Answer:(b)
Explanation:
Optical fiber works on the principle of total internal reflection . Total internal reflection occurs when the following two conditions are met
(1) When the angle of incidence of light is greater than the critical angle and,
(2) When a ray of light travels from a optically denser medium to optically rarer medium .
We know that light bend away from normal when it travels from denser to rarer medium. If we we choose such an angle that the refraction angle is [tex]90^{\circ}[/tex] then the incident angle is called critical angle.
So, the refractive index of cladding must be must be less than the refractive index of core material.
The noise level coming from a pig pen with
199 pigs is 74.3 dB. The intensity of the sound
coming from the pig pen is proportional to the
number of pigs.
Assuming each of the remaining pigs squeal
at their original level after 61 of their compan-
ions have been removed, what is the decibel
level of the remaining pigs?
Answer in units of dB.
The decibel level of the remaining pigs is 34.02dB.
To solve this problem, we need to understand the relationship between the number of pigs and the intensity of sound. Since the intensity is proportional to the number of pigs, we can use the formula:
[tex]\[ \text{Intensity} = k \times \text{Number of pigs} \][/tex]
where ( k ) is a constant of proportionality.
Given that the noise level from 199 pigs is 74.3 dB, we can set up the equation:
[tex]\[ 74.3 = k \times 199 \][/tex]
First, let's solve for ( k ):
[tex]\[ k = \frac{74.3}{199} \][/tex]
Now, we can use this value of ( k ) to find the intensity when only 138 pigs remain:
[tex]\[ \text{Intensity}_{\text{new}} = k \times 138 \][/tex]
Finally, we can convert this intensity back into decibels using the formula:
[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(\text{Intensity}) \][/tex]
Now, let's calculate:
[tex]\[ k = \frac{74.3}{199} \][/tex]
[tex]\[ k \approx 0.3726 \][/tex]
[tex]\[ \text{Intensity}_{\text{new}} = 0.3726 \times 138 \][/tex]
[tex]\[ \text{Intensity}_{\text{new}} = 51.36 \][/tex]
[tex]\[ \text{Noise level (dB)} = 10 \log_{10}(51.36) \][/tex]
[tex]\[ \text{Noise level (dB)} = 34.02 \][/tex]
So, the decibel level of the remaining pigs is 34.02dB.
3. Identify each statement as true or false. Current equals the product of voltage and resistance for an ohmic device.The greater the length of a wire, the higher the resistance of the wire.The potential difference produced by a battery varies depending on the circuit in which it is used.The power produced by a battery varies depending on the circuit in which it is used.The thicker an electrical wire, the higher the resistance of the wire. True False
Answer:
A) Current equals the product of voltage and resistance for an ohmic device. False (current is voltage divided by resistance)
B) The greater the length of a wire, the higher the resistance of the wire. True (Resistivity is directly proportional to lenght and inversely proportional to the cross sectional area)
C) The potential difference produced by a battery varies depending on the circuit in which it is used. False (the potential difference of a battery is constant, the current drawn is what varies with circuit)
D) The thicker an electrical wire, the higher the resistance of the wire. False ( Resistivity is inversely proportional to the cross sectional area of the conductor).
Final answer:
Current is the voltage divided by resistance, not the product. The resistance of a wire increases with its length and decreases with a larger cross-sectional area. Both the voltage and power output from a battery can vary with the circuit.
Explanation:
True or False: Electrical Principles
Let's evaluate each statement provided:
Current equals the product of voltage and resistance for an ohmic device. False. Current equals the voltage divided by the resistance (I = V/R), according to Ohm's Law.
The greater the length of a wire, the higher the resistance of the wire. True. Resistance is directly proportional to the length of the conductor (R = ρL/A).
The potential difference produced by a battery varies depending on the circuit in which it is used. True. The actual voltage across a battery can vary due to internal resistance and the characteristics of the external circuit.
The power produced by a battery varies depending on the circuit in which it is used. True. Power depends on both the current (I) and the voltage (P = IV), which can be affected by the circuit configuration.
