Explanation:
Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :
[tex]h=-\dfrac{1}{2}t^2+6t-9[/tex]
When the tide first be at 6 ft, put h = 6 ft in above equation as :
[tex]-\dfrac{1}{2}t^2+6t-9=6[/tex]
[tex]-t^2+12t-18=0[/tex]
On solving the above equation to find the value of t. It is equal to :
t = 3.551 seconds
or
t = 8.449 seconds
So, the tide of 6 ft is at 3.551 seconds and 8.449 seconds. Hence, this is the required solution.
A 440.0-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of 3 Hz is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?
Answer:
The original frequency of the guitar string is 437.0 Hz
Explanation:
It is an interesting phenomenon that when two waves of slightly different frequencies (f1 and f2) interfere they produce beats, and the frequency (fb) of those beats is:
[tex] f_{b}=\mid f_{1}-f_{2}\mid [/tex] (1)
Note that because the equation takes the absolute value of the difference between f1 and f2 we have two possible results of [tex] f_{1}-f_{2} [/tex] that satisfy the equation, those are:
[tex] 440.0-f_{2}=3\Longrightarrow f_{2}=437.0\,Hz [/tex]
[tex] -(440.0-f_{2})=3\Longrightarrow f_{2}=443.0\,Hz [/tex]
Because the frequency of the beat is heard to decrease when the frequency of the guitar string is increased, that means the guitar string was at first under the pitch, so the original frequency of the guitar string is 437.0 Hz
Final answer:
The original frequency of the guitar string was 437 Hz, since the string was out of tune and produced a beat frequency of 3 Hz with a 440 Hz tuning fork, and the beat frequency decreased when the string was tightened.
Explanation:
The original frequency of the guitar string can be determined using the concept of beat frequency. The beat frequency is the absolution difference between the frequencies of two sounds. In this case, the beat frequency was 3 Hz when the guitar string was sounded with the 440.0 Hz tuning fork.
Since the beat frequency decreased when the string tension was increased, the original frequency of the string must have been lower than the tuning fork. Therefore, the original frequency of the guitar string could be either 437 Hz or 443 Hz. However, since tightening the string increases its frequency and resulted in a decrease in the beat frequency, this indicates that the original frequency was 437 Hz (440 Hz - 3 Hz).
Express the angular velocity ω of the wheel in terms of the displacement d, the magnitude F of the applied force, and the moment of inertia of the wheel Iw, if you've found such a solution. Otherwise, following the hints for this part should lead you to express the angular velocity ω of the wheel in terms of the displacement d, the wheel's radius r, and α.
The angular velocity (ω) of a wheel can be expressed in terms of the force applied (F), wheel's radius (r), and its moment of inertia (I), all of which factor into the wheel's torque and resulting angular acceleration. Through the rotational analog of Newton's second law and kinematic equations, one can derive a formula for ω that connects these physical quantities
To express the angular velocity (ω) of the wheel in terms of displacement (d), magnitude of the force (F), and the moment of inertia of the wheel (Iw), we use physical principles of rotational motion. We can derive this using the definition of torque and the rotational version of Newton's second law, which states that torque (τ) is equal to the moment of inertia (I) times the angular acceleration (α): τ = Iα. For a wheel, torque is also equal to the force applied (F) times the radius (r): τ = Fr.
By equating the two expressions for torque, we get Fr = Iα, and therefore α = F × r/I. With the angular acceleration and knowing the time the force is applied, we can find the final angular velocity using the kinematic equation for rotational motion: ωfinal = ωinitial + α × t. Assuming that initial angular velocity (ωinitial) is zero and time (t) can be deduced from the linear displacement (d) and the linear velocity (v), given that v = ω × r, we can express the final angular velocity in terms of displacement (d), wheel's radius (r), and angular acceleration (α).
To further involve the moment of inertia in this expression, we would substitute the earlier derived expression for α into our final angular velocity equation: ω = (F × r/I) × t. Now, if we relate time (t) to the linear displacement (d) such that the linear velocity (v) = d/t, and since v = ω×r, we can isolate ω to derive a relationship that includes the moment of inertia.
An archer puts a 0.4 kg arrow to the bowstring. An average force of 190.4 N is exerted to draw the string back 1.47 m. The acceleration of gravity is 9.8 m/s². Assuming no frictional loss, with what speed does the arrow leave the bow? Answer in units of m/s. If the arrow is shot straight up, how high does it rise? Answer in units of m.
Answer:
v = 37.4 m/s , h = 71.39m
Explanation:
To find the velocity given:
m = 0.4 kg
F =190.4 N
d = 1.47 m
g = 9.8 m/s^2
So use the equation of work to solve the kinetic energy
W = F *d = 190.4 N * 1.47m
W = 279.88 J
Ke = 1 / 2 * m* v^2
v = √2*Ke / m =√ 2 *279.88 / 0.4 kg
v = 37.4 m/s
Now to find the high to rise can use the conserved law so:
Ke = Pe
279.88 = m*g*h
Solve to h'
h = 279.88 / 0.4kg * 9.8m/s^2
h =71.39 m
Final answer:
The arrow leaves the bow with a speed of approximately 37.49 m/s and, when shot straight up, rises to a maximum height of about 71.4 m.
