Final answer:
The question is about the angles of maxima and minima in a double-slit interference pattern for two identical loudspeakers placed 1.00 m apart.
Explanation:
The distance between two identical loudspeakers placed 1.00 m apart is given as 'd'.
In the given situation, the listener stands 4.00 m from the wall directly in front of one of the speakers. This distance can be considered as 'L'.
The formula to calculate the distance between adjacent maxima and minima in a double-slit interference pattern is:
d*sin(theta) = m*lambda
- Where 'd' is the distance between the speakers
- 'theta' is the angle at which the maxima or minima occur
- 'm' is the order of the maxima or minima
- 'lambda' is the wavelength of the sound
Applying this formula, we can calculate the angles at which the maxima and minima occur for the given situation.
The phase difference between the waves reaching the observer is approximately 0.66 radians. The frequency closest to 300 Hz for minimal sound due to destructive interference is approximately 286 Hz.
To solve the problem, we need to find the phase difference and the frequency closest to 300 Hz for minimal sound.
a.) The speed of sound, v, is typically about 343 m/s in air. Using the formula for wavelength, λ = v/f, we get:
λ = 343 m/s / 300 Hz λ = 1.143 mFor the listener standing 4.00 m from one speaker and 4.12 m from the second speaker, the difference in the path length, ΔL, is:
ΔL = sqrt((4.00 m)² + (1.00 m)²) - 4.00 m ΔL = 4.12 m - 4.00 m = 0.12 mThe phase difference, φ, is given by:
φ = (2π/λ) * ΔL φ = (2π/1.143 m) * 0.12 m φ ≈ 0.66 radiansb.) For destructive interference, the path difference should be half-integer multiples of the wavelength:
(2n + 1)λ/2 = ΔLSimplifying, we find the frequency:
0.12 m = (2n + 1) * (343 m/s) / (2f)Solving for f when n = 0 (closest to 300 Hz):
f = 343 m/s / (2 * 0.12 m) f ≈ 286 HzThus, the frequency closest to 300 Hz for minimal sound is approximately 286 Hz.
complete question:
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.
(a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.)
rad
(b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?in Hz
A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 15 seconds. How much friction force does the brake pad apply to the shaft?
Answer:
frictional force = 0.52 N
Explanation:
diameter of turn table (D1) = 30 cm = 0.3 m
mass of turn table (M1) = 1.2 kg
diameter of shaft (D2) = 1.2 cm = 0.012 m
mass of shaft (M2) = 450 g = 0.45 kg
time (t) = 15 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
radius of turn table (R1) = 0.3 / 2 = 0.15 m
radius of shaft (R2) = 0.012 / 2 = 0.006 m
total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft
I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}
I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}
I = 0.0135 + 0.0000081 = 0.0135081
ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s
α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}
torque = I x α
torque = 0.0135081 x (-0.23) = - 0.00311 N.m
torque = frictional force x R2
- 0.00311 = frictional force x 0.006
frictional force = 0.52 N
The amount of friction force that the brake pad applied to the shaft is 0.52 N
To determine how much friction force does the brake pad applied to the shaft, we need to first know the moment of inertia of the solid turntable, followed by the angular acceleration of the turntable.
The moment of inertia can be computed by using the formula:
[tex]\mathbf{I = \dfrac{1}{2} MR^2}[/tex]
where;
diameter = 0.3 mR = radius = (0.3/2) m[tex]\mathbf{I = \dfrac{1}{2} \times (1.2 \ kg) (\dfrac{0.3}{2})^2}[/tex]
I = 0.0135 kgm²
The angular acceleration of the solid turntable is also estimated by using the formula:
[tex]\mathbf{\alpha = \dfrac{\omega _f - \omega _i}{t}}[/tex]
where;
initial angular velocity = 33 rpmfinal angular veocity = 0∴
[tex]\mathbf{\alpha = \dfrac{0 -33 rpm \times( \dfrac{2 \pi \ rad/s}{60 rpm})}{15\ s}}[/tex]
[tex]\mathbf{\alpha = -0.23 rad/s^2}[/tex]
Finally, determining the friction force by using the equation of torque;
[tex]\mathbf{\sum \tau = I \times \alpha}[/tex]
From dynamics of rotational motion;
[tex]\mathbf{r \times f= I \times \alpha}[/tex]
[tex]\mathbf{ f= \dfrac{I \times \alpha}{r}}[/tex]
where;
r = radius of the distance from the pivot point = 1.2 cm/2[tex]\mathbf{ f= \dfrac{0.0135 \ kg.m^2 \times 0.23 \ rad/s^2}{(\dfrac{1.2 }{2} \times 10^{-2} m)}}[/tex]
f ≅ 0.52 N
Therefore, we can conclude that the amount of friction force that the brake pad applied to the shaft is 0.52 N.
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Two people with a combined mass of 127 kg hop into an old car with worn-out shock absorbers. This causes the springs to compress by 9.10 cm. When the car hits a bump in the road, it oscillates up and down with a period of 1.66 s. Find the total load supported by the springs.
Answer:
Total load = 2999.126 kg
Explanation:
Let the spring constant of the shock absorber be k.
We know that the force applied on a spring is directly proportional to elongated length and the constant of proportionality is called spring constant.