The thicker an electrical wire, the higher the resistance of the wire. False. A thicker wire has a larger cross-sectional area, which reduces resistance (R = ρL/A).
A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the stick and a 320.0 gram object is hung at the 75.0 cm mark. Drawing is approximate. The meter stick is unbalanced. Determine the cm-mark on the meterstick that a 400 gram object needs to be hung to achieve equilibrium. A) 10.0 B) 40.0 C) 60.0 D) 90.0 E) none of the above is within 10% of my answer
Answer:C
Explanation:
Given
mass [tex]m_1=400\ gm[/tex] is at [tex]x=20\ cm[/tex] mark
mass [tex]m_2=320\ gm[/tex] is at [tex]x=75\ cm[/tex] mark
Scale is Pivoted at [tex]x=50\ cm mark[/tex]
For scale to be in equilibrium net torque must be equal to zero
Taking ACW as positive thus
[tex]T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)[/tex]
[tex]T_{net}=0.12g-0.08g=0.04g[/tex]
Therefore a net torque of 0.04 g is required in CW sense which a mass [tex]400\ gm[/tex] can provide at a distance of [tex]x_o[/tex] from pivot
[tex]0.04g=0.4\times g\times x_o [/tex]
[tex]x_o=0.1\ m[/tex]
therefore in meter stick it is at a distance of [tex]x=60\ cm[/tex]
Beats: A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with a frequency of 39.6 kHz. A car is approaching him at a speed of 35.0 m/s. The wave is reflected by the car and interferes with the emitted sound producing beats. What is the frequency of the beats when the speed of sound in air is 343 m/s?
Answer:
The frequency of the beats is 43.6408 kHz
Explanation:
Given:
f = frequency = 39.6 kHz
vc = speed of the car = 35 m/s
vs = speed of the sound = 343 m/s
Question: What is the frequency of the beats, f' = ?
As the car moves towards the source, the frequency of the beats
[tex]f'=f*(1+\frac{v_{c} }{v_{s} } )=39.6*(1+\frac{35}{343} )=43.6408kHz[/tex]
A coil is wrapped with 140 turns of wire on the perimeter of a square frame of sides 35 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 1.46 . A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to −0.426 Wb/m2 in a time of 1.19 s, find the magnitude of the induced emf in the coil while the field is changing. Answer in units of V.
The magnitude of the induced emf in the coil, calculated using Faraday's Law of Electromagnetic Induction, is 6.18 V when a uniform magnetic field changes linearly over a time of 1.19 seconds.
To find the magnitude of the induced emf in the coil while the magnetic field is changing, we will apply Faraday's Law of Electromagnetic Induction which states that the emf induced in a circuit is proportional to the rate of change of the magnetic flux through the circuit. Given that the coil has 140 turns (N) and each side of the square frame is 35 cm (which we need to convert to meters to maintain SI units, 0.35 m), we can calculate the area (A) of the square frame as:
A = side x side = 0.35 m x 0.35 m = 0.1225 m2
The magnetic field (B) changes from 0 to -0.426 Wb/m2 over a time (t) of 1.19 s. The rate of change of magnetic field (dB/dt) is:
dB/dt = - ( 0 - (-0.426) ) / 1.19 s = 0.358 Wb/m2/s
Now using Faraday’s Law, the induced emf (ε) is given by:
ε = -N x (dΦ/dt) = -N x A x (dB/dt)
Plugging in the values we get:
ε = -140 x 0.1225 m2 x 0.358 Wb/m2/s = -6.18 V
The magnitude of the induced emf is therefore 6.18 V (taking the absolute value as the question asks for magnitude).