Explanation:
To determine the speed with which the arrow leaves the bow, we apply the work-energy principle, which states that work done on the arrow is converted into its kinetic energy. The work done W by the bow can be calculated by multiplying the force F exerted by the distance d over which the force is applied: W = F × d. It is given that F is 190.4 N and d is 1.47 m; thus, W = 190.4 N × 1.47 m = 279.888 J.
The kinetic energy KE of the arrow can be given by KE = ½ mv², where m is the mass of the arrow and v is its velocity. Since work done equals the kinetic energy, we get 279.888 J = ½ × 0.4 kg × v². Solving for v gives us a velocity of approximately 37.49 m/s.
To find how high the arrow goes if it's shot straight up, we use the conservation of energy, where the initial kinetic energy is converted to gravitational potential energy at the highest point. The potential energy PE at maximum height can be given by PE = mgh, where g is the acceleration due to gravity (9.8 m/s²) and h is the height. Setting KE equal to PE, we have 279.888 J = 0.4 kg × 9.8 m/s² × h. Solving for h gives us a maximum height of approximately 71.4 m.
The ability to move or change an object or what a wave carries
Waves can transfer energy and momentum from one place to another without a permanent shift in the mass. They propagate via different oscillation modes and can impact the propagation direction when they encounter an interface between two media.
Explanation:The ability to move or change an object or what a wave carries, essentially describes the energy transfer capability of waves. According to Big Idea 6, waves can transfer energy and momentum from one place to another without permanently relocating the mass. This aspect characterizes waves and differentiates them from other phenomena.
Waves can propagate through various oscillation modes, such as transverse and longitudinal. For instance, when you shake one end of a rope, the perturbation moves in the form of a wave up to the other end of the rope, thus, facilitating energy transfer without a physical shift of the rope's parts.
Another important aspect to remember about waves is their ability to affect the path of propagation when they encounter an interface between two different media, as articulated by Enduring Understanding 6.E. This property explains various natural phenomena, such as the bending of light when it moves from air to water.
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An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?A. 72.2 m/sB. 12.6 m/sC. 47.3 m/sD. 34.2 m/sE. 6.45 m/s
Answer:
The terminal speed of this object is 12.6 m/s
Explanation:
It is given that,
Mass of the object, m = 80 kg
The magnitude of drag force is,
[tex]F_{drag}=12v+4v^2[/tex]
The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.
[tex]F_{drag}=mg[/tex]
[tex]12v+4v^2=80\times 9.8[/tex]
[tex]4v^2+12v=784[/tex]
On solving the above quadratic equation, we get two values of v as :
v = 12.58 m/s
v = -15.58 m/s (not possible)
So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.
A player passes a 0.600-kg basketball downcourt for a fast break. The ball leaves the player’s hands with a speed of 8.30 m/s and slows down to 7.10 m/s at its highest point.
a. Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?
b. How would doubling the ball’s mass affect the result in part a.? Explain.
Answer:
Explanation:
Decrease in kinetic energy of the ball
= 1/2 m ( v² - u² )
= 1/2 x .6 ( 8.3² - 7.1² )
= 5.544 J
This energy will be converted into potential energy
mgH = 5.544 ( H is maximum height attained )
H = 5.544 /( .6 x 9.8 )
= .942 m
= 94.2 cm
b ) In a ) we used the relation
mgH = 1/2 m ( v² - u² )
H = ( 1/2g ) x ( v² - u² )
Here we see H or height attained does not depend upon mass as m cancels out on both sides.
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50.00 cm mark. (a) The period of oscillation is 2.50 s. Find d. (b) If you moved the pivot 5.00 cm closer to the 50.00 cm mark, what would the period of oscillation be
Answer:
(a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
Explanation:
Given that,
Length = 50.00 cm
Time period = 2.50 s
Time period of pendulum is defined as the time for one complete cycle.
The period depends on the length of the pendulum.
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{I}{mgh}}[/tex]
Where, I = moment of inertia
We need to calculate the value of d
Using parallel theorem of moment of inertia
[tex]I=I_{cm}+md^2[/tex]
For a meter stick mass m , the rotational inertia about it's center of mass
[tex]I_{cm}-\dfrac{mL^2}{12}[/tex]
Where, L = 1 m
Put the value into the formula of time period
[tex]T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}[/tex]
[tex]T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})[/tex]
Multiplying both sides by d
tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]
[tex](\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0[/tex]
Put the value of T, L and g into the formula
[tex]4.028d^2-6.25d+0.336=0[/tex]
[tex]d = 0.056\ m, 1.496\ m[/tex]
The value of d is 0.056 cm and 1.496 cm.
