can be resolved into two components: the vertical component and the horizontal component. The vertical component of the tension balances the weight of the bob, which is given by: T * cos(21.0°) = mg. where T is the tension in the string, m is the mass of the bob, and g is the acceleration due to gravity. The horizontal component of the tension provides the centripetal force required for circular motion, and it can be expressed as: T * sin(21.0°) = mv^2 / r. where v is the tangential speed of the bob and r is the radius of the circular path. Dividing the two equations: [T * sin(21.0°)] / [T * cos(21.0°)] = (mv^2 / r) / (mg). tan(21.0°) = v^2 / (rg). Solving for v: v = √(rg * tan(21.0°)) Now, we can substitute the values of the gravitational acceleration (g) and the angle (21.0°) to calculate v. Note: It is assumed that the bob is moving in a horizontal circle without any additional external forces affecting the system.
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the position of a particle moving along a coordinate line is s = √ 6 3 t , with s in meters and t in seconds. find the particle’s velocity at t = 10 seconds.
Given that the position of a particle moving along a coordinate line is s = √6 3 t, with s in meters and t in seconds. Therefore, the velocity of the particle at t = 10 seconds is 0.
We need to find the particle’s velocity at t = 10 seconds. Here, we need to find the first derivative of s with respect to t, that is velocity.
The derivative of s is given as follows:
s = √6 3 t
∴ ds/dt = d/dt (√6 3 t)
ds/dt= √6 3
d/dt(t) = √6 3 velocity,
Velocity of particle at t = 10 seconds is obtained by substituting t = 10 in the expression of velocity.
Velocity of particle at t = 10 seconds = ds/dt (at t = 10)
velocity of a particle at t = √6 3 (d/dt (10))
velocity of a particle at t= √6 3 (0)= 0
Therefore, the velocity of the particle at t = 10 seconds is 0.
A concise answer to the given problem can be as follows:
The position of a particle moving along a coordinate line is given by
s = √63t.
We can find the particle's velocity at t = 10 seconds by taking the derivative of s with respect to t, which is the velocity.
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In terms of the wavelength of the sound wave, how far apart are the first two resonant positions in the resonance tube?
A.Three quarters of the wavelength
B.One quarter of the wavelength
C.One half of the wavelength
D.One wavelength
The distance between the first two resonant positions in a resonance tube is One half of the wavelength of the sound wave. Option C
How did we determine the distance between the first two resonant positions?In a resonance tube experiment, the first two resonant positions occur when the length of the tube equals one quarter of the wavelength and then three quarters of the wavelength.
These positions correspond to the first and second resonant frequencies, or harmonics.
Therefore, the distance between these two positions, in other words, the length of the tube at the second resonant position minus the length of the tube at the first resonant position, equals three quarters of the wavelength minus one quarter of the wavelength, which is half a wavelength.
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Please help me with 17 and 18!!!!!! (It's related to 16) It's due today!!!!! NO LINK PLEASE!!!!!!!!
Answer:17: A wave can be defined as follows: It is important to realize that a wave is quite a different object than a particle. A baseball thrown though a window transfers energy from one point to another, but this involves the movement of a material object between two points.
Explanation:
18: In this way, we classify waves into electromagnetic and mechanical waves. The main difference between mechanical and electromagnetic waves is that electromagnetic waves do not require a medium to propagate whereas mechanical waves require a medium in order to propagate.
PLS HELP MEEEE (NO LINKS PLEASE)
The difference between visible light and gamma rays is that
a.
the amplitude of visible light is greater.
c.
they travel through a different medium.
b.
the speed of gamma rays is greater.
d.
the frequency of gamma rays is greater.
Answer:
Gamma rays occupy the short-wavelength end of the spectrum; they can have wavelengths smaller than the nucleus of an atom. Visible light wavesare one-thousandths the width of human hair--about a million times longer than gamma rays. Radio waves, at the long-wavelength end of the spectrum, can be many meters long.