Where does sheering often occur
In the development of atomic models, it was realized that the atom is mostly empty. Consider a model for the hydrogen atom where its nucleus is a sphere with a radius of roughly 10−15 m, and assume the electron orbits in a circle with a radius of roughly 10−10 m. In order to get a better sense for the emptiness of the atom, choose an object and estimate its width. This object will be your "nucleus". How far away would the "electron" be located away from your "nucleus"? Choose an object and estimate its width. Give an answer between 0.0001 m and 100 m. m Using your estimated width as the diameter of the "nucleus" and calculate the radius of the "electron's" orbit. m
Answer:
Check the explanation
Explanation:
a) in solving the first question, we will be choosing a spherical ball of radius 1 m, Therefore the width of the object is the diameter = 2m.
b)Given that and according to the question, the radius of the electron's orbit = 1x105 x radius of the object = 1x105 x1 = 1x105 m
Two speakers produce waves of the same wavelength that are in phase. 1) At a point midway between the speakers, you would expect to hear: Louder sound Softer sound Alternating louder and softer sounds Louder or softer sounds depending on the wavelength No interference You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. (Survey Question) 2) Briefly explain your answer to the previous questio
Answer:
louder
Explanation:
Since the two speakers producing same wavelength that are in phase,at the midpoint, the waves travel the same distance and hence path difference is zero
hence constructive interference takes place , due to this a louder sound is observed .
hence the answer is a) louder
When two speakers produce waves of the same wavelength that are in phase at a point midway between the speakers, constructive interference occurs. This results in the overlapping and combination of the waves to form a wave with higher amplitude, creating a louder sound.
Explanation:The subject of this question involves the principle of wave interference in physics. This is a phenomenon that occurs when two waves come together while traveling through the same medium. At a point midway between the speakers, when two speakers produce waves of the same wavelength that are in phase, taking into account the path lengths traveled by the individual waves, you would expect to hear a louder sound.
This is a case of constructive interference, where the two sound waves, being in phase and of the same wavelength, will overlap and combine to form a wave with a greater amplitude, leading to a louder sound. This is explained in Figure 17.17 and 16.36, where the difference in the path lengths is one wavelength, resulting in total constructive interference and a resulting amplitude equal to twice the original amplitude.
However, it is worthy to note that in the real world recognition of this increased amplitude or louder sound will depend on the specific frequency of the sound, as sonic perception can vary with frequencies. This explanation is in reference to a single tone or frequency. When discussing music which is composed of many frequencies, the actual perception might be a bit more complex.
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g The steam above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature (in °C) of 240 g of hot coffee initially at 80.0°C if 2.50 g evaporates from it? The coffee is in a Styrofoam cup, and so other methods of heat transfer can be neglected. Assume that coffee has the same physical properties as water.
Answer:
the final temperature = 74.33°C
Explanation:
Using the expression Q = mcΔT for the heat transfer and the change in temperature .
Here ;
Q = heat transfer
m = mass of substance
c = specific heat
ΔT = the change in temperature
The heat Q required to change the phase of a sample mass m is:
Q = m[tex]L_v[/tex]
where;
[tex]L_v[/tex] is the latent heat of vaporization.
From the question ;
Let M represent the mass of the coffee that remains after evaporation is:
ΔT = [tex]\frac{mL_v}{MC}[/tex]
where;
m = 2.50 g
M = (240 - 2.50) g = 237.5 g
[tex]L_v[/tex] = 539 kcal/kg
c = 1.00kcal/kg. °C
ΔT = [tex]\frac{2.50*539 \ kcal /kg}{237.5 g *1.00 \ kcal/kg . ^0C}[/tex]
ΔT = 5.67°C
The final temperature of the coffee is:
[tex]T_f = T_i -[/tex] ΔT
where ;
[tex]T_I[/tex] = initial temperature = 80 °C
[tex]T_f[/tex] = (80 - 5.67)°C
[tex]T_f[/tex] = 74.33°C
Thus; the final temperature = 74.33°C
Answer:
Final temperature of the hot coffee, [tex]\theta_{f} = 85.67^0 C[/tex]
Explanation:
Mass of coffee remaining after evaporation of 2.50 g, M = 240 - 2.50
M = 237.5 g
Mass of evaporated coffee, m = 2.5 g
Initial temperature of hot coffee, [tex]\theta_{i} = 80^{0} C[/tex]
Initial temperature of hot coffee, [tex]\theta_{f} =[/tex] ?
