The location of the mass at t = 5.47s can be determined by the equation for the position associated with SHM. To ascertain the direction of motion at the same time, the velocity function (the derivative of the position function) can be used. If the velocity value is positive, the mass moves in the positive x direction; if negative, it moves in the negative x direction.
Explanation:The problem describes the motion of a mass attached to a spring - specifically, it's simple harmonic motion (SHM). This can be understood in terms of the mass moving back and forth around a central, or equilibrium, position.
a) To determine the location of the mass at a specific time, t = 5.47s, we need to use the equation for the position x(t) associated with SHM: x(t) = A * cos(2 * π * t / T), where A is the amplitude (0.0360m in this case), π is the mathematical constant Pi, t is the time, and T is the period (2.25s in this case). Substituting the given values into this equation will give the position of the mass at t = 5.47s.
b) To determine the direction of motion at that time, we use the derivative of the position function, which gives us the velocity function: v(t) = dx(t) / dt = -A * (2 * π / T) * sin(2 * π * t / T). If this value is positive, then the mass is moving in the positive x direction; if negative, then it's moving in the negative x direction.
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Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will
a. fall.
b. rise.
c. neither rise nor fall.
From the definition of equilibrium, at the moment in which both children are sitting facing each other at a certain distance the torque within the seesaw will be zero.
However, if one of the children approaches the pivot, the center of mass will move towards the end of the other child, which will immediately cause the child to rise.
Another way of observing this problem is considering the distance between the two, the distance is proportional to the Torque,
[tex]\tau_1 = \tau_2[/tex]
[tex]F*d_1 = F*d_2[/tex]
Therefore by decreasing the distance - when walking towards the pivot - the torque of the child sitting at the other end will be greater because it keeps its distance.
Being said higher torque will cause the approaching child to rise.
The correct answer is B.
The child's side will rise because of shifting of weight of the child from its place.
How the child's side rise?Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will rise because of shifting of weight of the child from its place. Due to movement from its place, the weight of other child increases which leads to lowering of its side.
So we can conclude that the child's side will rise because of shifting of weight of the child from its place.
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In a series RCL circuit the generator is set to a frequency that is not the resonant frequency.
This nonresonant frequency is such that the ratio of the inductive reactance to the capacitive reactance of the circuit is observed to be 6.72. The resonant frequency is 240 Hz.
What is the frequency of the generator?
Answer: 624 Hz
Explanation:
If the ratio of the inductive reactance to the capacitive reactance, is 6.72, this means that it must be satified the following expression:
ωL / 1/ωC = 6.72
ω2 LC = 6.72 (1)
Now, at resonance, the inductive reactance and the capacitive reactance are equal each other in magnitude, as follows:
ωo L = 1/ωoC → ωo2 = 1/LC
So, as we know the resonance frequency, we can replace LC in (1) as follows:
ω2 / ωo2 = 6.72
Converting the angular frequencies to frequencies, we have:
4π2 f2 / 4π2 fo2 = 6.72
Simplifying and solving for f, we have:
f = 240 Hz . √6.72 = 624 Hz
As the circuit is inductive, f must be larger than the resonance frequency.
A space shuttle in orbit around the Earth carries its payload with its mechanical arm. Suddenly, the arm malfunctions and releases the payload. What will happen to the payload?
a. It will fall straight down and hit the Earth.
b. It will follow a curved path and eventually hit the Earth.
c. It will drift out into deep space.
d. It will remain in the same orbit with the shuttle.
We must remember that a body in Earth orbit is under perpetual free fall, that is, that the body moves under a constant speed in constant around the earth.
The moment one of the charges is released, its displacement speed will be equal to that of the ship, so it will tend to continue rotating in the orbit in which the ship was traveling.
The correct answer is D. The payload will remain in the same orbit with the space shuttle.
A man fell out of an airplane and barely survived. He was moving at a speed of 100m/s just before landing in deep snow on a mountain side. Experts estimated that the average net force on him was 600 N as he plowed through the snow for 10 s. What was his mass, in kg
Answer:
Mass in kg will be 60 kg
Explanation:
We have given initial speed of the man u = 100 m/sec
And final speed v = 0 m/sec
Time is given, t = 10 sec
So acceleration [tex]a=\frac{v-u}{t}=\frac{0-100}{10}=-10m/sec^2[/tex]
Force is given , F= 600 N
From Newton's law we know that F = ma
So [tex]600=m\times 10[/tex]
mass = 60 kg
So mass in kg will be 60 kg
A fisherman’s scale stretches 3.3 cm when a 2.6-kg fish hangs from it. The fish is pulled down 2.5 cm more and released so that it oscillates up and down with a simple harmonic motion (SHM). ? (a) What is the spring stiffness constant? (b) What is be the amplitude? (c) What is the period and frequency of oscillations?
a) The spring constant is 772.1 N/m
b) The amplitude is 0.025 m
c) The period is 0.365 s, the frequency is 2.74 Hz
Explanation:
a)
At equilibrium, the weight of the fish hanging on the spring is equal to the restoring force of the spring. Therefore, we can write:
[tex]mg=kx[/tex]
where
(mg) is the weight of the fish
kx is the restoring force
m = 2.6 kg is the mass of the fish
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
k is the spring constant
x = 3.3 cm = 0.033 m is the stretching of the spring
Solving for k, we find:
[tex]k=\frac{mg}{x}=\frac{(2.6)(9.8)}{0.033}=772.1 N/m[/tex]
b)
Later, the fish is pulled down by 2.5 cm (0.025 m), and the system starts to oscillate.