(b). Given that,
L = 50-5 = 45 cm
We need to calculate the time period
Using formula of period
[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}[/tex]
[tex]T=1.35\ sec[/tex]
Hence, (a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. The sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density = 0.0180 kg / m ) sustains a tension of 350 N because the wire is stretched between two poles that are 17.43 m apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number n that could be responsible for the "moaning" sound?
Answer:
N = 5 harmonics
Explanation:
As we know that frequency of the sound is given as
[tex]f = \frac{N}{2L}\sqrt{\frac{T}{\mu}}[/tex]
now we have
[tex]T = 350 N[/tex]
[tex]\mu = 0.0180 kg/m[/tex]
L = 17.43 m
now we have
[tex]f = \frac{N}{2(17.43)}\sqrt{\frac{350}{0.0180}}[/tex]
[tex]f = 4 N[/tex]
if the lowest audible frequency is f = 20 Hz
so number of harmonics is given as
[tex]20 = 4 N[/tex]
N = 5 harmonics
Plane a flies at a constant speed from new york to los angeles along a route which is 2000 miles. Plane b flies in the opposite direction at a constant speed which is 100 mph faster than plane a. Plane b takes off one hour after plane a. They land at the same moment. How far are they from los angeles when they pass?
Answer:
from los angeles distance plan a = 1111.08 mi
from los angeles distance plan b = 888.92 mi
Explanation:
given data
new york to los angeles distance = 2000 miles
Plane b speed = 100 mph faster than plane a
Plane b takes off time = 1 hour after plane a
to find out
How far are they from los angeles when they pass
solution
we consider speed of plan a is = x mph
so speed of plan b will be = x + 100 mph
and we know plan b take here 1 hour less time than plan a so it mean time is distance divide speed i.e
[tex]\frac{2000}{x} - 1 =\frac{2000}{x+100}[/tex]
solve it we get x = 400 mph
it mean here
plan a speed is 400 mph
and plan b speed is 500 mph
and
now we consider they meet at time = t hour after a take off
then plan a travel = 400 t
and plan b travel = 500 ( t - 1 )
add both distance that is equal to 2000 mi
so 400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t-1) = 2000
solve we get
t = 2.777
so total distance travel plan a = 400 × 2.777 = 1111.08 mi
total distance travel plan b = 2000 - 1111.08
total distance travel plan b = 888.92 mi
Determine the length of a wind instrument, assuming that it is modeled as a closed tube and that the lowest sound that it can play has a frequency of 99 Hz. The speed of sound in air is 343m/s.
Answer:
Length, l = 0.866 meters
Explanation:
Given that,
Frequency of sound, f = 99 Hz
Speed of sound in air, v = 343 m/s
To find,
The length of a wind instrument.
Solution,
The standing wave will gets formed in wind instrument. For the closed tube, the closed tube the frequency is given by :
[tex]f=\dfrac{v}{4l}[/tex]
Where
l is the length of the instrument
[tex]l=\dfrac{v}{4f}[/tex]
[tex]l=\dfrac{343\ m/s}{4\times 99\ Hz}[/tex]
l = 0.866 meters
So, the length of a wind instrument is 0.866 meters. Hence, this is the required solution.
A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface of length 2 m with a coefficient of friction 0.16, then compresses the second spring of constant 2 N/m. The acceleration of gravity is 9.8 m/s2.
How far will the second spring compress in order to bring the mass to a stop?
Answer:
[tex]X_2=25.27m[/tex]
Explanation:
Here we will call:
1. [tex]E_1[/tex]: The energy when the first spring is compress
2. [tex]E_2[/tex]: The energy after the mass is liberated by the spring
3. [tex]E_3[/tex]: The energy before the second string catch the mass
4. [tex]E_4[/tex]: The energy when the second sping compressed
so, the law of the conservations of energy says that:
1. [tex]E_1 = E_2[/tex]
2. [tex]E_2 -E_3= W_f[/tex]
3.[tex]E_3 = E_4[/tex]
where [tex]W_f[/tex] is the work of the friction.
1. equation 1 is equal to:
[tex]\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2[/tex]
where K is the constant of the spring, x is the distance compressed, M is the mass and [tex]V_2[/tex] the velocity, so:
[tex]\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2[/tex]
Solving for velocity, we get:
[tex]V_2[/tex] = 65.319 m/s
2. Now, equation 2 is equal to:
[tex]\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd [/tex]
where M is the mass, [tex]V_2[/tex] the velocity in the situation 2, [tex]V_3[/tex] is the velocity in the situation 3, [tex]U_k[/tex] is the coefficient of the friction, N the normal force and d the distance, so:
[tex]\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2) [/tex]
Volving for [tex]V_3[/tex], we get:
[tex]V_3 = 65.27 m/s[/tex]
3. Finally, equation 3 is equal to:
[tex]\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2[/tex]
where [tex]K_2[/tex] is the constant of the second spring and [tex]X_2[/tex] is the compress of the second spring, so:
[tex]\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2[/tex]
solving for [tex]X_2[/tex], we get:
[tex]X_2=25.27m[/tex]
The second spring will compress by 25.27 meters to bring the mass to a stop.