Let the specific heat capacity of the coffee, c = 1 kcal/kg
Latent heat of vaporization of coffee, [tex]L_{v} = 539 kcal/kg[/tex]
The heat energy due to temperature change:
[tex]Q = Mc \triangle \theta[/tex]
[tex]Q = 237.5 * 1 * \triangle \theta[/tex]...........(1)
The heat energy due to change in state
[tex]Q = mL_{v}[/tex]
Q = 2.5 * 539
Q = 1347.5..........(2)
Equating (1) and (2)
[tex]1347.5 = 237.5 \triangle \theta\\\triangle \theta = 1347.5/237.5\\\triangle \theta =5.67^{0} C[/tex]
[tex]\triangle \theta = \theta_{f} - \theta_{i} \\5.67 = \theta_{f} - 80\\\theta_{f} = 80 + 5.67\\\theta_{f} = 85.67^0 C[/tex]
Which side of the electromagnetic spectrum has shorter wavelengths?
Which side of the electromagnetic spectrum has higher energy?)
Which side of the electromagnetic spectrum has higher frequencies?
Gamma rays have the smallest wavelengths and the most energy of any wave in the electromagnetic spectrum. They are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.
What is wavelength ?
"Distance between corresponding points of two consecutive waves." “Corresponding points” refers to two points or particles in the same phase—i.e. points that have completed identical fractions of their periodic motion.
What is frequency?
"Tthe number of waves that pass a fixed point in unit time. "It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
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Answer:
1.blue
2.blue
3.blue
Explanation:
I took the quiz
A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at point A. The angle of incidence is 70 degrees. The depth of the lake is 4.3 m. On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point B would the laser beam strike the lakebottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?
Answer:
A) d = 11.8m
B) d = 4.293 m
Explanation:
A) We are told that the angle of incidence;θ_i = 70°.
Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;
tan 70° = d/4.3m
Where d is the distance from point B at which the laser beam would strike the lakebottom.
So,d = 4.3*tan70
d = 11.8m
B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)
So,
n1*sinθ_i = n2*sinθ_r
Thus; sinθ_r = (n1*sinθ_i)/n2
sinθ_r = (1 * sin70)/1.33
sinθ_r = 0.7065
θ_r = sin^(-1)0.7065
θ_r = 44.95°
Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;
d = 4.3 tan44.95
d = 4.293 m
Which resistors in the circuit must always have the same current?
Answer:
Resistors in series in the circuit must always have the same current
Explanation:
Resistors are said to be connected in series if they are connected one after another.
The total resistance in the circuit with resistors connected in series is equal to the sum of individual resistances.
Individual resistors in series do not get the total source voltage. Total source voltage divide among them.
Answer:
it's a and d
Explanation:
because it all goes around if that makes sense.. and if u guys are here from apex
Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that drivers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long exposure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 10-3 T, oscillating sinusoidally at 72.5 Hz. If the diameter of a red blood cell is 7.60 µm, determine the maximum emf that can be generated around the perimeter of a cell in this field.
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
A researcher with the Ministry of Transportation is commissioned to study the drive times to work (one-way) for U.S. cities. The underlying hypothesis is that average commute times are different across cities. To test the hypothesis, the researcher randomly selects six people from each of the four cities and records their one-way commute times to work. Refer to the below data on one-way commute times (in minutes) to work. Note that the grand mean is 36.625. Houston Charlotte Tucson Akron 45 25 25 10 65 30 30 15 105 35 19 15 55 10 30 10 85 50 10 5 90 70 35 10 x¯i 74.167 36.667 24.833 10.833 s2i 524.167 436.667 82.167 14.167 The competing hypotheses about the mean commute times are ______________ a) H0: μ1 = μ2 = μ3, HA: Not all population means are equal H0: b) Not all population means are equal, HA: μ1 = μ2 = μ3 H0: μ1 = μ2 = μ3 = μ4, HA: c) Not all population means are equal H0: d) Not all population means are equal, HA: μ1 = μ2 = μ3 = μ4
Answer:
d) H0: Not all population means are equal,
HA: μ1 = μ2 = μ3 = μ4
Explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
Therefore, for the case above;
Since the underlying hypothesis is that average commute times are different across cities.
