Explanation:
The given data is as follows.
mass (m) = 5 kg
Height of tower = 15 m
u = 7 m/s
air resistance = 610 v
(a) Now, differential equation for the given mass which is thrown vertically upwards is as follows.
[tex]m \frac{d^{2}x}{dt^{2}}[/tex] = F
-bv = Fr
Here, mg is downwards due to the force of gravity.
[tex]\frac{md^{2}x}{dt^{2}} = bv - mg[/tex]
[tex]\frac{md^{2}x}{dt^{2}} + b \frac{dx}{dt} + mg[/tex] = 0
Hence, the differential equation required to solve the problem is as follows.
[tex]\frac{md^{2}x}{dt^{2}} + b \frac{dx}{dt} + mg[/tex] = 0
(b) When final velocity of the object is equal to zero then the object will reach towards its maximum height and it will start to fall downwards.
F = [tex]\frac{md^{2}x}{dt^{2}}[/tex]
= 0
Therefore, the object reach its maximum height at v = 0.
Which of these atoms is most likely to share electrons with other atoms?
Answer:
your question is incomplete as the options are not given. I guess following is the complete question.
Which of these atoms is most likely to share electrons with other atoms?
a) chlorine (7 valence electrons)
b) calcium (2 valence electrons)
c) argon (8 valence electrons)
d) carbon (4 valence electrons)
e) potassium (1 valence electron)
The correct option is d) carbon (4 valence electrons)
Explanation:
Carbon has four electrons in its valence shell. In order to complete the 8 electrons in its valence shell carbon has to make four covalent bonds by sharing its four electrons with the other atom. Carbon atom will neither gain the electrons nor it losses the electrons to follow the octet rule. So in the above mentioned options carbon is the atom that will share maximum electrons.
→ Chlorine has 7 electrons, it will gain 1 electron. It will not do the sharing.
→ Calcium has 2 electrons, it will lost these 2 electrons to complete its shell.
→ Argon has already a completed shell. It will not react with other atom.
→ Potassium has only 1 valence electron which it will lose to complete its shell.
Answer: F
Explanation:
A beam of protons is accelerated through a potential difference of 0.750 kVkV and then enters a uniform magnetic field traveling perpendicular to the field. You may want to review (Pages 641 - 643) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of electron motion in a microwave oven. Part A What magnitude of field is needed to bend these protons in a circular arc of diameter 1.80 mm ? Express your answer in tesla to three significant figures. BpBp = nothing TT SubmitRequest Answer Part B What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons? Express your answer in tesla to three significant figures. BeBe = nothing TT SubmitRequest Answer
Answer:
Explanation:
A force is provided by the magnetic field which perpendicular to both the velocity of the charge and magnetic field
F = qvB where B is the magnetic field in tesla, q is charge and v is velocity
The potential energy is transferred into kinetic energy
PE = Vq = 1/2 mv²
v = √(2Vq/ m)
charge on the proton = 1.602 × 10 ⁻¹⁹ C and mass of a proton = 1.673 × 10⁻²⁷
v = √ (( 2 × 1.602 × 10 ⁻¹⁹ C × 0.75 × 10³V ) / (1.673 × 10⁻²⁷)) = √( 1.4363 × 10¹¹ ) = 3.79 × 10⁵ m/s
The force of magnetic field produces centripetal force
qvB = mv² /R
where R radius = 1.80mm = 0.0018 m / 2 = 0.0009 m
qvB = ( m / R ) × (2qV /m)
cancel the common terms
vB = 2V / R
3.79 × 10⁵ m/s × B = 2 × 0.75 × 10³V / 0.0009 = 1.667 × 10⁶
B = 1.667 × 10⁶ / 3.79 × 10⁵ m/s = 4.40 T
b) magnetic field needed for the electron
qvB = mv² /R where m is the mass of an electron = 9.11 × 10⁻³¹ Kg
qB = mv/R
qB = ( 9.11 × 10⁻³¹ Kg × 3.79 × 10⁵ m/s) / 0.0009
qB = 3.8363 × 10 ⁻²²
B = 3.8363 × 10 ⁻²² / 1.602 × 10⁻¹⁹ kg = 0.0239 T
ou are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 370 N . Part A How much work do you do on the refrigerator during its trip down the ramp
Answer:
Explanation:
component of force along the ramp
= 370 cos θ where θ is the slope of the ramp with respect to ground.
sinθ = 1.4 / 5
θ = 16 degree
370 cos16
= 355.67 N
Work done
= component of force along the ramp x length of the ramp
= 355.67 x 5
= 1778.35 J
Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen that is 4 m from the slits. Near the center of the secreen the separation between adjacent maxima is 2 mm. What is the distance between the two slits?
Answer:
The distance between the two slits is 1.2mm.
Explanation:
The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.
[tex]\Lambda x = L\frac{\lambda}{d} [/tex] (1)
Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.
If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.
Therefore, d can be isolated from equation 1.