To solve this problem, we will go step-by-step considering energy conservation and the work done by friction.
1) Calculate the potential energy stored in the first spring:
The spring constant k is 20 N/m and the compression x is 8 m. The potential energy (PE) in the spring is calculated using the formula:
PE = (1/2) * k * x²
PE = (1/2) * 20 N/m * (8 m)²= 640 J
When the spring is released, this potential energy will convert to kinetic energy (KE) of the mass:
KE = 640 J
Calculate the work done by friction (Wf):
The frictional force (f) is calculated by:
f = μ * m * g
f = 0.16 * 0.3 kg * 9.8 m/s² = 0.4704 N
The work done by friction over a distance (d) of 2 m is:
Wf = f * d
Wf = 0.4704 N * 2 m = 0.9408 J
The kinetic energy after overcoming friction is:
Remaining KE = 640 J - 0.9408 J = 639.0592 J
Calculate the compression of the second spring (x₂):
The remaining kinetic energy will convert into potential energy in the second spring with a constant of 2 N/m.
PE = (1/2) * k₂ * x₂²
We set the remaining KE equal to this potential energy:
639.0592 J = (1/2) * 2 N/m * x₂²
639.0592 J = x₂²
x₂ = √639.0592 = 25.27 m
Therefore, the second spring will compress by 25.27 meters.
Oppositely charged objects attract each other. This attraction holds electrons in atoms and holds atoms to one another in many compounds. However, Ernest Rutherford’s model of the atom failed to explain why electrons were not pulled into the atomic nucleus by this attraction. What change to the atomic model helped solve the problem seen in Rutherford’s model?
Answer:
A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.
Explanation:
The change to the atomic model that helped solve the problem seen in Rutherford's model was the discovery of the strong nuclear force.
Explanation:Rutherford's model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. However, Rutherford's model failed to explain why electrons were not pulled into the atomic nucleus by this attraction. The change to the atomic model that helped solve this problem was the discovery of the strong nuclear force, which is much stronger than electrostatic interactions and holds the protons and neutrons together in the nucleus.
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On your first trip to Planet X you happen to take along a 210 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 26.9 cm . You then pull the mass down 7.00 cm and release it. With the stopwatch you find that 9.00 oscillations take 14.3 s .Can you now satisfy your curiosity?what is the new g
Answer:
4.20m/s^2
Explanation:
The mass of 210g stretch the spring by 26.9cm, 9 oscillation took 14.3 s.
Period (T) = time of oscillation / number of oscillation = 14.3/ 9 = 1.59s
Using the formula for period of a spring;
T = 2π√(m/k)
Divide both side by 2π
T/2π = √(m/k)
Square both side
T^2/ (2π)^2 = m/k
Make k subject of the formula
K = 4π^2 * m/ T^2
Also using Hooke's law
Since the spring was suspended from the ceiling,
F (mg) = k * DL ( where g is the new gravity in m/s^2, k is the force constant of the spring in N/m and DL is the change in length in meter
Make k subject of the formula
Mg/DL = k
Since both equation equal to K then
4π^2 * m/ T^2 = mg/DL
Cancel m on both side
4π^2 * DL / T^2 = g
DL = 26.9cm = 26.9/100 m = 0.269 m
T = 1.59s and π = 3.142
Substitute these into the equation
g = ( 4 * 3.142 * 3.142* 0.269) / (1.59^2) = 4.20 m/s^2
Final answer:
To determine the acceleration due to gravity on Planet X, the period T of the spring-mass system's oscillations is used along with the mass m and the spring constant k, which is calculated using the initial stretch of the spring.
Explanation:
To find the acceleration due to gravity (g) on Planet X, we can use the properties of simple harmonic motion for a spring-mass system. The formula for the period T of a mass m attached to a spring with spring constant k is T = 2π√(m/k). Since the spring stretches 26.9 cm when the 210 g mass is attached, the force due to gravity must be equal to the force the spring exerts at that stretch, which is F = kx, where x is the stretch distance. Thus, mg = kx, and we can solve for k using k = mg/x (with g as the gravity on Earth, 9.81 m/s²). Then we can calculate the period using the measured time of oscillations. From the period, we can solve for the gravity on Planet X.
Considering 9 oscillations took 14.3 seconds, the period T for one oscillation is 14.3 s / 9. With T and m known, one can solve for k and then use that to find g. Since the initial stretch was caused by the force of gravity, the spring constant k can be found using Earth's gravity, and subsequently used in the equation T = 2π√(m/g) rearranged to solve for g on Planet X.