Thus
Force, F = kx
where,
x = elongation = 9.1 cm 0.091 m
mass of the people, m = 127 kg
F = weight of the people = mg = 127 x 9.8 = 1244.6 N
substituting these values in the first equation,
1244.6 = k x 0.091
thus, k = 13,676.923 N/m
Now we know that the time period, T of an oscillating spring with a load of mass m is
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
[tex]\frac{m}{k} = \frac{T^{2} }{4\pi ^{2} }[/tex]
thus,
[tex]m = k\frac{T^{2} }{4\pi ^{2}}[/tex]
T = 1.66s
substituting these values in the equation,
[tex]m =13,676.923\frac{1.66^{2} }{4\pi ^{2} }[/tex]
m = 2999.126 kg
The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used to construct the can, what must be the height, in inches, of the can?
The required height of the tin can is given by the height at the minimum value of the surface area function of the tin can
The height of the tin can must be 4 inchesReason:
The given information on the tin can are;
Volume of the tin can = 16·π in.³
The amount of tin to be used = Minimum amount
The height of the can in inches required
Solution;
Let h, represent the height of the can, we have;
The surface area of the can, S.A. = 2·π·r² + 2·π·r·h
The volume of the can, V = π·r²·h
Where;
r = The radius of the tin can
h = The height of the tin can
Which gives;
16·π = π·r²·h
16 = r²·h
[tex]h = \dfrac{16}{r^2}[/tex]
Which gives;
[tex]S.A. = 2 \cdot \pi \cdot r^2 + 2 \cdot \pi \cdot r \cdot \dfrac{16}{r^2} = 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r}[/tex]
When the minimum amount of tin is used, we have;
[tex]\dfrac{d(S.A.)}{dx} =0 = \dfrac{d}{dx} \left( 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r} \right) = \dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2}[/tex]
Therefore;
[tex]\dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2} = 0[/tex]
4·π·(r³ - 8) = r² × 0
r³ = 8
r = 2
The radius of the tin can, r = 2 inches
The
[tex]h = \dfrac{16}{2^2} = 4[/tex]
The height of the tin can, h = 4 inches
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The required height of can is 4 inches.
Given data:
The volume of cylindrical tin with top and bottom is, [tex]V = 16 \pi \;\rm in^{3}[/tex].
The volume of tin is,
[tex]V = \pi r^{2}h\\16 \pi = \pi r^{2}h\\[/tex]
here, r is the radius of can and h is the height of can.
[tex]16 =r^{2}h\\\\h=\dfrac{16}{r^{2}}[/tex]
The surface area of tin is,
[tex]SA = 2\pi r(r+h)\\SA = 2\pi r(r+\dfrac{16}{r^{2}})\\SA = 2\pi r^{2}+\dfrac{32 \pi}{r})[/tex]
For minimum amount of tin used, we have,
[tex]\dfrac{d(SA)}{dr} =0\\\dfrac{ d(2\pi r^{2}+\dfrac{32 \pi}{r})}{dr} = 0\\4\pi r-\dfrac{32 \pi}{r^{2}}=0\\4\pi r=\dfrac{32 \pi}{r^{2}}\\r^{3}=8\\r =2[/tex]
So, height of can is,
[tex]h=\dfrac{16}{2^{2}}\\h=4[/tex]
Thus, the required height of can is 4 inches.
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Oceanic lithosphere is lighter than continental lithosphere. True or False
Answer:
FALSE
Explanation:
The earth's lithosphere is divided into 2 types, namely the continental and the oceanic crust. The continental crust is comprised of lighter rock minerals such as silicate minerals, alkali and potash feldspar, and some oxide minerals, whereas, the oceanic crust is comprised of olivine, pyroxene and some feldpars that are comparatively denser than the minerals present in the continental crust. Due to this composition, the oceanic lithosphere is denser than the continental lithosphere.
During the convergent plate motion, the oceanic lithosphere (plate) subducts below the continental lithosphere due to its greater density.
Thus, the above given statement is False.
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well. Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg*m^2 and for arms and legs in is 0.80 kg*m^2. If she starts out spinning at 5.0 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
[tex]w_f=1.143\frac{rev}{s}[/tex]
Explanation:
1) Notation
[tex]I_{i}=0.8kgm^2[/tex] (Inertia with arms and legs in)
[tex]I_{f}=3.5kgm^2[/tex] (Inertia with arms out and one leg extended)
[tex]w_{i}=5\frac{rev}{s}[/tex]
[tex]w_{f}=?[/tex] (variable of interest)
2) Analysis of the situation
For this case we can assume that there are no external forces acting on the skater, so based on this assumption we don’t have any torque from outside acting on the system. And for this reason, we can consider the angular momentum constant throughout the movement.
On math terms then the initial angular momentum would be equal to the final angular momentum.
[tex]L_i =L_f[/tex] (1)
The angular momentum of a rigid object is defined "as the product of the moment of inertia and the angular velocity and is a vector quantity"
3) Formulas to use
Using this definition we can rewrite the equation (1) like this:
[tex]I_{i}w_{i}=I_{f}w_{f}[/tex] (2)
And from equation (2) we can solve for [tex]w_f[/tex] like this:
[tex]w_f=\frac{I_i w_i}{I_f}[/tex] (3)
And replacing the values given into equation (3) we got:
[tex]w_f=\frac{0.8kgm^2 x5\frac{rev}{s}}{3.5kgm^2}=1.143\frac{rev}{s}[/tex]
And that would be the final answer [tex]w_f=1.143\frac{rev}{s}[/tex].
Angular speed ( in rev/s ) when arms out and one leg open extended outward is 1.14 rev/s
What is moment of Inertia of a rotating body?moment of inertia, I is the measure of distibution of the mass of the body along the axis of rotatiton.