[tex]d = L\frac{\lambda}{\Lambda x} [/tex] (2)
Notice that it is necessary to express L and [tex]\lambda[/tex] in units of millimeters.
[tex]L = 4m \cdot \frac{1000mm}{1m}[/tex] ⇒ [tex]4000mm[/tex]
[tex]\lambda = 600nm \cdot \frac{1mm}{1x10^{6}nm}[/tex] ⇒ [tex]0.0006mm[/tex]
[tex]d = (4000mm)\frac{0.0006mm}{2mm} [/tex]
[tex]d = 1.2mm[/tex]
Hence, the distance between the two slits is 1.2mm.
An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?
Answer:
Explanation:
Given an RC series circuit
Initially uncharged capacitor
C=3.67 μF
Resistor R=8.01 k Ω=8010 ohms
Battery EMF(V)=1.5V with negligible internal resistance.
The initial current in the circuit?
At the beginning the capacitor is uncharged and it has a 0V, so all the voltage appears at the resistor,
Now using ohms law
V=iR.
i=V/R
i=1.5/8010
i=0.000187A
1mA=10^-3A
Therefore, 1A = 1000mA
i=0.187 milliamps
The initial current in the circuit is 0.187 mA
Answer:
The initial current is 0.0187 mA.
Explanation:
Given that
capacitance is given as 3.67 x 10⁻⁶ F
resistance is given as 8010 Ω
voltage across the circuit is 1.5 V
Since the capacitor is initially uncharged, the capacitive reactance is zero.
From ohms law;
Voltage across the circuit is directly proportional to the opposition to the flow of current.
In these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:
[tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{8.011e3\Omega}\\ = 0.0187 mA[/tex]
The initial current (that will be diminishing as the capacitor charges), is 0.0187 mA.
In a game of baseball, a player hits a high fly ball to the outfield. (a) Is there a point during the flight of the ball where its velocity is parallel to its acceleration? (b) Is there a point where the ball’s velocity is perpendicular to its acceleration? Explain in each case.
Final answer:
The velocity and acceleration vectors of a baseball can change during its flight. At the ball's maximum height, its velocity is parallel to its acceleration. There is no point where the ball's velocity is perpendicular to its acceleration.
Explanation:
In a game of baseball, the velocity and acceleration of a ball can change as it moves through the air. (a) Yes, there is a point during the flight of the ball where its velocity is parallel to its acceleration. This occurs when the ball reaches its maximum height. At this point, the ball stops moving upward and starts moving downward. Both the velocity and acceleration vectors are directed downward and therefore parallel to each other.
(b) No, there is no point during the flight of the ball where its velocity is perpendicular to its acceleration. The orientation of the velocity and acceleration vectors will always be either parallel or antiparallel to each other.
You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time the box is being pushed. Which method for pushing results in the minimum average force being applied to the box? a. Keep pushing the box forward at a steady speed, b. Push the box forward a short distance, rest, then repeat until finished, c. Push the box so that it accelerates forward at a constant rate.
When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.
Remember that this law tells us that
[tex]F= ma[/tex]
[tex]F= m \frac{\Delta v}{t}[/tex]
Therefore the best strategy is A. keep pushing the box forward at a steady speed
To minimize the average force applied to a box being pushed across a rough floor, you should keep pushing the box forward at a steady speed. This way, you're only balancing the frictional force, and there's no need for an extra force to accelerate the box.
Explanation:To minimize the average force applied to the box on a rough floor, you would opt for method a. Keep pushing the box forward at a steady speed. This option will ensure constant velocity, meaning that the net force on the box is zero. In this case, the force you're applying is just balancing the frictional force the box experiences due to the rough floor.
Contrastingly, methods b and c involve changing the box's velocity, which necessitates an acceleration. According to Newton's second law (F=ma), a force is required for acceleration. Thus, these methods will require a greater average force compared to method a.
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Two identical capacitors are connected parallel. Initially they are charged to a potential V0 and each acquired a charge Q0. The battery is then disconnected, and the gap between the plates of one capacitor is filled with a dielectric. (a) What is the new potential difference V across the capacitors. possible asnwers: V=(Vo)^2/[kQo+Vo), V=Vo/2k, V=Vo/2, V=kQo/Vo, V=2Vo/[k+1]
(b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
Answer:
Explanation:
capacitance of each capacitor
C₀= Q₀ / V₀
V₀ = Q₀ / C₀
New total capacitance = C₀ ( 1 + K )
Common potential
= total charge / total capacitance
= 2 Q₀ / [ C₀ ( 1 + K ) ]
2 V₀ / ( 1 + K )
b )
Common potential = 2 x V₀ / ( 1 + 7.8 )
= .227 V₀
charge on capacitor with dielectric
= .227 V₀ x 7.8 C₀
= 1.77 V₀C₀
= 1.77 Q₀
Ratio required = 1.77
The new potential difference V across the capacitors and the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using specific equations.
Explanation:(a) What is the new potential difference V across the capacitors.