Tia needs to produce a solenoid that has an inductance of 2.07 μ H . She constructs the solenoid by uniformly winding 1.19 m of thin wire around a tube. How long, in centimeters, should the tube be?
Answer:
[tex]l=0.068 m[/tex]
Explanation:
given,
inductance of solenoid = 2.07 μ H
winding of the wire = 1.19 m
Using formula of inductance
[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]
L is the inductance
N is number of turns of the coil
μ₀ is permeability of free space
L is length of winding
N (2π r) = 1.19
squaring both side
4π(N²(πr²))=1.19²
N² A = 0.113
now
[tex]2.07 \times 10^{-6}= \dfrac{4\pi\times 10^{-7}\times 0.113}{l}[/tex]
[tex]l= \dfrac{4\pi\times 10^{-7}\times 0.113}{2.07 \times 10^{-6}}[/tex]
[tex]l=0.068 m[/tex]
This question involves the concepts of the inductance of an inductor and the length of the winding.
The tube should be "6.84 cm" long.
The inductance of an inductor can be given by the following formula:
[tex]L=\frac{N^2\mu_oA}{l}[/tex]
where,
L = inductance = 2.07 μH = 2.07 x 10⁻⁶ H
N = No. of turns in coil = [tex]\frac{length\ of\ wire}{circumference\ of\ tube} = \frac{1..19\ m}{2\pi r}[/tex]
μ₀ = permeability of free space = 4π x 10⁻⁷ Wb/A.m
A = Area of tube = πr²
l = length of tube = ?
Therefore,
[tex]2.07\ x\ 10^{-6}\ H = \frac{(\frac{1.19\ m}{2\pi r})^2(\pi r^2)(4\pi\ x\ 10^{-7}\ Wb/A.m)}{l}\\\\l = \frac{1.4161\ x\ 10^{-7}}{2.07\ x\ 10^{-6}}\\\\[/tex]
l = 0.0684 m = 6.84 cm
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The attached picture shows the inductance formula.
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the stick a distance D from the middle of the stick. Before the collision, the puck was moving with speed v0 in a direction perpendicular to the stick, as indicated in the figure. The collision is completely inelastic, and the puck remains attached to the stick after the collision.
What is the angular momentum Lcm of the system before the collision, with respect to the center of mass of the final system?
Express Lcm in terms of the given variables.
The angular momentum of the hockey stick-puck system before the collision, with respect to the center of mass of the final system, is given by Lcm = mp * v0 * D, where mp is the puck's mass, v0 is its initial velocity, and D is the distance from the center of the stick.
Angular Momentum of Hockey Stick-Puck System
To find the angular momentum Lcm of the system before the collision with respect to the center of mass of the final system, we follow these steps:
Before the collision, the puck is moving with velocity v0. Since the collision is completely inelastic, the stick and puck move as a single system after the collision.The angular momentum L before the collision is given by: This expression simplifies to:Therefore, the angular momentum of the system before the collision, with respect to the center of mass of the final system, is Lcm = mp * v0 * D.
A student standing in a canyon yells "echo", and her voice produces a sound wave of frequency of f = 0.61 kHz. The echo takes t = 3.9 s to return to the student. Assume the speed of sound through the atmosphere at this location is v = 322 m/s.
Answer:
[tex]d=627.9\ m[/tex] is the distance from the obstacle of reflection.
wavelength [tex]\lamb=0.5279\ m [/tex]
Explanation:
Given that:
frequency of sound, [tex]f=610\ Hz[/tex]time taken for the echo to be heard, [tex]t=3.9\ s[/tex]speed of sound, [tex]v=322\ m.s^{-1}[/tex]We know,
[tex]\rm distance = speed \times time[/tex]
During an echo the sound travels the same distance back and forth.
[tex]2d=v.t[/tex]
[tex]2d=322\times 3.9[/tex]
[tex]d=627.9\ m[/tex] is the distance from the obstacle of reflection.
Now the wavelength of sound waves:
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{322}{610}[/tex]
[tex]\lamb=0.5279\ m [/tex]
Planetary orbits around a star can be modeled with the following potentialU(r) =ar+br2(1)(a) Show that the equilibrium position for this potential is equal tore= 2b=a.(b) Use the Taylor expansion on the potential about the equilibrium position to show thatthe \spring" constant of small oscillations around this equilibrium position isa4=8b3
Answer:
a) r eq = -a/(2b)
b) k = a/r eq = -2b
Explanation:
since
U(r) = ar + br²
a) the equilibrium position dU/dr = 0
U(r) = a + 2br = 0 → r eq= -a/2b
b) the Taylor expansion around the equilibrium position is
U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!