I = angular momentum, L / angular velocity, ω
L = I * ω
L1 = L2
L1 = angular momentum arms and leg in
L2 = angular momentum arms out and one leg extended
I1 * ω1 = I2 * ω2 conservation of angular momentum
0.8 * 5 = 3.5 * ω2
ω2 = 4 / 3.5
ω2 = 1.14 rev/s
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A boat moves at 10.8 m/s relative to the water. If the boat is in a river where the current is 2.00 m/s, how long does it take the boat to make a complete round trip of 1 100 m upstream followed by a 1 100-m trip downstream?
336 s
250 s
172 s
297 s
211 s
Answer:
Option E is the correct answer.
Explanation:
Velocity of boat = 10.8 m/s
Velocity of river = 2 m/s
Relative velocity upstream = 10.8 - 2 = 8.8 m/s
Displacement = Velocity x Time
1100 = 8.8 x t₁
Time in upstream, t₁ = 125 s
Relative velocity downstream = 10.8 + 2 = 12.8 m/s
Displacement = Velocity x Time
1100 = 12.8 x t₂
Time in downstream, t₂ = 86 s
Total time = t₁ + t₂ = 125 + 86 = 211 s
Option E is the correct answer.
Final answer:
To find the total time for a complete round trip, we calculate the time taken for both the upstream and downstream trips. The effective speeds for both directions are determined by considering the boat's speed and the river current's speed. The total time for the round trip is 211 seconds.
Explanation:
To determine the time taken for the boat to make a complete round trip in the river, we need to calculate the time taken for each leg of the journey (upstream and downstream) separately and then sum those times. The boat's speed relative to the water is 10.8 m/s, and the river current's speed is 2.00 m/s.
For the upstream trip, the boat's effective speed is reduced by the current, so the boat's speed relative to the riverbank is 10.8 m/s - 2.00 m/s = 8.8 m/s. The distance for the upstream trip is 1100 m, so the time taken upstream is:
Time_upstream = Distance / Speed_upstream = 1100 m / 8.8 m/s = 125 seconds.
For the downstream trip, the boat's effective speed is increased by the current, so the boat's speed relative to the riverbank is 10.8 m/s + 2.00 m/s = 12.8 m/s. The distance for the downstream trip is also 1100 m, so the time taken downstream is:
Time_downstream = Distance / Speed_downstream = 1100 m / 12.8 m/s = 85.9375 seconds, which can be rounded to 86 seconds.
The total time for a complete round trip is the sum of the upstream and downstream times:
Total time = Time_upstream + Time_downstream = 125 seconds + 86 seconds = 211 seconds.
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s). If a particular disk is spun at 646.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds, what is the magnitude of the average angular acceleration of the disk?
Answer:
Angular acceleration will be [tex]1135.5rad/sec^2[/tex]
Explanation:
We have given initial angular speed of a particular disk [tex]\omega _i=646.1rad/sec[/tex]
And it finally comes to rest so final angular speed [tex]\omega _f=0rad/sec[/tex]
Time is given as t = 0.569 sec
From third equation of motion we know that
[tex]\omega _f=\omega _i+\alpha t[/tex]
So angular acceleration [tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-646.1}{0.569}=1135.5rad/sec^2[/tex]
Based on discoveries to date, which of the following conclusions is justified?
a) Most stars have one or more terrestrial planets orbiting within their habitable zones.
b) Planets are common, but planets as small as Earth are extremely rare.
c) Planetary systems are common and planets similar in size to Earth are also common.
d) Although planetary systems are common, few resemble ours with terrestrial planets near the Sun and jovian planets far from the Sun.
Most stars have one or more terrestrial planets orbiting within their habitable zones is the correct answer.
Explanation:
According to the science, the habitable zones are the region around the stars where one or more terrestrial planets can orbit. A terrestrial planet orbits inhabitable zone is called potentially habitable zone which have roughly comparable conditions to those of earth.
This situation is a proof stating that most of the stars have one or more terrestrial planets orbiting within their habitable zone.
An object is dropped from 26 feet below the tip of the pinnacle atop a 702 ft tall building. The height h of the object after t seconds is given by the equation h=-16t^2+676. Find how many seconds pass before the object reaches the ground.
Answer:
6.5 seconds.
Explanation:
Given: h=-16t²+679
When the object reaches the ground, h=0.
∴ 0=-16t²+679
collecting like terms,
⇒ 16t²=679
Dividing both side of the equation by the coefficient of t² i.e 16
⇒ 16t²/16 = 679/16
⇒ t² = 42.25
taking the square root of both side of the equation.
⇒ √t² =√42.25
⇒ t = 6.5 seconds.
A man is marooned at rest on level frictionless ice. In desperation, he hurls his shoe to the right at 15 m/s. If the man weighs 720 N and the shoe weighs 4.0 N, the man moves to the left at approximately______.A) 0B) 2.1 x 102 m/s.C) 8.3 x 102 m/s.D) 15 m/s.E) 2.7 x 10-3 m/s 19.
Answer:
A 0.083 m/s approx 0
Explanation:
mass of the man is 720 N, the mass of the shoe is 4 N, and the man and the shoe were initially at rest. After throwing the shoe, the shoe had a velocity of 15 m/s. Using conservation of momentum:
since the man and the shoe were initially at rest, their initial momentum is zero
0 = M1V1 + M2V2 where M1 is the mass of the shoe ( 4 / 9.81), V1 is the velocity of the shoe 15m/s, M2 is mass of the man ( 720 / 9.81), V2 is the velocity of the man
MAKE V2 subject of the formula
- M1V1 = M2V2
- M1V1 / M2
substitute the values into the equation
- ((4/9.8) × 15) /( 720 / 9.81) = V2
V2 = - 0.0833 m/s approx 0
The man moves to the left at approximately 0 m/s.