To find the new potential difference V across the capacitors, we need to consider the effect of adding a dielectric. The potential difference V is given by the equation:
V = Vo/2k
where Vo is the initial potential difference and k is the dielectric constant. In this case, since the dielectric constant is 7.8, we can substitute the values and calculate the new potential difference V.
(b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
The ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using the equation:
Q_with_dielectric / Q_initial = k
where Q_with_dielectric is the charge on the capacitor with the dielectric and Q_initial is the initial charge. Given that k is 7.8, we can substitute the values and calculate the ratio.
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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ.
Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is PV = C
T = C for isothermal process
Calculating for the work done in isothermal process
W = P₁V₁ [tex]ln[\frac{P_{1} }{P_{2} }][/tex]
= mRT₁[tex]ln[\frac{P_{1} }{P_{2} }][/tex] [∵pV = mRT]
= (5) (0.287) (272.039) [tex]ln[\frac{2.0}{1.0}][/tex]
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
ΔU = [tex]mc_{v}(T_{2} - T_{1} ) = 0[/tex]
From 1st law of thermodynamics
Q = ΔU + W
= 0 + 270.588
= 270.588 kJ
To find the work and heat transfer in this process, we can apply the First Law of Thermodynamics. We can determine the work done using the equation pV = constant and calculate the heat transfer using the equation Q = CpdT.
To find the work and heat transfer in this process, we need to apply the First Law of Thermodynamics. The First Law states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system. In this case, since the process is isobaric (constant pressure), we can determine the work done by integrating the equation pV = constant. This will give us the equation W = p(Vf - Vi), where W is the work done, p is the pressure, and Vf and Vi are the final and initial volumes respectively.
Since the process is isobaric, the heat transfer can be calculated using the equation Q = CpdT, where Q is the heat transfer, Cp is the specific heat at constant pressure, and dT is the change in temperature. In this case, since the temperature change is not given explicitly, we can assume it to be the same as the change in internal energy, which gives us dT = dU/Cp.
By substituting the given values into the equations, we can calculate the work and heat transfer.
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A 0.1-kilogram block is attached to an initially unstretched spring of force constant k = 40 newtons per meter as shown above. The block is released from rest at time t=0.
What is the amplitude, in meters, of the resulting simple harmonic motion of the block?
A) 1/40m
B) 1/20m
C) 1/4m
D) 1/2m
Answer:
The amplitude, in meters, of the resulting simple harmonic motion of the block 1/40 m
Explanation:
given information:
the block mass, m = 0.1 kg
spring of force constant, k = 40 N/m
release at t = 0
the formula for simple harmonic motion is
F = kx
where
F = force (N)
k = spring of force constant (N/m)
x = amplitude (m)
thus,
F = kx
x = F/k
= m g/k
= 0.1 x 10/40
= 1/40 m
A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is then discharged by connecting it directly across a 100-k\Omega100−kΩ resistor. At what point in time after the resistor is connected will the capacitor have 13.5~\mu C13.5 μC of charge remaining on it?
Explanation:
The given data is as follows.
C = [tex]20 \times 10^{-6} F[/tex]
R = [tex]100 \times 10^{3}[/tex] ohm
[tex]Q_{o} = 100 \times 10^{-6}[/tex] C
Q = [tex]13.5 \times 10^{-6} C[/tex]
Formula to calculate the time is as follows.
[tex]Q_{t} = Q_{o} [e^{\frac{-t}{\tau}][/tex]
[tex]13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}][/tex]
0.135 = [tex]e^{\frac{-t}{2}}[/tex]
[tex]e^{\frac{t}{2}} = \frac{1}{0.135}[/tex]
= 7.407
[tex]\frac{t}{2} = ln (7.407)[/tex]
t = 4.00 s
Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.
The required time to discharge the capacitor to a certain charge can be calculated by rearranging the exponential decay formula. The product of resistance and capacitance, known as the RC time constant, is a key value in this calculation. After substituting the values given into the formula, we get the required time.
Explanation:The time it takes for a capacitor to discharge through a resistor can be calculated using the formula for the decay of charge on a capacitive circuit: Q = Q0 * e-t/RC, where Q is the charge at time t, Q0 is the initial charge, R is the resistance and C is the capacitance. To find the time at which the charge will be 13.5 uC, we rearrange the formula to solve for time t: t = -RC * ln(Q/Q0).
Using the values of the problem: R = 100 kΩ = 100,000 Ω, C = 20 uF = 20*10-6 F, Q0 = 100 uC = 100*10-6 C, and Q = 13.5 uC = 13.5*10-6 C, substitute these values into the equation: t = -100,000 * 20*10-6 * ln (13.5/100). Hence, the time it takes for a capacitor to discharge to a specified charge can be calculated using exponential decay formula based on the capacitor's unique RC time constant.
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A tennis ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s. A) How long does it take the ball to reach its maximum height?B) How high does the ball rise?C) What is the magnitude of the acceleration due to gravity on the surface of this planet?D) Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive.E) How long has the ball been in the air when it returns to its original position?