,where Un(a) is the nth derivative of U respect with r , evaluated in a
Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative
U(r) = U(r eq) + dU/dr(r eq) (r- r eq) + d²U/dr²(r eq) (r- r eq)² /2
since dU/dr(r eq)=0
U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2
comparing with an energy balance of a spring around its equilibrium position
U(r) - U(r eq) = 1/2 k (r-r eq)² → U(r) = U(r eq) + 1/2 k (r-r eq)²
therefore we can conclude
k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq
thus
k= a/r eq
A box is moved a distance of 10 meters with a force of 5 newtons. The amount of work done is:
Answer:
50 J
Explanation:
Work = force × distance
W = Fd
W = (5 N) (10 m)
W = 50 J
The amount of work done is 50 J (Joule)
How to find the amount of work done?To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.
By applying the formula, we get:
Work = force × distance
Force = 5 newtons ...(given)
distance = 10 meters ...(given)
⇒ W = Fd
⇒ W = (5 N) (10 m)
⇒ W = 50 J
What is the work done?
Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.
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Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? Ignore air resistance.
Answer:before throwing and after catching the ball
Explanation:
When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.
After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.
The point in time where the zero net force acted on the ball is before throwing and after catching the ball
The point at which Net force acted on the ball:At the time of basketball is in the hand of the player, the net force should be zero since holding force is canceled via gravity Force. At the time of overall motion, gravitational force should be acted on the ball which is acting downward. After catching the ball net force on it zero since holding force should be canceled via gravity force and ball should be continue in stationary motion.
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Two asteroids in outer space collide, and stick together. The mass of each asteroid, and the velocity of each asteroid before the impact, are known. To find the momentum of the stuck-together asteroids after the impact, what approach would be useful?
(1) Use the Energy Principle.
(2) Use the Momentum Principle.
(3) It depends on whether or not the speed of the asteroids was near the speed of light.
(4) Use the relationship among velocity, displacement, and time.
(5) It depends on whether the collision was elastic or inelastic.
Answer: (2) Use the Momentum Principle.
Explanation:
In fact, it is called the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] of the asteroids before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision, no matter if the collision was elastic or inelastic (in which the kinetic energy is not conserved).
In this sense, the linear momentum [tex]p[/tex] of a body is defined as:
[tex]p=mV[/tex]
Where [tex]m[/tex] is the mass and [tex]V[/tex] the velocity.
Therefore, the useful approach in this situation is option (2).
A gust of wind blows an apple from a tree. As the apple falls, the gravita- tional force on the apple is 2.25 N downward, and the force of the wind on the apple is 1.05 N to the right.
Find the magnitude and direction of the net force on the apple.
Answer:
The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical
Explanation:
Consider the gravitational force as [tex]\vec{F_g}=2.25\ N[/tex]
Consider the wind force as [tex]\vec{F_w}=1.05\ N[/tex]
The angle between the gravitational force and wind force is 90°
From the parallelogram law we have
[tex]F=\sqrt{F_g^2+F_w^2+2F_gF_wcos\theta}\\\Rightarrow F=\sqrt{2.25^2+1.05^2+2\times 2.25\times 1.05\times cos90}\\\Rightarrow F=2.48294\ N[/tex]
The angle between the resultant and the gravitational force would be
[tex]tan^{-1}\frac{1.05}{2.25}=25.02^{\circ}[/tex]
The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical
A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does the carrier lengthen when it is traveling in the warm Mediterranean Sea at a temperature of 21 °C?
Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature
[tex]\Delta T=T_{f}-T_{i}[/tex]
[tex]\Delta T=21-2.0[/tex]
[tex]\Delta T=19^{\circ}C[/tex]
We need to calculate the carrier lengthen
Using formula of length
[tex]\Delta L=\alpha_{steel}\times L_{0}\times\Delta T[/tex]
Put the value into the formula
[tex]\Delta L=1.2\times10^{-5}\times370\times19[/tex]
[tex]\Delta L=0.08436\ m[/tex]
Hence, The carrier lengthen is 0.08436 m.
At a rock concert, a dB meter registered 128 dB when placed 2.9 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2. Determine the following:
A. What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
B. How far away would the sound level be 82 dB?
Answer:
A)
665.5 W
B)
575.5 m
Explanation:
A)
[tex]S[/tex] = Sound level registered = 128 dB
[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²
[tex]I[/tex] = Intensity at the location of meter
Sound level is given as
[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]
[tex]128 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\12.8 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{12.8} = \frac{I}{1\times10^{-12}} \right \\I = 6.3[/tex]
[tex]P[/tex] = power output of the speaker
[tex]r[/tex] = distance from the speaker = 2.9 m
Power output of the speaker is given as
[tex]P = I(4\pi r^{2} )\\P = (6.3) (4) (3.14) 2.9^{2}\\P = 665.5 W[/tex]
B)
[tex]S[/tex] = Sound level = 82 dB
[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²
[tex]I[/tex] = Intensity at the location of meter
Sound level is given as
[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]
[tex]82 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\8.2 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{8.2} = \frac{I}{1\times10^{-12}} \right \\I = 1.6\times10^{-4}Wm^{-2}[/tex]
Power of the speaker is given as
[tex]P = I(4\pi r^{2} )\\665.5 = (1.6\times10^{-4}) (4) (3.14) r^{2}\\r = 575.5 m[/tex]
The power output of the speaker at the rock concert was approximately 137 Watts. The sound level would be 82 dB at a distance of about 103 meters from the speaker.