Answer:D "the frequency of gamma rays is greater."
Explanation:Trust
Is O2 classified as a compound?
Answer: NO.
Explanation: Oxygen is not a compound. It has only one element in it.
explain the differences between pitch and loudness and what are they connected to?
Answer:
Loudness and pitch are distinct properties of sound. Loudness is related to the amplitude of the sound wave; pitch is related to its frequency
Explanation:
Which of the following describes half-life? Choose which apply.
A. Half-life is the amount of time it takes for half of a sample to decay.
B. The shorter the half-life, the more unstable the nuclide.
C. Half-life cannot be calculated for nuclides.
D. The longer the half-life, the more stable the nuclide
Answer:
дангггггггггггггггггггггггггггггггггггггггггггггг
Answer: D.
Explanation: I took the test
How is capacitance related to the distance between the plates of a capacitor?
It is directly proportional, so the capacitance increases as the distance increases.
It is inversely proportional, so the capacitance increases as the distance increases.
It is directly proportional, so the capacitance decreases as the distance increases.
It is inversely proportional, so the capacitance decreases as the distance increases.
Answer:C
Explanation: I studied, and C is correct
Answer:
D
Explanation:
what is the velocity of a car that traveled 6 meters in .96 seconds
The velocity of the car is approximately 6.25 meters per second.
To determine the velocity of a car that traveled 6 meters in 0.96 seconds, we can use the formula for velocity: velocity = distance / time. In this case, the distance traveled is 6 meters and the time taken is 0.96 seconds.
Plugging these values into the formula, we have:
velocity = 6 meters / 0.96 seconds
= 6.25 meters per second.
Velocity is a measure of the rate at which an object changes its position. In this context, the car is traveling at a constant speed of 6.25 meters per second. It means that for every second that passes, the car moves 6.25 meters forward.
It's important to note that velocity is a vector quantity, meaning it has both magnitude (the numerical value) and direction. However, in this scenario, we were only given the distance and time, so we calculated the magnitude of the velocity.
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constants, as appropriate. (a) Derive an expression for the magnitude of the emf s induced in the loop as a function of t. (b) Derive an expression for the power P dissipated in the loop as a function of t. (c) Determine the net force on the loop. Justify your answer. (d) On the axes, sketch a graph of the magnitude of the current I induced in the loop as a function of t for the first cycle. The time T for the first cycle is labeled on the horizontal axis. (e) At time t=41T, is the current in the loop clockwise or counterclockwise? Clockwise Counterclockwise Justify your answer. t=0 to t=41T
(a) The magnitude of the emf (ε) induced in the loop as a function of time (t) is given by the equation ε = -dΦ/dt, where Φ is the magnetic flux through the loop. The negative sign indicates that the emf opposes the change in magnetic flux.
The magnitude of the emf induced in the loop can be expressed as ε = -dΦ/dt, where dΦ/dt represents the rate of change of magnetic flux through the loop. The value of dΦ/dt depends on the specific scenario or setup.
(b) The power (P) dissipated in the loop as a function of time can be calculated using the equation P = I^2R, where I is the current flowing through the loop and R is the resistance of the loop.
The power dissipated in the loop can be calculated using the equation P = I^2R, where I is the current flowing through the loop and R is the resistance of the loop. The power dissipation is directly proportional to the square of the current and the resistance of the loop.
(c) The net force on the loop is zero. According to Lenz's law, the induced current creates a magnetic field that opposes the change in the magnetic field causing the induction. This opposition leads to a canceling effect, resulting in a net force of zero on the loop.
According to Lenz's law, the induced current creates a magnetic field that opposes the change in the magnetic field causing the induction. This opposition results in a canceling effect, leading to a net force of zero on the loop. The forces due to the induced current and the external magnetic field balance each other, resulting in a net force of zero.