Null hypothesis H0: is that Not all population means are equal.
Alternative hypothesis Ha: μ1 = μ2 = μ3 = μ4
What is the most common and powerful agent of erosion? wind ice water animal behavior
Answer:
water
Explanation:
Answer:water
Explanation:because it shaped the grand canyon
In order for a current to be maintained through a conductor a difference in ________________________ must be maintained from one end of the conductor to the other end.
Answer:
Potential
Explanation:
In order for a current to be maintained through a conductor a difference in potential must be maintained from one end of the conductor to the other end. Due to potential difference and emf is induced in it. As a result current starts to flow.
A 0.415-kg mass suspended from a spring undergoes simpleharmonic oscillations with a period of 1.4 s. How much mass, inkilograms, must be added to the object to change the period to2.2 s?
Answer:
m=0.893kg
Explanation:
time period of oscillations is given by= 2π√(m/k)
m: mass of the object
k: spring constant
when T=1.5 and m=0.415
1.5= 2π√(0.415/k)
k= 7.27 N/m
when T= 2.2s
2.2= 2π√(m/7.27)
m=0.893kg
The mass in kilograms that must be added to the object to change the period to 2.2 s is 1.025kg
The formula for calculating the period of a simple pendulum is expressed as:
[tex]T= 2\pi \sqrt{\frac{m}{k} }[/tex]
m is the mass of the spring
k is the spring constant
Given the following parameters
m = 0.415kg
T = 1.4 secs
Get the spring constant
[tex]1.4=2(3.14)\sqrt{\frac{0.415}{k} } \\1.4=6.28\sqrt{\frac{0.415}{k} }\\ 0.2229=\sqrt{\frac{0.415}{k} }\\ \frac{0.415}{k} =0.0497\\0.0497k=0.415\\k=\frac{0.415}{0.0497}\\k= 8.35N/m[/tex]
Given the period is 2.2secs, the mass of the spring will be expressed as:
[tex]2.2=2 (3.14)\sqrt{\frac{m}{8.35} } \\2.2=6.28\sqrt{\frac{m}{8.35} }\\\sqrt{\frac{m}{8.35} }=0.3503\\\frac{m}{8.35} =0.1227\\m=1.025kg[/tex]
Hence the mass in kilograms that must be added to the object to change the period to 2.2 s is 1.025kg
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Nitrogen is stored in a large chamber under conditions of 450 K and 1.5 × 105 N/ m2 . The gas leaves the chamber through a convergent-only nozzle whose outlet area is 30 cm2 . The ambient room pressure is 1 × 105 N/m2 and there are no losses. (a) What is the velocity of the nitrogen at the nozzle exit? (b) What is the mass flow rate? (c) What is the maximum flow rate that could be obtained by lowering the ambi- ent pressure?
Answer:
a) 319.56 m/s
b) 0.8057 kg /s
c) 0.840 kg /s
Explanation:
Check the pictures attached for the whole explanation
Final answer:
The velocity of nitrogen at the nozzle exit and the mass flow rate depend on the conditions and properties of the gas, using principles such as the Bernoulli equation and the ideal gas law. Without specific constants, these values cannot be determined precisely. The maximum flow rate would be obtained by reducing the ambient pressure below the critical pressure ratio, causing choked flow.
Explanation:
To answer the student's question about the velocity of the nitrogen gas at the nozzle exit, we need to apply the principles of fluid dynamics. Under the provided conditions, without any losses and assuming adiabatic expansion, the velocity can be calculated using the Bernoulli equation and the equation of continuity. However, given that we only have the inlet pressure, outlet pressure, and temperature, we need additional equations like the ideal gas law to find the density at the exit. Unfortunately, without the specific heat ratio or gas constant for nitrogen, a precise calculation cannot be provided.
The mass flow rate can then be calculated by multiplying the outlet velocity by the outlet area and the density of nitrogen under the given conditions. An approximation can be made, but with the given data, it is not possible to provide an exact value since the outlet velocity is not known.