Explanation:This problem involves the use of the formula for calculating sound levels in decibels (dB). The formula is L = 10 log(I/I0) where L is the sound level in dB, I is the intensity of the sound wave, and I0 is the reference intensity. Now, let's solve both parts.
A. At the dB meter location, reverse calculate the intensity I: I = I0 * 10^(L/10) = 1.0 10^−12 * 10^(128/10) = 1.26 W/m^2. Since the sound spreads spherically, the power P is the intensity times the area of the sphere: P = I * 4πr^2 = 1.26 * 4π*(2.9)^2 = 137 Watts.
B. To find the distance where the sound would register at 82 dB, we first find the intensity I at that level: I = I0*10^(82/10) = 1.0x10^−12 * 10^8.2 = 6.31x10^-5 W/m^2. With I, we can solve for r: r = sqrt(P/(4πI)) = sqrt(137/(4π*(6.31x10^-5))) = 103 meters.
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An object is hanging from a rope. When it is held in air the tension in the rope is 8.86 N, and when it is submerged in water the tension is 7.84 N. What is the density of the the object
Answer:
8684.2 kg/m³
Explanation:
Tension in the rope as a result of the weight = 8.86 N
Tension in the rope when submerge in water = 7.84
upthrust = 8.86 - 7.84 =1.02 N = mass of water displaced × acceleration due to gravity
Mass of water displaced = 1.02 / 9.81 = 0.104 kg
density of water = mass of water / volume of water
make volume subject of the formula
volume of water displaced = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³
volume of the object = volume of water displaced
density of the object = mass of the object / volume of the object = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³
One end of a string is attached to a ball, with the other end held by a student such that the ball is swung in a horizontal circular path of radius R at a constant tangential speed. At a later time, the tension force exerted on the ball remains constant, but the length of the string is decreased to R4. What is the new tangential speed of the ball?
a. four times the original speed
b. two times the original speed
c. Half the original speed
d. One-fourth the original speed
The new tangential speed of the ball is four times the original speed.
Explanation:When the length of the string is decreased to R/4, the radius of the circular path becomes R/4. The tension force exerted on the ball remains constant because it only depends on the mass of the ball and the gravitational force acting on it. The tangential speed is given by the formula v = ΩR, where Ω is the angular velocity. Since the radius is now 1/4 of the original radius, the new tangential speed is 4 times the original speed (option a).
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The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?
A. 10.36 m/s
B. 3960 m/s
C. 219.3 m
D. 0.0965 m/s
Answer:
A
Explanation:
v = change of X / change of T
v = 200/19.3
Answer:
A. 10.36 m/s
Explanation:
The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?
A. 10.36 m/s
B. 3960 m/s
C. 219.3 m
D. 0.0965 m/s
speed is the change in distance per time
speed is scalar quantity and hence as no direction but only magnitude. time and distance are also scalar quantity
200/19.3
speed=10.36m/s
How long does it take a 750-W coffeepot to bring to a boil 0.65 L of water initially at 13 ∘C? Assume that the part of the pot which is heated with the water is made of 360 g of aluminum, and that no water boils away. Ignore the heat loss to the surrounding environment. The value of specific heat for water is 4186 J/kg⋅C∘ and for aluminum is 900 J/kg⋅C∘.
Answer:
353 s
Explanation:
Energy = energy needed to raise the temperature of pot to 100°C +
required energy needed to raise the temperature of water to 100°C
= MCΔT + mcΔT
= 0.65 kg * 4186 J/kg-°C * 87 °C + 0.36 kg * 900 J/kg-°C * 87 °C
= 264 906.3 J
Energy required = energy delivered * time
264 906.3 J = 750 J/s * time
time = 353 s
The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.
The correct answer is [tex]5787*10^{-6}[/tex]
What is rate law?The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.The balanced reaction is:-
[tex]2H_2O_2(aq) ----> 2H_2O(l) + O_2(g).[/tex]
The rate law of decomposition:-
[tex]rate =-\frac{1}{2}\frac{d[H_2O_2]}{dt}[/tex]
Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:
[tex]rate= -\frac{1}{2} \frac{0.5}{2.16*10^4}[/tex]
rate =[tex]1157*10^{-5}[/tex]
As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s
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A jet transport has a weight of 1.87 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel, and the plane's center of gravity is 12.0 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.
Answer:
(a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].