Explanation:To solve this problem, we can use the principle of conservation of momentum. The man and the shoe are initially at rest on the frictionless ice, so the initial momentum is zero. When the man throws the shoe to the right, it gains momentum in that direction. According to the conservation of momentum, the man must gain an equal and opposite momentum in the left direction. Since the man weighs more than the shoe, his velocity will be significantly less than the shoe's velocity. Therefore, the man moves to the left at approximately 0 m/s.
A chef sanitized a thermometer probe and then checked the temperature of minestrone soup being held in a hot-holding unit. the temperature was 120 which did not meet the operation's critical limit of 135. the chef recorded the temperature in the log and reheated the soup to 165 for 15 seconds. which was the corrective action?
A) reheating the soup
B) checking the critical limit
C) sanitizing the thermometer probe
D) recording the temperature in the log
Answer: Option (A) is the correct answer.
Explanation:
A corrective action is defined as the action with the help of which a person can avoid a difficulty or problem that he/she was facing earlier.
For example, when the chef checked the temperature of soup using thermometer then it was 120 but his operation's critical limit was 135.
So, to avoid this problem he heated the soup to 165 at 15 seconds following which he got the result as desired.
Therefore, reheating the soup was his corrective action.
Thus, we can conclude that reheating the soup was the corrective action.
In this context, the corrective action taken by the chef was reheating the soup because it was not meeting the operation's critical safe temperature limit.
Explanation:In the provided scenario, the corrective action referring to steps taken to rectify a situation that does not meet specified standards, is reheating the soup. When the chef figured out the soup was below the operation's critical limit, prompt correction was made to ensure food safety by reheating the soup to 165 for 15 seconds. This measure ensures that any potential harmful microorganisms in the food are eliminated, thus making the food safe to consume.
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Which is the correct order for these steps in cell signaling?
A. signal transduction, receptor activation, response response,
B. signal transduction, termination receptor activation,
C. response, signal transduction receptor
D. activation, signal transduction, response
Answer:
D. activation, signal transduction, response
Explanation:
The correct sequence for cell signaling given as
1.Signal
2.Reception
3.Transduction
4.Response
In the diagram form
Signal ⇒ Reception ⇒ Transduction ⇒ Response
These should be in the above sequence , except option D all other option does not have a proper sequence.
Therefore option D is correct.
D. activation, signal transduction, response
The correct order of steps in cell signaling is receptor activation, signal transduction, and response.
Explanation:In the process of cell signaling, the correct order of steps is: receptor activation, signal transduction, and response. This process begins when a signaling molecule binds to a receptor on the cell surface, activating it (receptor activation).
The signal is then converted, or transduced, into a form that can bring about a specific cellular response (signal transduction). Finally, the cell responds to the signal (response). Each of these steps is crucial for proper communication between cells.These sequential steps are fundamental to the proper functioning of cell signaling and are integral to various physiological processes, ensuring cells can appropriately perceive and respond to extracellular signals.
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A child twirls his yo-yo horizontally about his head rather than using it properly. The yo-yo has a mass of 0.200 kg and is attached to a string 0.800 m long. (a) If the yo-yo makes a complete revolution each second, what tension must exist in the string?
Answer:
Tension in the string, F = 6.316 N
Explanation:
It is given that,
Mass of the Yo - Yo, m = 0.2 kg
Length of the string, l = 0.8 m
It makes a complete revolution each second, angular velocity, [tex]\omega=2\pi\ rad[/tex]
Let T is the tension exist in the string. The tension acting on it is equal to the centripetal force acting on it. Its expression is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]F=m\omega^2 r[/tex]
[tex]F=0.2\times (2\pi )^2 \times 0.8[/tex]
F = 6.316 N
So, the tension must exist in the string is 6.316 N. Hence, this is the required solution.
The African and South American continents are separating at a rate of about 3 cm per year, according to the ideas of plate tectonics. If they are now 5000 km apart and have moved at a constant speed over this time, how long is it since they were in contact?
Answer:
166 666 666.7 years
Explanation:
We start the question by making the units uniform. We are told that the continents move at 3 cm/year = 0.03 m/year.
We are also told that the continents are now 5000 km = 5 000 000 m apart
So to calculate the time it took for them to be this far apart
t = distance/speed
t = 5 000 000 m/(0.03 m/year) = 166 666 666.7 years
A proton is traveling to the right at 2.0 * 107 m/s. It has a headon perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?
Answer:
Speed of proton = 1.69 x 10⁷ m/s to the left
Speed of carbon = 3.08 x 10⁶ m/s to the right
Explanation:
Elastic collision means momentum and kinetic energy are conserved.
Mass of carbon = 12 x Mass of proton.
Initial velocity of proton = 2 x 10⁷ m/s
Initial velocity of Carbon = 0 m/s
Final velocity of proton = u
Final velocity of carbon = v
Initial momentum = Final momentum.
m x 2 x 10⁷ + 12m x 0 = m x u + 12m x v
u + 12v = 2 x 10⁷
u = 2 x 10⁷ - 12v
Initial K.E = Final K.E
0.5 x m x (2 x 10⁷)² + 0.5 x 12m x 0² = 0.5 x m x u² + 0.5 x 12m x (v)²
u² + 12v² = 4 x 10¹⁴
(2 x 10⁷ - 12v)² + 12v² = 4 x 10¹⁴
4 x 10¹⁴ - 48 x 10⁷v + 144v² + 12v² = 4 x 10¹⁴
156v² = 48 x 10⁷v
v = 3.08 x 10⁶ m/s
u + 12 x 3.08 x 10⁶ = 2 x 10⁷
u = -1.69 x 10⁷ m/s
Speed of proton = 1.69 x 10⁷ m/s to the left
Speed of carbon = 3.08 x 10⁶ m/s to the right
In a head-on perfectly elastic collision between a proton and a carbon atom, the speed of each particle remains the same after the collision, but their directions are reversed.