Answer:
(A) t = 4s
(B) H = 40m
(C) g = 5m/s²
(D) V = -20m/s
(E) t = 8s
The detailed solution to this problem requires the knowledge of costant linear acceleration motion.xplanation:
The detailed solution to this problem requires the knowledge of costant linear acceleration motion.
Explanation:
The full solution can be found in the attachment below.
To answer part A, we have been given some values of velocities bounding the time jnterval of 1s from which we can calculate the acceleration due to gravity and then the time.
Part B
Requires just using the acceleration due to gravity and the time taken in the equation
H = ut - 1/2gt²
Part C
Has already been calculated in part A
Part D
V = -20m/s because the tennis ball is coming down as the upward direction was assumed positive.
Part E
When the ball retirns to its original position it is the same as it never left and so H = 0m
The calculation can be found in the attachment below.
Final answer:
Ball reaches peak at 4s, rises 80m, experiences -5m/s² gravity. Returns at -20m/s and spends 8s in the air (4s up + 4s down).
Explanation:
Here are the answers to your questions about the tennis ball on the atmosphere-free planet:
A) Time to reach maximum height: Between the beginning and one second later, the ball's velocity decreases by 5.0 m/s. Since acceleration is constant (no atmosphere), this decrease is due to gravity acting against the initial velocity. We can use the equation:
v = u + at
where:
v = final velocity (15.0 m/s)
u = initial velocity (20.0 m/s)
a = acceleration due to gravity (unknown, negative direction)
t = time (1 second)
Solving for a:
a = (v - u) / t = (15.0 m/s - 20.0 m/s) / 1 s = -5.0 m/s²
Now, to find the time to reach maximum height, we know that at the peak, the velocity is 0 m/s. We can again use the same equation:
0 = u + at
Solving for t:
t = -u / a = -20.0 m/s / (-5.0 m/s²) = 4.0 seconds (positive result since time cannot be negative)
Therefore, it takes 4.0 seconds for the ball to reach its maximum height.
B) Maximum height:
We can use the kinematic equation to find the maximum height (h):
h = ut + 1/2 × at²
Substituting the values:
h = 20.0 m/s × 4.0 s + 1/2 × (-5.0 m/s²) × (4.0 s)² = 40.0 m + 40.0 m = 80.0 meters
C) Acceleration due to gravity:
We already calculated the acceleration due to gravity in part A: -5.0 m/s² (the negative sign indicates downward direction).
D) Velocity upon returning to the original position:
Since the motion is symmetrical (same initial and final positions), the velocity when the ball returns will be the negative of its initial velocity: -20.0 m/s.
E) Time spent in the air:
The total time in the air is the sum of the time to reach the maximum height and the time to fall back to the original position. Since the motion is symmetrical, the time to fall back is also 4.0 seconds. Therefore, the total time in the air is:
4.0 seconds (upward) + 4.0 seconds (downward) = 8.0 seconds
(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the battery? V (b) If the same capacitor is connected to another battery and 192.0 µC of charge is stored on the capacitor, what is the voltage of the battery?
Answer:
a.6 V
b.24 V
Explanation:
We are given that
a.[tex]C=8\mu F=8\times 10^{-6} F[/tex]
[tex]1\mu =10^{-6} [/tex]
Q=[tex]48\mu C=48\times 10^{-6} C[/tex]
We know that
[tex]V=\frac{Q}{C}[/tex]
Using the formula
[tex]V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V[/tex]
b.[tex]Q=192\mu C=192\times 10^{-6} C[/tex]
[tex]V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V[/tex]
A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they have covered only 137 km and their direction is not south but 15.0∘ east of south. What is the wind velocity?
Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.
Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)
In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:
132.33km/h - 200km/h = -67.69km/h
km and the position in "y" is equal to 137km*sin(15°) = 35.4km
This means that the velocity of the air in the y-axis is 35.4km/h
So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.
A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motion, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.Find the coefficients of static and kinetic friction from this information.
Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8
Explanation:
in the attachment
Water from a fire hose is directed horizontally against a wall at a rate of 53.9 kg/s and a speed of 38.9 m/s. Calculate the magnitude of the force exerted on the wall (in N), assuming the water's horizontal momentum is reduced to zero.
Answer:
[tex]F_{avg}=2096.71N[/tex]
Explanation:
As from the given data that:
m/Δt=53.9kg/s
The numerator is the mass and denominator is the change in time
Solve for change in velocity we have:
Δv=v₂-v₁
[tex]=0-38.9m/s\\=-38.9m/s[/tex]
Δv= -38.9m/s
From Newtons second law we know that:
F=ma
We can write this equation as:
Favg=(m/Δt)Δv
Substitute the given values
So
[tex]F_{avg}=(53.9kg/s)(-38.9m/s)\\F_{avg}=-2096.71N[/tex]
Using the average velocity formula.The answer will be positive, the negative only implies that force is coming away from the wall
[tex]F_{avg}=2096.71N[/tex]
Imagine you derive the following expression by analyzing the physics of a particular system: a=gsinθ−μkgcosθ, where g=9.80meter/second2. Simplify the expression for a by pulling out the common factor.
a.) a = g sinθ − μk g cosθ (no simplification should be performed on the expression in this situation)
b.) a = g (sinθ−μk cosθ)
c.) a = (9.80meter/second^2)sinθ −μk (9.80 meter/second^2)cosθ
Answer:
Explanation:
Analysis of structure gives
a=gsinθ−μkgcosθ
Notice that all the expression are right but we want to know of we can simplify the expression further.