Regarding the maximum flow rate, it is related to the critical pressure ratio of the gas. If the ambient pressure is decreased below a certain critical point, the flow becomes choked, and the mass flow rate reaches a maximum. This maximum is a function of the chamber conditions and the specific properties of nitrogen.
Using this formula , What is the weight, on Earth, of an object with a mass
of 72.4kg?*
716.52N
659.52N
739.52N
709.52N
Answer:
W = 709.52 N
Explanation:
It is given that,
The mass of an object on Earth is 72.4 kg.
We need to find the weight of the object on Earth.
The formula of the weight of an object is given by :
W = mg
g is acceleration due to gravity on Earth, g = 9.8 m/s²
So,
[tex]W = 72.4\times 9.8\\\\W=709.52\ N[/tex]
So, the weight of the object is 709.52 N.
Answer:
So, the weight of the object is 709.52 N.
Explanation:
Submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1 400 Hz. The speed of sound in the water is 1 533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. What frequency is detected by an observer riding on sub B as the subs approach each other?
Answer:
1385 hz
Explanation:
Yes if it’s right pls mark the brainliest
An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses the ammeter to measure the current I and calculates the power P dissipated by the resistor using I and e. The student later discovers that the ammeter is not an ideal ammeter. Which of the following correctly compares the actual power dissipated by the resistor to the value of power P calculated by the student and explains why? (A) The actual power dissipated by the resistor is greater than P because the ammeter had some resistance (B) The actual power dissipated by the resistor is less than P because the ammeter had some resistance. (C) The actual power dissipated by the resistor is greater than P because the ammeter should have been connected in parallel to the resistor. (D) The actual power dissipated by the resistor is less than P because the ammeter should have been connected in parallel to the resistor. (E) The actual power dissipated by the resistor is equal to P because the ammeter was connected properly in series.
Answer:
B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Explanation:
Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.
Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.
So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.
Also,option D is ruled out as ammeter is connected in series.
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
a. Is angular momentum conserved?
b. Find the change in kinetic energy of the block, in J.
c. How much work was done in pulling the cord? in J.
Answer:
W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]
Explanation:
a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.
b) By the law of conservation of energy angular momentum
[tex]L_1=L_2[/tex]
[tex]I_1\omega_1=I_2\omega_2[/tex]
[tex]mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1[/tex]
[tex]\omega_2=(\frac{0.3}{0.15})^2\times2.85[/tex]
[tex]\omega_2=5.7\text{ rad/sec}[/tex]
c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)
[tex]K_1=\frac{1}{2} mr_1^2\omega_1^2[/tex]
[tex]K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2[/tex]
[tex]=9.12\times10^{-3} J[/tex]
Now,
[tex]K_2=\frac{1}{2} mr_2^2\omega_2^2[/tex]
[tex]K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2[/tex]
[tex]K_2=18.24\times10^{-3}[/tex] J
Therefore, Work done W= [tex]K_2-K_1==9.12\times10^{-3} J[/tex]
a. Angular momentum is conserved in this system. b. The change in kinetic energy of the block can be calculated. c. No work is done in pulling the cord.
Explanation:a. Angular momentum is conserved when there are no external torques acting on the system. In this case, since the block is revolving on a frictionless surface, and there is no mention of any external torques, we can assume that angular momentum is conserved.
b. The change in kinetic energy of the block can be calculated using the equation ΔKE = KE_final - KE_initial. Since the block is modeled as a particle, its kinetic energy is given by KE = 1/2 * m * v^2, where m is the mass and v is the linear velocity. As the radius is changed, the linear velocity changes, and we can calculate the change in kinetic energy.
c. The work done in pulling the cord can be calculated using the equation W = ΔKE + ΔPE, where W is the work done, ΔKE is the change in kinetic energy, and ΔPE is the change in potential energy. In this case, since the block is on a frictionless surface, there is no change in potential energy, and we only need to calculate the change in kinetic energy.