Explanation:An elastic collision between a proton and a carbon atom can be analyzed using the conservation of momentum and kinetic energy.
First, we need to determine the initial momentum of the proton and the carbon atom. The momentum of an object is calculated by multiplying its mass by its velocity.
Given that the mass of the carbon atom is 12 times the mass of the proton, the initial momentum of the proton is equal to the momentum of the carbon atom.
After the collision, the total momentum of the system must still be conserved. Since the collision is head-on and elastic, only the directions of the velocities of both particles will change. The speed of each particle will remain the same.
Therefore, after the collision, the speed of the proton will still be 2.0 * 10^7 m/s, but its direction will be reversed to the left. The carbon atom will also have a speed of 2.0 * 10^7 m/s, but it will be traveling in the right direction.
A 97 kg man holding a 0.365 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 11.3 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall.
Answer:
[tex]v = 0.085 m/s[/tex]
Explanation:
First when man throws the ball then the speed of the man is given as
[tex]m_1v_1 = m_2v_2[/tex]
[tex]97 v = 0.365 \times 11.3[/tex]
[tex]v = 0.042 m/s[/tex]
now ball will rebound after collision with the wall
so speed of the ball will be same as initial speed
now again by momentum conservation we will have
[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]
[tex]97 \times 0.042 + 0.365 \times 11.3 = (97 + 0.365) v[/tex]
[tex]v = 0.085 m/s[/tex]
According to Guinness, the tallest man to have ever lived was Robert Pershing Wadlow of Alton, Illinois. He was last measured in 1940 to be 2.72 meters tall (8 feet, 11 inches). Determine the speed which a quarter would have reached before contact with the ground if dropped from rest from the top of his head.
Answer:
7.30523 m/s
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 2.72 m
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 2.72+0^2}\\\Rightarrow v=7.30523\ m/s[/tex]
The speed which a quarter would have reached before contact with the ground is 7.30523 m/s
Final answer:
The speed at which the quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head is approximately 7.3 m/s.
Explanation:
To determine the speed at which a quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head, we can use the principle of conservation of mechanical energy.
Let's denote the height of Robert Wadlow's head as h and the mass of the quarter as m. We are given that h = 2.72 meters.
The potential energy (PE) of the quarter at the top of Robert Wadlow's head is:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The kinetic energy (KE) of the quarter just before it hits the ground is:
KE = (1/2) * m * v^2
where v is the speed of the quarter just before it hits the ground.
According to the principle of conservation of mechanical energy, the total mechanical energy at the top of Robert Wadlow's head is equal to the total mechanical energy just before the quarter hits the ground:
PE = KE
Substituting the expressions for PE and KE, we get:
m * g * h = (1/2) * m * v^2
Solving for v^2, we get:
v^2 = 2 * g * h
Substituting the given values for g and h, we get:
v^2 = 2 * 9.8 m/s^2 * 2.72 m
v^2 ≈ 53.2 m^2/s^2
Taking the square root of both sides, we get:
v ≈ 7.3 m/s
So, the speed at which the quarter would have reached before contact with the ground if dropped from rest from the top of Robert Wadlow's head is approximately 7.3 m/s.
You are working in the large animal ward at the hospital. You are receiving a horse that needs immediate treatment for dehydration. The vet asks you to calculate the fluid deficit for this horse. He is a 950-pound horse that is about 7% dehydrated. What is his fluid deficit?
A) 63 L
B) 15 L
C) 5 L
D) 30 L
Answer:
Your answer is D) 30 L
Explanation:
(Here's the explanation that I found on a website so Dont copy). You can estimate the fluid deficit by. Body weight in kg x% dehydration = Liters needed to re-establish hydration. 950 pounds / 2.2 pounds per kilogram ( pounds cancel out) = 431.8 kg ( will round up to 432 kg) Then multiply by percent dehydrated: 432kg x 0.07 ( which 7%) = 30 Liters.
A box is sitting on the ground and weighs 100 kg and the coefficient of friction is 0.23. Is it easier to push by applying the force?
Answer:
-No
Explanation:
Given that
Mass of box = 100 kg
Coefficient of friction ,μ= 0.23
We know that friction force depends on the normal force acts on the box
Fr= μ N
When we pull the box then :
Normal force N= mg
Friction force Fr= μ mg
When we push the box :
Lets take pushing force = F
θ =Angle make by pushing force from the vertical line
Normal force
N = mg + F cosθ
Fr= μ ( mg + F cosθ )
The friction force is more when we push the box.That is why this is not easier to push the box.
Therefore answer is ---No
4. ""Giving in"" during a catching action in baseball More than one answer is correct A. reduces the magnitude of the force required to stop the ball. B. increases the magnitude of the force required to stop the ball. C. increases the contact time of the hand with the ball. D. does neither increase nor decrease the impulse required to stop the ball.