We want to analyse if we can still further simplify the expression,
Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.
So option a is wrong because the expression can be simplified further to
a=g(sinθ−μkcosθ)
Option b is right and the best option.
Since we are given that, g=9.8m/s²
We can as well substitute that to option a
So we will have
a=9.8metre/second²(sinθ−μkcosθ)
Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first
So it will have been the best option if it was written as
a=9.8metre/second²(sinθ−μkcosθ)
So the best option is B.
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 29 cm.
What is the wavelength of the sound?
Express your answer using two significant figures.
\lambda ={\rm cm}
Part B
If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?
Express your answer using two significant figures.
Answer:
Explanation:
The the observer must not be stationed between speakers . Let its position be d cm from one of the speaker. When sound has maximum intensity , the the two sounds reaching his ear must have constructive interference and during zero intensity , they must have destructive interference. Distance between sources will be equal to path difference.
For constructive interference
n λ = path diff
= 19 cm
For destructive interference
(2n+1) λ /2= path diff
= 29 cm
n λ + λ /2 = 29
19 + λ /2 = 29
λ = 20 cm
B )
The path difference of 19 cm corresponds to constructive interference or maximum sound intensity . Wave length is 20 cm . Therefore , next path difference that can create condition of constructive interference will be 19 + 20 = 39 cm . Present path difference is 29 cm. So the distance between the speaker must be increased to 39 cm , so that path difference becomes 39 cm and constructive interference takes place.
ANS 39 cm .
A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time of 4.50 s .(a) What is the mass of the block of ice?
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
70.31kg
Explanation:Step I: Consider Newton's second law of motion which states that;
∑F = m x a;
Where;
∑F = net force acting on a body
m = the mass of the body
a = acceleration due to the force on the body.
Step II: Now to the question;
Since frictional force is negligible and the only force acting on the block of ice is the applied force by the dockworker, the net force on the body (block of ice) is the constant horizontal force. i.e
∑F = 90.0N
Also;
the block starts from rest and moves a distance (s) of 13.0m in a time (t) of 4.50s. Here, we can get the acceleration in that duration of time using one
the equations of motion as follows;
s = ut + [tex]\frac{1}{2}[/tex]at² ------------------------------(ii)
Where;
s = distance covered = 13.0m
u = initial velocity = 0 [since the block starts from rest]
t = time taken to cover the distance = 4.50s
a = acceleration of the body.
Substitute these values into equation (ii) as follows;
13.0 = 0(4.5) + [tex]\frac{1}{2}[/tex](a)(4.50)²
13.0 = 0 + [tex]\frac{1}{2}[/tex](a)(20.25)
13.0 = [tex]\frac{1}{2}[/tex](a)(20.25)
13.0 = 10.125a
Solve for a;
a = [tex]\frac{13.0}{10.125}[/tex]
a = 1.28m/s²
Step III: Now substitute the values of a = 1.28m/s² and ∑F = 90.0N into equation (i) as follows;
90.0 = m x 1.28
m = [tex]\frac{90.0}{1.28}[/tex]
m = 70.31
Therefore, the mass of the block of ice is 70.31kg
A 1.50-V battery supplies 0.204 W of power to a small flashlight. If the battery moves 8.33 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used
Answer:
981.41 secs
Explanation:
Parameters given:
Voltage, V = 1.5V
Power, P = 0.204W
Number of electrons, n = 8.33 * 10^20
First, we calculate the current:
P = I*V
I = P/V
I = 0.204/1.5 = 0.136A
The total charge of 8.33 * 10^20 electrons is:
Q = 8.33 * 10^20 * 1.6023 * 10^(-19)
Q = 133.47 C
Current, I, is given as:
I = Q/t
=> t = Q/I
t = 133.47/0.136 = 981.41 secs
In physics, power is calculated using the formula P = IV, where P is power, I is current, and V is voltage. By applying this formula, along with the relationship between charge, current, and time, the time the flashlight was used can be calculated.
Power is determined by the rate at which energy is transferred, calculated as P = IV where P is power, I is current, and V is voltage. In this scenario, the power supplied is 0.204 W from a 1.50-V battery. The formula P = IV can be rearranged to find the current flowing, which is 0.136 A.
To determine the time the flashlight was used, we can utilize the relationship between charge, current, and time: Q = It. Given 8.33 x 10^20 electrons moved, we need to convert this to Coulombs by recognizing that 1 electron has a charge of approximately 1.6 x 10^-19 C. By dividing the total charge by the current, we find the time t to be approximately 1.23 x 10^19 seconds.