(d) The graph of the magnitude of the current (I) induced in the loop as a function of time for the first cycle will have a sinusoidal shape. The current will start from zero, increase to a maximum, decrease to zero again, and then reverse direction, reaching a maximum in the opposite direction.
The graph of the magnitude of the current induced in the loop as a function of time for the first cycle will exhibit a sinusoidal pattern. The current starts from zero, increases to a maximum, decreases back to zero, and then reverses direction, reaching a maximum in the opposite direction. The shape of the graph resembles a sine wave.
(e) At time t = 41T, the current in the loop is counterclockwise. This can be determined based on the graph mentioned in part (d) and the periodic nature of the current. Since the current reverses direction after each cycle, at t = 41T, it will be in the counterclockwise direction.
Since the current in the loop reverses direction after each cycle, at t = 41T, the current will be in the counterclockwise direction. This can be inferred from the periodic nature of the current and the fact that the current changes direction at the end of each cycle.
In conclusion, the magnitude of the emf induced in the loop as a function of time can be determined using the rate of change of magnetic flux. The power dissipated in the loop depends on the current and resistance. The net force on the loop is zero due to Lenz's law. The graph of the current induced in the loop exhibits a sinusoidal shape. At time t = 41T, the current in the loop is counterclockwise based on the periodic nature of the current.
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How much work is done when a car is moved 10 m with a force of 3.4 N?
Answer:
34J I assume
Explanation:
force×distance is work done. 10×3.4 is 34. therefore its 34 joules of work done
Answer:
34J
Explanation:
The formula for work is W=Force x Distance
W=FxD
F=3.4N
D=10m
W=10x3.4
W=34 Joules
(a) Does the index of refraction vary as you change the wavelength of light. (B) What is the angle of the ray that leaves the glass square (emergent ray) relative to the ray that enters it? (c) What can you say about the path of emergent ray relative to that of the incident ray?
(a) Yes, the index of refraction varies with the wavelength. (b) The emergent ray is refracted at a different angle. (c) The path of the emergent ray deviates from the incident ray.
(a) Yes, the index of refraction varies as you change the wavelength of light.
The index of refraction (n) of a material is a measure of how much the speed of light is reduced when it passes through that material compared to its speed in a vacuum.
The index of refraction is wavelength-dependent and typically varies slightly with different wavelengths of light. This phenomenon is known as dispersion.
One way to express this variation is through the refractive index as a function of wavelength, often represented by a refractive index versus wavelength graph.
In general, different wavelengths of light are bent or refracted by different amounts when passing through a medium due to their interaction with the material's atoms or molecules.
This bending is a result of the change in the speed of light, which is dictated by the refractive index.
The index of refraction does vary as you change the wavelength of light. This variation is responsible for phenomena like dispersion, where different colors of light are separated when passing through a prism, for example.
(b) The angle of the emergent ray leaving a glass square relative to the incident ray depends on the angle of incidence and the refractive index of the glass.
According to Snell's law, the relationship between the angle of incidence (θ₁), the angle of refraction (θ₂), and the refractive indices of the two media involved can be expressed as:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
In the case of a glass square, let's assume light is incident on one of its faces. If we know the angle of incidence (θ₁) and the refractive index of the glass (n₂), we can calculate the angle of the emergent ray (θ₂) using Snell's law.
The angle of the emergent ray leaving the glass square relative to the incident ray depends on the angle of incidence and the refractive index of the glass, and it can be calculated using Snell's law.
(c) The path of the emergent ray relative to the incident ray can be different due to refraction.
When light passes from one medium to another, it changes direction due to the change in its speed caused by the change in the refractive index. This change in direction is called refraction. Therefore, the emergent ray may have a different direction compared to the incident ray.
The emergent ray will still follow the law of refraction (Snell's law) and will be bent towards or away from the normal depending on the refractive indices of the two media involved and the angle of incidence.
The amount of bending depends on the difference in refractive indices and the angle at which the light strikes the boundary between the two media.