Answer:
A. reduces the magnitude of the force required to stop the ball.
C. increases the contact time of the hand with the ball.
D. does neither increase nor decrease the impulse required to stop the ball.
Explanation:
As we know that the force required to stop the ball is given as
[tex]F = \frac{\Delta P}{\Delta t}[/tex]
here we know that
[tex]impulse = \Delta P[/tex]
so we have
[tex]impulse = m(v_f - v_i)[/tex]
now we know that time to stop the ball is increased due to which the force to stop the ball is decreased
so we have correct answer will be
A. reduces the magnitude of the force required to stop the ball.
C. increases the contact time of the hand with the ball.
D. does neither increase nor decrease the impulse required to stop the ball.
At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s, and an 83 kg person feels a 560N force pressing against his back. What is the radius of a chamber? a)1.2 m b) 1.44 m c) 1.5 m d) 1.65
Answer:
Radius, r = 1.5 meters
Explanation:
It is given that,
When an object is moving in an amusement park it will exert centripetal force. The centripetal force acting on the person, F = 560 N
Mass of the person, m = 83 kg
Speed of the wall, v = 3.2 m/s
The centripetal force acting on the person is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv^2}{F}[/tex]
[tex]r=\dfrac{83\times (3.2)^2}{560}[/tex]
r = 1.51 meters
or
r = 1.5 meters
So, the radius of the chamber is 1.5 meters.
When an automobile rounds a curve at high speed, the loading (weight distribution) on the wheels is markedly changed. For suffi- ciently high speeds the loading on the inside wheels goes to zero, at which point the car starts to roll over. This tendency can be avoided by mounting a large spinning flywheel on the car. (a) In what direction should the flywheel be mounted, and what should be the sense of rotation, to help equalize the loading
Answer:
Explanation:
The tendency of the automobile running on a circular path at high speed to turn towards left or right around one of its wheels , is due to torque by centripetal force acting at its centre of mass about that wheel .
Suppose the automobile tends to turn in clockwise direction about its wheel on the outer edge . A rotational angular momentum is created . To counter this effect , we can take the help of a rotating wheel or flywheel . We shall have to keep its rotation in anticlockwise direction so that it can create rotational angular momentum in direction opposite to that created by centrifugal force on fast moving automobile. Each of them will nullify the effect of the other . In this way , rotating flywheel will save the automobile from turning upside down .
Final answer:
To counteract the tendency for a car to roll over when rounding curves at high speed, a flywheel should be mounted horizontally and perpendicular to the travel direction, with a sense of rotation that produces gyroscopic precession force downward on the inside wheels.
Explanation:
When an automobile rounds a curve at high speeds, the loading on the wheels changes, with the tendency to lighten the load on the inside wheels, which can lead to the car rolling over. To combat this, a large spinning flywheel can help equalize the loading on the wheels. The flywheel should be mounted such that its axis of rotation is horizontal and perpendicular to the direction of the car's travel. The sense of rotation should be such that when the car turns, the gyroscopic precession of the flywheel produces a force that pushes down on the inside wheels. This force counteracts the effect of the weight transfer to the outside wheels, reducing the risk of the car tipping over.
This concept exploits the conservation of angular momentum and the phenomenon known as gyroscopic precession. In a closed system, any attempt to change the direction of the axis of rotation of the spinning flywheel produces a force perpendicular to the direction of the applied force (precession). If the top of the flywheel is tilted outward, the precession force acts downward on the side of the flywheel closest to the inside of the turn, creating a stabilizing effect on the inside wheels.
Air at 1 atm and 20°C is flowing over the top surface of a 0.2 m 3 0.5 m-thin metal foil. The air stream velocity is 100 m/s and the metal foil is heated electrically with a uniform heat flux of 6100 W/m2. If the friction force on the metal foil surface is 0.3 N, determine the surface temperature of the metal foil. Evaluate the fluid properties at 100°C.
Answer:[tex]180.86^{\circ}C[/tex]
Explanation:
Properties of Fluid at [tex]100^{\circ}C[/tex]
[tex]P_r=0.711[/tex]
[tex]\rho =0.9458 kg/m^3[/tex]
[tex]c_p=1009 J/kg/k[/tex]
[tex]Flux =6100 W/m^2[/tex]
Drag force [tex]F_d=0.3 N[/tex]
[tex]A=0.2\times 0.5=0.1 m^2[/tex]
drag force is given by
[tex]F_d=c_f\cdot A\rho \frac{v^2}{2}[/tex]
[tex]c_f=\frac{2F_d}{\rho Av^2}[/tex]
[tex]c_f=\frac{2\times 0.3}{0.9458\times 0.1\times 100^2}[/tex]
[tex]c_f=\frac{0.6}{945.8}[/tex]
[tex]c_f=0.000634[/tex]
we know average heat transfer coefficient is
[tex]h=\frac{c_f}{2}\times \frac{\rho vc_p}{P_r^{\frac{2}{3}}}[/tex]
[tex]h=\frac{0.000634}{2}\times \frac{0.9458\times 100\times 1009}{(0.711)^{\frac{2}{3}}}[/tex]
[tex]h=37.92 W/m^2-K[/tex]
Surface Temperature of metal Foil
[tex]\dot{q}=h(T_s-T{\infty })[/tex]
[tex]T_s=\frac{\dot{q}}{h}[/tex]
[tex]T_s[/tex] is the surface temperature and T_{\infty }[/tex] is ambient temperature
[tex]T_s=\frac{6100}{37.92}+20=180.86^{\circ}C[/tex]
A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium?