A projectile is launched with a launch angle of 55° with respect to the horizontal direction and with initial speed 78 m/s. How long does it remain in flight?
Answer:
The projectile is in air for 13.03 seconds.
Explanation:
Given that,
Angle of projection of the projectile, [tex]\theta=55^{\circ}[/tex]
Initial speed of the projectile, u = 78 m/s
To find,
We need to find the time of flight of the projectile.
Solution,
It is defined as the time taken by the projectile when it is in air. It is given by the formula as :
[tex]T=\dfrac{2u\ \sin\theta}{g}[/tex]
[tex]T=\dfrac{2\times 78\ \sin(55)}{9.8}[/tex]
T = 13.03 seconds
So, the projectile is in air for 13.03 seconds.
During a walk on the Moon, an astronaut accidentally drops his camera over a 20.0-m cliff. It leaves his hands with zero speed, and after 2.00 s it attains a velocity of 3.40 m/s downward. How far has the camera fallen after 9.10 s
Answer:
Δy=70.3885 downwards
Explanation:
First we get the acceleration on moon by substituting the values we have(v=0m/s at t=0s and v=-3.40m/s at t=2.0 s) into equation of simple motion:
[tex]v_{f}-v_{i}=at\\-3.40m/s-0=(2s)a\\a=-1.7m/s^{2}[/tex]
As we know that
Δy=vit+(1/2)at²
Substitute the values of acceleration and time t=9.10s
So
Δy=vit+(1/2)at²
[tex]=(0)(9.10s)+(1/2)(-1.7m/s^{2})(9.10s)^{2}\\ =-70.3885m[/tex]
Δy=70.3885 downwards
The camera has fallen approximately 402.5 meters after 9.10 seconds.
Explanation:To find the distance the camera has fallen after 9.10 s, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance fallen
u = initial velocity (0 m/s)
t = time (9.10 s)
a = acceleration (due to gravity, -9.8 m/s^2)
Plugging the values into the equation, we get:
s = 0 + (1/2)(-9.8)(9.10)^2
Simplifying, we find that the camera has fallen approximately 402.5 meters after 9.10 seconds.
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A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. Calculate the work done in J
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
[tex]V_1[/tex] = initial volume = 5.68 L
[tex]V_2[/tex] = final volume = 2.35 L
Putting values in above equation, we get:
[tex]W=-0.96atm\times (2.35-5.68)L=3.20L.atm[/tex]
To convert this into joules, we use the conversion factor:
[tex]1L.atm=101.33J[/tex]
So, [tex]3.20L.atm=3.20\times 101.3=324J[/tex]
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
Given values:
Pressure, [tex]P = 732 \ torr \ or \ 0.96 \ atm[/tex]Initial volume, [tex]V_1 = 5.68 \ L[/tex]Final volume, [tex]V_2 = 2.35 \ L[/tex]We know the equation,
→ [tex]W = - P\Delta V[/tex]
[tex]= -P(V_2-V_1)[/tex]
By substituting the values,
[tex]= -0.96\times (2.35-5.68)[/tex]
[tex]= 3.20 \ L.atm[/tex]
By converting it in "J",
[tex]= 3.20\times 101.3[/tex]
[tex]= 324 \ J[/tex]
Thus the above answer is right.
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A 2.0 m ordinary lamp extension cord carries a 3.0 A current. Such a cord typically consists of two parallel wires carrying equal currents in opposite directions.
a. Find the magnitude of the force that the two segments of this cord exert on each other. (You will need to inspect an actual lamp cord at home and measure or reasonably estimate the quantities needed to do this calculation.)
b. Find the direction (attractive or repulsive) of the force that the two segments of this cord exert on each other. (Opposite or same directional currents, and do the wires attract or repel?)
Our values are given are,
[tex]l = 2m[/tex]
[tex]I = 3A[/tex]
[tex]r = 3.5mm = 3.5*10^{-3} m[/tex]
Electromagnetic force can then be defined as,
[tex]F = \frac{\mu_0 I^2 l}{2\pi r}[/tex]
Here,
[tex]\mu_0[/tex] = Permeability free space
I = Current
l = Length
r = Distance between them
Replacing,
[tex]F = \frac{(4\pi *10^{-7})(3)^2(2)}{(2\pi)(3.5*10^{-3})}[/tex]
[tex]F = 0.001028N[/tex]
Therefore the magnitude of the force that two segments of this cord exert on each other is 0.001028N.
Since the magnitude of the force is positive, the direction of both is attractive.
a) The magnitude of the force that the two segments of the cord exert on each other is [tex]\(1.2 \times 10^{-4} \, \text{N}\)[/tex]
b) The force is repulsive.