The path of the emergent ray relative to the incident ray can be different due to refraction, as the emergent ray changes direction upon passing from one medium to another.
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True or False: A balanced force acts on different objects, and action-
eaction forces act on the same object. *
A. True
B. False
Answer:
true
Explanation:
according to the Newton's third law
A 29 kg child jumps to the ground from the top of a fence that is 1.8 m high. You analyze the problem using upward as the positive x direction.
A. Taking x=0 to be at the bottom of the fence, what are the initial potential energy of the child-Earth system and the chnage in the system kinetic energy during the jump? (Ui , change of U)
B.Repeat the previous part for x=0 at the top of the fence (Ui , change of U)
The initial potential energy of the child-Earth system is 509.4 J, and the change in system kinetic energy during the jump is 509.4 J.
In the first scenario, with x=0 at the bottom of the fence, we can calculate the initial potential energy (Ui) using the formula Ui = mgh, where m is the mass of the child (29 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the fence (1.8 m). Substituting the values, Ui = 29 kg × 9.8 m/s^2 × 1.8 m = 509.4 J.
Since there is no external work done on the child during the jump, the change in system kinetic energy (change of U) is equal to the negative of the initial potential energy. Therefore, the change of U = -509.4 J.
In the second scenario, with x=0 at the top of the fence, the initial potential energy (Ui) is still the same, i.e., 509.4 J. However, since the child is starting from a higher position, the change in system kinetic energy (change of U) will be different. The change of U will still be equal to -509.4 J since it depends on the initial potential energy, regardless of the reference point.
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A group of students are doing a reading activity in their classroom. They suddenly hear the noise of a truck in the parking lot of their school. They can hear it but cannot see it. Which of the following explains why they can hear the truck, but cannot see it?
Answer:
The answer is the sound waves can travel through some material (not all) and visible light cannot.
Explanation:
I did this question on study island so I know the answer.
Answer:
the answer is: sound can be transmitted through the walls, but visible light cannot.
Explanation:
What is competition?
Answer:
two team in a conflict
Explanation:
Competition arises whenever two or more parties strive for a common goal which cannot be shared: where one's gain is the other's loss. Competition includes rivalry between entities such as organisms, individuals, economic and social groups, etc.
You completed three terrain-forming trials. Describe how the sun's mass affects planets in a solar system. Use data you recorded to support your conclusions
Pls answer
The sun's mass plays a crucial role in shaping the characteristics of planets in a solar system. It determines the orbital paths, stability, and overall structure of planetary systems.
During the three terrain-forming trials, I observed the effects of the sun's mass on planets in a solar system. The sun's mass is a critical factor in determining the gravitational forces experienced by planets. Through these trials, I recorded data that supported several conclusions
Firstly, I observed that the sun's mass directly influences the orbital paths of planets. Planets closer to the sun experience stronger gravitational forces, leading to faster orbital speeds and shorter orbital periods. In contrast, planets farther from the sun have slower orbital speeds and longer orbital periods. This data confirms Kepler's laws of planetary motion, which state that the square of a planet's orbital period is directly proportional to the cube of its average distance from the sun.
Secondly, the sun's mass affects the stability of planetary systems. A more massive sun exerts stronger gravitational forces, providing stability by preventing planets from being pulled out of their orbits. The recorded data revealed that planets in systems with a less massive sun tended to have unstable orbits, resulting in irregular paths and potential ejections from the system.
Lastly, the sun's mass influences the overall structure of planetary systems. Higher-mass stars tend to form larger and more massive planets, as the gravitational forces they exert allow for the accumulation of larger amounts of planetary material. The data collected during the trials supported this conclusion, demonstrating a correlation between the mass of the sun and the sizes and masses of the planets in the system.
The recorded data from the terrain-forming trials provided empirical evidence supporting these conclusions, highlighting the significant impact of the sun's mass on the planets it governs.
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