Answer:
Equilibrium temperature will be [tex]T=52.2684^{\circ}C[/tex]
Explanation:
We have given weight of the lead m = 2.61 gram
Let the final temperature is T
Specific heat of the lead c = 0.128
Initial temperature of the lead = 11°C
So heat gain by the lead = 2.61×0.128×(T-11°C)
Mass of the water m = 7.67 gram
Specific heat = 4.184
Temperature of the water = 52.6°C
So heat lost by water = 7.67×4.184×(T-52.6)
We know that heat lost = heat gained
So [tex]2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)[/tex]
[tex]0.334T-3.67=1688-32.031T[/tex]
[tex]T=52.2684^{\circ}C[/tex]
By equating the heat gained by the lead to the heat lost by the water, the final temperature is calculated to be approximately 41.5°C.
To determine the final temperature of the lead weight and water, we can use the principle of calorimetry, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.
Given:
Mass of lead[tex](m_l_e_a_d) = 2.61 g[/tex]
Specific heat capacity of lead [tex](c_l_e_a_d)[/tex] = 0.128 J/g°C
Initial temperature of lead[tex](T_l_e_a_d_i_n_i_t_i_a_l)[/tex]) = 11.1°C
Mass of water [tex](m_w_a_t_e_r)[/tex]= 7.67 g
Specific heat capacity of water[tex](c_w_a_t_e_r)[/tex] = 4.184 J/g°C
Initial temperature of water [tex](T_w_a_t_e_r_i_n_i_t_i_a_l)[/tex] = 52.6°C
Let T_final be the final equilibrium temperature.
Heat gained by lead [tex](Q_lead) = m_lead * c_lead * (T_final - T_lead_initia[/tex]l)
Heat lost by water [tex](Q_water) = m_water * c_water * (T_water_initial - T_final)[/tex]
Since the system is thermally insulated, Q_lead = -Q_water
[tex]m_lead * c_lead * (T_final - T_lead_initial) = - m_water * c_water * (T_water_initial - T_final)[/tex]
Solving for T_final:
[tex]T_final = (m_lead * c_lead * T_lead_initial + m_water * c_water * T_water_initial) / (m_lead * c_lead + m_water * c_water)[/tex]
Plugging in the values:
T_final = (2.61 g * 0.128 J/g°C * 11.1°C + 7.67 g * 4.184 J/g°C * 52.6°C) / (2.61 g * 0.128 J/g°C + 7.67 g * 4.184 J/g°C)
T_final ≈ 41.5°C
Therefore, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 41.5°C.
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Maxwell’s theory of Electromagnetism in 1865 was the first "unified field theory" _________ no further theory has united the electroweak field with either the Strong (Hadronic) force or Gravity.A. with Electroweak in 1961 only being the other, becauseB. with Electroweak in 1961 being the only other, becauseC. and Electroweak in 1961 only was the other, becauseD. and Electroweak in 1961 only being the other, asE. and Electroweak in 1961 the only other, as
Answer: E. and Electroweak in 1961 the only other, as
Explanation:
This is more an English grammar question than a physics question, so taking that perspective, one should look for the answer that best completes the sentence.
Based on the word "first" in the sentence, implying the need for a conjunction to join the two theories and the last part of the sentence does not give a reason but further supports the determination of the theory being the first unified theory.
So, to complete the sentence, the best option is;
Maxwell’s theory of Electromagnetism in 1865 was the first "unified field theory" and Electroweak in 1961 the only other, as no further theory has united the electroweak field with either the Strong (Hadronic) force or Gravity.
A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds. express your answer in meters per second to three significant figures.
Answer:
The speed v of the particle at t=5.00 seconds = 43 m/s
Explanation:
Given :
mass m = 5.00 kg
force f(t) = 6.00t2−4.00t+3.00 N
time t between t=0.00 seconds and t=5.00 seconds
From mathematical expression of Newton's second law;
Force = mass (m) x acceleration (a)
F = ma
[tex]a = \frac{F}{m}[/tex] ...... (1)
acceleration (a) [tex]= \frac{dv}{dt}[/tex] ......(2)
substituting (2) into (1)
Hence, F [tex]= \frac{mdv}{dt}[/tex]
[tex]\frac{dv}{dt} = \frac{F}{m}[/tex]
[tex]dv = \frac{F}{m} dt[/tex]
[tex]dv = \frac{1}{m}Fdt[/tex]
Integrating both sides
[tex]\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt[/tex]
The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;
[tex]v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt[/tex] ......(3)
Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):
[tex]v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt[/tex]
[tex]v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}[/tex]
[tex]v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|[/tex]
[tex]v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |[/tex]
[tex]v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |[/tex]
[tex]v = \frac{1}{5} | 250 - 50 + 15 |[/tex]
[tex]v = \frac{215}{5}[/tex]
v = 43 meters per second
The speed v of the particle at t=5.00 seconds = 43 m/s
What are Nebulae and what are they made of?
Answer:
Nebulae are made of dust and gases—mostly hydrogen and helium. The dust and gases in a nebula are very spread out, but gravity can slowly begin to pull together clumps of dust and gas.
Explanation:
Answer:
See explanation.
Explanation:
Nebulae are interstellar cloud of dust, hydrogen, helium and other ionized gases.
Nebulae are the places where stars are born. Nebulae formed when portions of the interstellar medium undergo gravitational collapse.
Nebulae are made of dust and gases; mostly hydrogen(about seventy-five percent) and helium(about twenty-five percent).
The interstellar gases are very dispersed and the amount of matter adds up over the vast distances between the stars.
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?