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Identify three pollutants released into the air when fossil fuels are burned.
Answer:
Carbon Monoxide (CO), Carbon Dioxide (CO2), and Sulfur Dioxide (SO2).
Answer:Carbon dioxide (CO2), Carbon monoxide (CO), Methane (CH4)
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A large, 36.0 kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell’s center of mass is 0.55 m below the pivot. The bell’s moment of inertia about an axis at the pivot is 36.0 kg ·m2 . The clapper is a small, 2.8 kg mass attached to one end of a slender rod of length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length L of the clapper rod for the bell to ring silently — that is, for the period of oscillation for the bell to equal that of the clapper?
Answer:
Length of the rod is 0.043m
Explanation:
Mass of the bell (M) = 36kg
Distance of the center of mass from pivot = 0.55m
Moment of inertia (I) = 36.0kgm²
Mass of clipper (m) = 2.8kg
Length of the bell to ring = L
The period of the pendulum with small amplitude of oscillation is
T = 2π √(I / mgd)
Where g = acceleration due to gravity = 9.8 m/s²
T = 2π √(36 / 36*9.8*0.55)
T = √(0.1855)
T = 0.43s
The period of the pendulum is 0.43s
To find the length of the Clapper rod for which the bell ring slightly,
T = 2π√(L / g)
T² = 4π²L / g
T² *g = 4π²L
L = (T² * g) / 4π²
L = (9.8* 0.43²) / 4π²
L = 1.7287 / 39.478
L = 0.043m
The length of the Clapper rod for the bell to ring slightly is 0.043m
A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of radius R. The small and large cars have accelerations as and aL respectively, as they travel around the curve. The magnitude of as is twice of that of aL. How does the speed of the small car VS compare to the speed of the large car VL as they move around the curve
Answer:
[tex]v_S=\sqrt{2}v_L[/tex]
Explanation:
The acceleration experimented while taking a curve is the centripetal acceleration [tex]a=\frac{v^2}{r}[/tex]. Since [tex]a_S=2a_L[/tex], we have that: [tex]\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}[/tex]
They take the same curve, so we have: [tex]r_S=r_L=R[/tex]
Which means: [tex]v_S^2=2v_L^2[/tex]
And finally we obtain: [tex]v_S=\sqrt{2}v_L[/tex]
You are in the lab and are given two rods set up so that the top rod is directly above the bottom one. The two straight rods 50-cm long are separated by a distance of 1.5-mm each with 15 A of current passing through them. Assume that the bottom rod is fixed and that the top rod is somehow free to move vertically but not horizontally. How much must the top rod weigh so that the system is at equilibrium
Answer:
Explanation:
The magnetic force due to lower rod must be equal to weight of upper rod for equilibrium .
magnetic field due to lower rod on upper rod
= ( μ₀ / 4π ) x(2i / r ) , i is current , r is distance between rod
= 10⁻⁷ x 2 x 15 / 1.5 x 10⁻³
= 20 x 10⁻⁴ T
force on the upper rod
= B i L , B is magnetic field , i is current in second rod and L is its length
= 20 x 10⁻⁴ x 15 x .50
= 150 x 10⁻⁴ N
= .015 N
This force can balance a wire having weight equal to .015 N .
= .00153 kg
= 1.53 g .
wire should weigh 1.53 g .
what is the the wavelength if a wave cycles up and down three times per second and the distance between each wave is 1.7m?
Answer:5.1 meters
Explanation:
wavelength=3 x1.7
Wavelength=5.1 meters
g I can test a new wheel design by rolling it down a test ramp. I release a wheel of mass m=1.6 kg and radius r=0.37 m from rest at an initial height of h=6.7 m at the top of a test ramp. It rolls smoothly to the bottom without sliding. I measure the linear speed of the wheel at the bottom of the test ramp to be v=4.7 m/s. What is the rotational inertia of my wheel?