The force per unit length between two parallel wires carrying currents [tex]\(I_1\)[/tex] and [tex]\(I_2\)[/tex] separated by a distance d is given by Ampère's law:
[tex]\[F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}\][/tex]
Where:
F is the force,
L is the length of the wires,
[tex]\(\mu_0\)[/tex] is the permeability of free space [tex](\(4 \pi \times 10^{-7} \, \text{N/A}^2\))[/tex],
[tex]\(I_1\)[/tex] and [tex]\(I_2\)[/tex] are the currents through the wires (in this case, both are 3.0 A but in opposite directions),
d is the distance between the wires.
a. First, assume the distance d between the wires in the extension cord. A reasonable estimate for the distance between the wires in a standard extension cord might be around 3 mm (0.003 m).
It is given that:
[tex]\(I_1 = I_2 = 3.0 \, \text{A}\),[/tex]
[tex]\(L = 2.0 \, \text{m}\),[/tex]
[tex]\(d \approx 0.003 \, \text{m}\).[/tex]
Using the formula for the force per unit length:
[tex]\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}\][/tex]
Substitute the known values:
[tex]\[\frac{F}{L} = \frac{(4 \pi \times 10^{-7} \, \text{N/A}^2) \times (3.0 \, \text{A})^2}{2 \pi \times 0.003 \, \text{m}}\][/tex]
[tex]\[\frac{F}{L} = \frac{4 \pi \times 10^{-7} \times 9}{2 \pi \times 0.003}\][/tex]
[tex]\[\frac{F}{L} = \frac{36 \times 10^{-7}}{0.006}\][/tex]
[tex]\[\frac{F}{L} = 6 \times 10^{-5} \, \text{N/m}\][/tex]
Now, multiply by the length L to find the total force:
[tex]\[F = \left(6 \times 10^{-5} \, \text{N/m}\right) \times 2.0 \, \text{m}\][/tex]
[tex]\[F = 1.2 \times 10^{-4} \, \text{N}\][/tex]
b. Since the currents in the two parallel wires are equal but in opposite directions, the magnetic force between them will be repulsive. This is because currents flowing in opposite directions repel each other according to the right-hand rule for magnetic fields and forces between current-carrying wires.
A short circuit is a circuit containing a path of very low resistance in parallel with some other part of the circuit. Discuss the effect of a short circuit on the portion of the circuit it parallels. Use a lamp with a frayed line cord as an example.
Answer:
note:
please find the attachment
Explanation:
A frayed cord means that the insulation of the cord is worn out which exposes the cord and when it touches any other conductor a short circuit happens which is basically the flow of very high amount of current.
If we talk about the example of frayed cord, when such short circuit happens the lamp will not turn on and a massive amount of fault current will flow throughout the path of short circuit.
Now what happens?
There are two possibilities;
If there exists a circuit breaker, then no substantial amount of damage would be done since the circuit breaker will trip itself and open the circuit that eventually halts the fault current.If there is no circuit breaker, then this massive fault current will keep on accumulating that causes excessive heating of the conductors and also producing sparks which most probably would end up in a fire hazard.Sand falls from a conveyor belt at a rate of 30 m3m3/min onto the top of a conical pile. The height of the pile is always 3535 of the base diameter. Answer the following. a.) How fast is the height changing when the pile is 2 m high?
Explanation:
As the given data is as follows.
h = [tex]\frac{3}{5}d[/tex]
= [tex]\frac{3}{5} \times (2r)[/tex]
Also, we know that r = [tex]\frac{4}{3}h[/tex]
and Volume (V) = [tex]\frac{1}{3} \pir^{2}h[/tex]
= [tex]\frac{1}{3} \pi (\frac{4}{3}h)^{2} h[/tex]
= [tex]\frac{16}{27} \pi h^{3}[/tex]
And, [tex]\frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]
Putting the given values into the above formula as follows.
[tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]
[tex]30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = 1.343 m/min
or, = 134.3 cm/min (as 1 m = 100 cm)
thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.
A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.
A mass weighing 32 pounds stretches a spring 2 feet.
(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.
(b) How many complete cycles will the mass have completed at the end of 4 seconds?
Answer:
[tex]A = 1.803 ft[/tex]
Period = [tex]\frac{\pi}{2}[/tex] seconds
8 cycles
Explanation:
A mass weighing 32 pounds stretches a spring 2 feet;
it implies that the mass (m) = [tex]\frac{w}{g}[/tex]
m= [tex]\frac{32}{32}[/tex]
= 1 slug
Also from Hooke's Law
2 k = 32
k = [tex]\frac{32}{2}[/tex]
k = 16 lb/ft
Using the function:
[tex]\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0[/tex]
[tex]x(0) = -1[/tex] (because of the initial position being above the equilibrium position)
[tex]x(0) = -6[/tex] ( as a result of upward velocity)
NOW, we have:
[tex]x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)[/tex]
However;
[tex]x(0) = -1[/tex] means
[tex]-1 =c_1\\c_1 = -1[/tex]
[tex]x(0) =-6[/tex] also implies that:
[tex]-6 =4(c_2)\\c_2 = - \frac{6}{4}[/tex]
[tex]c_2 = -\frac{3}{2}[/tex]
Hence, [tex]x(t) =-cos4t-\frac{3}{2} sin 4t[/tex]
[tex]A = \sqrt{C_1^2+C_2^2}[/tex]
[tex]A = \sqrt{(-1)^2+(\frac{3}{2})^2 }[/tex]
[tex]A=\sqrt{\frac{13}{4} }[/tex]
[tex]A= \frac{1}{2}\sqrt{13}[/tex]
[tex]A = 1.803 ft[/tex]
Period can be calculated as follows:
= [tex]\frac{2 \pi}{4}[/tex]
= [tex]\frac{\pi}{2}[/tex] seconds
How many complete cycles will the mass have completed at the end of 4 seconds?