Answers:
a) [tex]W_{g}=mgh[/tex]
b) [tex]\Delta K=mgh[/tex]
c) [tex]K_{f}=\frac{1}{2}mV_{o}^{2}+mgh[/tex]
d) No
Explanation:
We have the following data:
[tex]m[/tex] is the mass of the projectile
[tex]V_{o}[/tex] is the initial speed of the projectile
[tex]h[/tex] is the height at which the projectile was fired
[tex]\theta=0\°[/tex] is the angle (it was fired horizontally)
[tex]g[/tex] is the acceleration due gravity
a) The Work when the applied force and the distance (height [tex]h[/tex]) are parallel is:
[tex]W_{g}=F_{g} hcos \theta[/tex] (1)
Where [tex]F_{g}=mg[/tex] is the force exerted by gravity on the projetile (its weight)
So:
[tex]W_{g}=mg h cos (0\°)[/tex] (2)
[tex]W_{g}=mgh[/tex] (3) This is the work done by gravity
b) According to Conservation of energy principle the initial total mechanical energy [tex]E_{o}[/tex] is equal to the final mechanical energy [tex]E_{f}[/tex]:
[tex]E_{o}=E_{f}[/tex] (4)
Where:
[tex]E_{o}=K_{o}+U_{o}[/tex] (5)
Being [tex]K_{o}=\frac{1}{2}mV_{o}^{2}[/tex] the initial kinetic energy and [tex]U_{o}=mgh[/tex] the initial gravitational potential energy
[tex]E_{f}=K_{f}+U_{f}[/tex] (6)
Being [tex]K_{f}=\frac{1}{2}mV_{f}^{2}[/tex] the final kinetic energy and [tex]U_{f}=mg(0)[/tex] the final gravitational potential energy
So: [tex]K_{o}+U_{o}=K_{f}+U_{f}[/tex] (7)
[tex]K_{f}-K_{o}=U_{o}-U_{f}[/tex] (8)
Where [tex]\Delta K=K_{f}-K_{o}[/tex] is the change in kinetic energy.
Hence:
[tex]\Delta K=U_{o}-U_{f}[/tex]
[tex]\Delta K=mgh-mg(0)[/tex]
[tex]\Delta K=mgh[/tex] (9) This is the change in kinetic energy
c) Isolating [tex]K_{f}[/tex] from (8):
[tex]K_{f}=U_{o}-U_{f}+K_{o}[/tex] (10)
[tex]K_{f}=mgh-mg(0)+\frac{1}{2}mV_{o}^{2}[/tex]
Hence:
[tex]K_{f}=mgh+\frac{1}{2}mV_{o}^{2}[/tex] (11) This is the final kinetic energy of the projectile
d)These results will not change if we change the angle, since these equations do not depend on the angle.
The work done by the force of gravity is m*g*h. The change in kinetic energy is also m*g*h. The final kinetic energy is 1/2*m*v0² + m*g*h. These answers don't change with an initial angle change.
Explanation:(a) The work done by the force of gravity on a projectile is equal to the force of gravity acting downwards (which is m*g, where m is the mass and g is gravity) times the distance over which this force acts, which is h (the initial height above the ground). Therefore the work done by the force of gravity is m*g*h.
(b) The change in kinetic energy is equal to the work done by the force (work-energy theorem). Therefore, the change in kinetic energy is also m*g*h.
(c) The projectile is initially fired horizontally so it only has horizontal kinetic energy. The vertical kinetic energy just before it hits the ground can be found using the energy equation where it is equal to the work done by gravity. The total final kinetic energy would therefore be the initial kinetic energy (1/2*m*v0²) plus the work done by gravity. So the final kinetic energy is 1/2*m*v0² + m*g*h.
(d) The answers to parts (a), (b) and (c) would not change if the initial angle is changed because the vertical and horizontal motions are independent of each other in a projectile motion.
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What is the smallest resistance you can make by combining them?
The equivalent resistance is calculated by taking the inverse of the sum of the reciprocals of each resistance.
The smallest resistance you can obtain by connecting a 36.0-ohm, a 50.0-ohm, and a 700-ohm resistor together can be found by connecting them all in parallel. When resistors are connected in parallel, the equivalent resistance (Requivalent) is smaller than the smallest individual resistance in the network. The formula for the equivalent resistance of parallel resistors is:
1/Requivalent = 1/R1 + 1/R2 + 1/R3
Substituting the given values gives us:
1/Requivalent = 1/36 + 1/50 + 1/700
Calculating the sum of inverses and then taking the inverse of that sum gives us the equivalent resistance for our parallel network. The smallest resistance value, which is Requivalent, can be calculated using this method.
A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at 24.0oC. Neglecting the heat capacity of the container, find the final equilibrium temperature (in oC) of the lead and water.
Answer:[tex]24.70 ^{\circ}C[/tex]
Explanation:
Given
mass of lead piece [tex]m_l=234 gm\approx 0.234 kg[/tex]
mass of water in calorimeter [tex]m_w=611 gm\approx 0.611 kg[/tex]
Initial temperature of water [tex]T_w=24^{\circ}C[/tex]
Initial temperature of lead piece [tex]T_l=24^{\circ}C[/tex]
we know heat capacity of lead and water are [tex]125.604 J/kg-k[/tex] and [tex]4.184 kJ/kg-k[/tex] respectively
Let us take [tex]T ^{\circ}C[/tex] be the final temperature of the system
Conserving energy
heat lost by lead=heat gained by water
[tex]m_lc_l(T_l-T)=m_wc_w(T-T_w)[/tex]
[tex]0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)[/tex]
[tex]86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)[/tex]
[tex]86-T=86.97T-2087.49[/tex]
[tex]T=\frac{2173.491}{87.97}=24.70^{\circ}C[/tex]