Answer:
The rotational inertia of my wheel is [tex]I =1.083 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the wheel is [tex]m = 1.6 \ kg[/tex]
The radius of the wheel is [tex]r = 0.37 \ m[/tex]
The height is [tex]h = 6.7 m[/tex]
The linear speed is [tex]v = 4.7 m/s[/tex]
According to the law of energy conservation
[tex]PE = KE + KE_R[/tex]
Where PE is the potential energy at the height h which is mathematically represented as
[tex]PE = mgh[/tex]
While KE is the kinetic energy at the bottom of height h
[tex]KE = \frac{1}{2} mv^2[/tex]
Where [tex]KE_R[/tex] is the rotational kinetic energy which is mathematically represented as
[tex]KE_R = \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
Where [tex]I[/tex] is the rotational inertia
So substituting this formula into the equation of energy conservation
[tex]mgh = \frac{1}{2} mv^2 + \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
=> [tex]I =[ \ mgh - \frac{1}{2} mv^2 \ ]* \frac{2 r^2}{v^2}[/tex]
substituting values
[tex]I =[ \ 1.6 * 9.8 * 6.7 - \frac{1}{2} * 1.6 *4.7^2 \ ]* \frac{2 * 0.37^2}{4.7^2}[/tex]
[tex]I =1.083 \ kg \cdot m^2[/tex]
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.59 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2190 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.91 V/m, (b) in the negative z direction and has a magnitude of 3.91 V/m, and (c) in the positive x direction and has a magnitude of 3.91 V/m?
Answer:
a) 1.88*10^-18 N
b) 6.32*10^-19 N
c) 1.9*10^-18 N
Explanation:
The total force over the electron is given by:
[tex]\vec{F}=q\vec{E}+q\vec{v}\ X\ \vec{B}[/tex]
the first term is the electric force and the second one is the magnetic force.
You have that the velocity of the electron and the magnetic field are:
[tex]\vec{v}=2190\frac{m}{s}\ \hat{j}\\\\\vec{B}=-5.39*10^{-3}T\ \hat{i}[/tex]
by using the relation j X (-i) =- j X i = -(-k) = k, you obtain:
[tex]\vec{v} \ X\ \vec{B}=(2190m/s)(3.59*10^{-3}T)\hat{k}=7.862\ T m/s[/tex]
a) For an electric field of 3.91V/m in +z direction:
[tex]\vec{F}=q[\vec{E}+\vec{v}\ X\ \vec{B}]=(1.6*10^{-19})[3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=1.88*10^{-18}N\hat {k}\\\\F=1.88*10^{-18}N[/tex]
b) E=3.91V/m in -z direction:
[tex]\vec{F}=(1.6*10^{-19})[-3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=6.32*10^{-19}N\hat {k}\\\\F=6.32*10^{-19}N[/tex]
c) E=3.91 V/m in +x direction:
[tex]\vec{F}=(1.6*10^{-19})[3.91\hat{i}+7.862\ \hat{k}]N\\\\\vec{F}=[6.25*10^{-19}\ \hat{i}+1.25*10^{-18}\ \hat{k} ]N\\\\F=\sqrt{(6.25*10^{-19})^2+(1.25*10^{-18})^2}N=1.9*10^{-18}\ N[/tex]
A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a distance L from the pivot. On the other side of the pivot point are two smaller children each of mass M/2. One is sitting a distance L/6 from the pivot. How far from the pivot must the other small child be sitting in order for the system to be balanced?
Answer:[tex]\frac{5L}{6}[/tex]
Explanation:
Given
Wooden board is pivoted at center and
Older child of mass [tex]M=32\ kg[/tex] is sitting at a distance of L from center
if two child of mass [tex]\frac{M}{2}[/tex] is sitting at a distance [tex]\frac{L}{6}[/tex] and [tex]x[/tex](say) from pivot then net torque about pivot is zero
i.e.
[tex]\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx[/tex]
as [tex]\tau_{net}=0[/tex]
Therefore
[tex]MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx[/tex]
[tex]L-\frac{L}{6}=x[/tex]
[tex]x=\frac{5L}{6}[/tex]
Therefore another child is sitting at a distance of [tex]\frac{5L}{6}[/tex]