At the end of 4 seconds, we have:
[tex]x* \frac{\pi}{2} = 4 \pi[/tex]
[tex]x \pi = 8 \pi[/tex]
[tex]x=8[/tex] cycles
You are testing a new amusement park roller coaster with an empty car with a mass of 120 kg. One part of the track is a vertical loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s. As the car rolls from point A to point B, how much work is done by friction?
Answer:
[tex]W_f=-62460\ J[/tex]
Explanation:
Given that
mass of the car ,m = 120 kg
Radius ,R= 12 m
Speed at the bottom , u = 25 m/s
Speed at top ,v= 8 m/s
We know that
Work done by all the forces = Change in the kinetic energy
Work done by gravity + Work done by friction =Change in the kinetic energy
By taking point A as reference
[tex]m g \times (2R) + W_f=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2[/tex]
Now by putting the values in the above equation we get
[tex]120\times 10\times 2\times 12+ W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2[/tex]
[tex]W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2-120\times 10\times 2\times 12\ J[/tex]
[tex]W_f=-62460\ J[/tex]
Therefore the work done by friction force will be -62460 J.
A block with a mass of 4 kg is sliding across a horizontal surface with an initial speed of 5 m/s. Because of kinetic friction, the energy of this system will decrease linearly with the distance traveled by the block.
What is the average power (in W) supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to a stop?
Explanation:
Below is an attachment containing the solution.
The average power supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to stop is -31.25 W.
What is Power?In physics, power is the amount of energy that is transferred or transformed in a given amount of time. The International System of Units uses the watt, or one joule per second, as the unit of power. Ancient literature frequently used the term "activity" to describe power. Powers are scalar variables.
According to the question, the given values are :
Block mass, m = 4 kg,
The initial speed of the block at the horizontal surface, v₁ = 5 m/s,
The final speed of the block will be, v₂ = 0 m/s.
Distance covered, d = 4 m
Let t be the time when the box is going to stop.
d = [(v₁+v₂)/2] × t
4 = (5/2) t
t = 1.6 sec.
The change in kinetic energy of the block will be ΔE.
ΔE = mv₂²/2 - mv₁²/2
ΔE = 0 - (4)(5)²/2
ΔE = -50 J
Average power supplied by the force of kinetic friction acting on the block = P(avg)
P(avg) = ΔE / t
P(avg) = -50 / 1.6
P(avg) = -31.26 W.
So, the average power supplied by the force of kinetic friction on the block will be -31.26 W.
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What is the critical angle θcritθcrittheta_crit for light propagating from a material with index of refraction of 1.50 to a material with index of refraction of 1.00?
Answer:
The critical angle is 41.8°.
Explanation:
The critical angle is [tex]\theta_1[/tex] for which the angle of refraction [tex]\theta_2[/tex] is 90°. From Snell's law we have
[tex]n_1sin (\theta_1 ) = n_2 sin(\theta_2)[/tex]
[tex]n_1sin (\theta_1 ) = n_2 sin(90^o)[/tex]
[tex]sin (\theta_1 ) = n_2/n_1,[/tex]
[tex]\theta_1 = sin^{-1}(\dfrac{n_2}{n_1} ).[/tex]
Putting in [tex]n_2 =1.00,[/tex] and [tex]n_1 = 1.50[/tex] we get:
[tex]\theta_1 = sin^{-1}(\dfrac{1.00}{1.500} ),[/tex]
[tex]\boxed{\theta_1 = 41.8^o}[/tex]
Thus, the critical angle is 41.8°.
Final answer:
The critical angle for light propagating from a material with an index of refraction of 1.50 to a material with an index of refraction of 1.00 is approximately 41.8 degrees.
Explanation:
The critical angle θcrit for light propagating from a material with an index of refraction (n1) of 1.50 to a material with an index of refraction (n2) of 1.00 is found using Snell's Law and the concept of total internal reflection. The formula to calculate the critical angle is θcrit = sin−1(n2/n1). Plugging in the values for the indices of refraction, we have θcrit = sin⁻¹(1.00/1.50).
Performing the calculation, we get:
θcrit = sin⁻¹(0.6667) ≈ 41.8°
Therefore, the critical angle for light traveling from a medium with an index of refraction of 1.50 to a medium with an index of refraction of 1.00 is approximately 41.8 degrees.