(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]
Explanation:
Given that,
Weight of jet [tex]W=1.87\times10^{6}\ N[/tex]
Distance = 16 m
Second distance = 12.0 m
We need to calculate the normal force exerted by the ground on the front wheel
Using formula of torque
[tex]\sum\tau=-Wl_{W}+F_{f}l_{f}=0[/tex]
Where, W = weight of jet
[tex]l_{f}[/tex]=lever arms for the forces [tex]F_{f}[/tex]
[tex]l_{w}[/tex]=lever arms for the forces W
Put the value into the formula
[tex]-(1.87\times10^{6})\times(16.0-12.0)+F_{f}\times16.0=0[/tex]
[tex]F_{f}=\dfrac{(1.87\times10^{6})\times(16.0-12.0)}{16.0}[/tex]
[tex]F_{f}=4.67\times10^{5}\ N[/tex]
(b). We need to calculate the normal force exerted by the ground on each of the two rear wheels
The sum of vertical forces equal to zero.
[tex]\sum F_{y}=F_{f}+2F_{r}-W=0[/tex]
We using 2 for two rear wheels
[tex]\sum F_{y}=0[/tex]
[tex]F_{f}+2F_{r}-W=0[/tex]
[tex]F_{r}=\dfrac{F_{f}-W}{2}[/tex]
Put the value into the formula
[tex]F_{r}=\dfrac{-4.67\times10^{5}+1.87\times10^{6}}{2}[/tex]
[tex]F_{r}=7.02\times10^{5}\ N[/tex]
Hence, (a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].
(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]
A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes the forces the objects exert on each other when the collide and provides the best explanation? (A) The force exerted by A on B will be twice as great as the force exerted by B on A, as great as the force exerted by B on A other are the same, because the product of because A has twice the mass of B (B) The force exerted by A on B will be half because A has half the speed of B (C) The forces exerted by each object on the mass and speed is the same for both objects. (D) The forces exerted by each object on the other are the same, because inter objects cannot exert forces of difi magnitude on each other.
Answer:D
Explanation:
Given
mass of A is twice the mass of B half the velocity of B
Suppose [tex]F_a[/tex] and [tex]F_b[/tex] be the average force exerted on A and B respectively
and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.
Thus [tex]F_a=-F_b[/tex] and option d is correct
According to Newton's third law of motion, the forces exerted by each object on the other are the same.
Explanation:The correct answer is (C) The forces exerted by each object on the other are the same because inter objects cannot exert forces of different magnitudes on each other.
According to Newton's third law of motion, whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force it exerts. This means that the forces exerted by object A on object B and by object B on object A are equal in magnitude.
Here, object A has twice the mass of object B, but because both objects have the same speed before the collision, the forces exerted on each other will be the same.
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A 32.5 g iron rod, initially at 21.8°C, is submerged into an unknown mass of water at 63.3°C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.9°C. What is the mass of the water?
Final Answer:
The mass of the water is approximately 48.4 grams.
Explanation:
1. To solve this problem, we can use the principle of conservation of energy, specifically the principle of heat exchange.
2. We can calculate the heat gained by the iron rod and the heat lost by the water, and equate them since there is no heat exchange with the surroundings.
3. The formula for heat exchange is given by: [tex]\[ Q = mc\Delta T \][/tex]
where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.
4. We can set up the equation for heat exchange for the iron rod and the water:
[tex]\[ mc_{\text{iron}}\Delta T_{\text{iron}} = mc_{\text{water}}\Delta T_{\text{water}} \][/tex]
5. Rearranging the equation and solving for the mass of water:
[tex]\[ m_{\text{water}} = \frac{m_{\text{iron}}c_{\text{iron}}\Delta T_{\text{iron}}}{c_{\text{water}}\Delta T_{\text{water}}} \][/tex]
6. Substituting the given values, we can calculate the mass of water to be approximately 48.4 grams.
Therefore, the mass of the water is approximately 48.4 grams.
The mass of the water is approximately 199.8 grams.
Here's how to find the mass of the water:
1. Define the variables:
m_iron = mass of iron rod = 32.5 g
c_iron = specific heat capacity of iron = 0.49 J/g°C
T_iron_initial = initial temperature of iron rod = 21.8°C
T_water_initial = initial temperature of water = 63.3°C
T_final = final temperature of the mixture = 58.9°C
m_water = mass of water (unknown)
c_water = specific heat capacity of water = 4.186 J/g°C
2. Apply the principle of heat transfer:
Heat lost by iron = Heat gained by water
3. Set up the equation:
m_iron * c_iron * (T_iron_initial - T_final) = m_water * c_water * (T_final - T_water_initial)
4. Substitute the known values and solve for m_water:
32.5 g * 0.49 J/g°C * (21.8°C - 58.9°C) = m_water * 4.186 J/g°C * (58.9°C - 63.3°C)
5. Solve for m_water:
m_water ≈ 199.8 g
Therefore, the mass of the water is approximately 199.8 grams.