The amplitude of a simple harmonic motion is equal to the maximum displacement of the system. In this case, when the fish is pulled down by 2.5 cm and then released, immediately after the releasing the fish moves upward, until it reaches the same displacement on the upper side (+2.5 cm).
This means that the amplitude of the motion corresponds also to the initial displacement of the fish when it is pulled down: therefore, the amplitude is
A = 2.5 cm = 0.025 m
c)
The period of oscillation of a mass-spring system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
Here we have
m = 2.6 kg
k = 772.1 N/m
So, the period is
[tex]T=2\pi \sqrt{\frac{2.6}{772.1}}=0.365 s[/tex]
And the frequency is given by the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}=\frac{1}{0.365}=2.74 Hz[/tex]
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Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass, the vine was hanging vertically from a tall tree limb and the end of the vine is directly level with his center of mass. Tarzan decides to grab the vine and swing which causes him to experience uniform circular motion. At the maximum height, Tarzan notices that the angle the vine makes with the vertical is 19 degrees. Tarzan mass is 65kg. Ignoring any elastic properties of the vine may have.a) Determine the magnitude of Tarzan's centripetal accelerationb) Determine the magnitude of the tension at the beginning of the swing
Answer:
a) [tex]a_c = 1.09m/s^2[/tex]
b) T = 720.85N
Explanation:
With a balance of energy from the lowest point to its maximum height:
[tex]m*g*L(1-cos\theta)-1/2*m*V_o^2=0[/tex]
Solving for [tex]V_o^2[/tex]:
[tex]V_o^2=2*g*L*(1-cos\theta)[/tex]
[tex]V_o^2=2.408[/tex]
Centripetal acceleration is:
[tex]a_c = V_o^2/L[/tex]
[tex]a_c = 2.408/2.21[/tex]
[tex]a_c = 1.09m/s^2[/tex]
To calculate the tension of the rope, we make a sum of forces:
[tex]T - m*g = m*a_c[/tex]
Solving for T:
[tex]T =m*(g+a_c)[/tex]
T = 720.85N
A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 62.0 Hz, amplitude 5.20 mm and wavelength 0.560 m.
(a) How long does it take the wave to travel a distance of 8.40 m along the length of the string?
(b) How long does it take a point on the string to travel a distance of 8.40 m, once the wave train has reached the point and set it into motion?
(c) In parts (a) and (b), how does the time change if the amplitude is doubled?
a) The time taken for the wave to travel a distance of 8.40 m is 0.24 s
b) The time taken for a point on the string to travel a distance of 8.40 m is 6.46 s
c) The time in part a) does not change; the time in part b) is halved (3.23 s)
Explanation:
a)
In order to solve this part, we have to find the speed of the wave, which is given by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the wave in this problem,
[tex]f = 62.0 Hz\\\lambda =0.560 m[/tex]
So its speed is
[tex]v=(62.0)(0.560)=34.7 m/s[/tex]
Now we want to know how long does it take for the wave to travel a distance of
d = 8.40 m
Since the wave has a constant velocity, we can use the equation for uniform motion:
[tex]t=\frac{d}{v}=\frac{8.40}{34.7}=0.24 s[/tex]
b)
In this case, we want to know how long does it take for a point on the string to travel a distance of
d = 8.40 m
A point on the string does not travel along the string, but instead oscillates up and down. The amplitude of the wave corresponds to the maximum displacement of a point on the string from the equilibrium position, and for this wave it is
A = 5.20 mm = 0.0052 m
The period of the wave is the time taken for a point on the string to complete one oscillation. It can be calculated as the reciprocal of the frequency:
[tex]T=\frac{1}{f}=\frac{1}{62.0}=0.016 s[/tex]
During one oscillation (so, in a time corresponding to one period), the distance covered by a point on the string is 4 times the amplitude:
[tex]d=4A=4(0.0052)=0.0208 m[/tex]
Since d is the distance covered by a point on the string in a time T, then we can find the time t needed to cover a distance of
d' = 8.40 m
By setting up the following proportion:
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0208}=6.46 s[/tex]
c)
As we can see from the equation used in part a), the speed of the wave depends only on its frequency and wavelength, not on the amplitude: therefore, if the amplitude is changed, the speed of the wave is not affected, therefore the time calculated in part a) remains the same.
On the contrary, the answer to part b) is affected by the amplitude. In fact, if the amplitude is doubled:
A' = 2A = 2(0.0052) = 0.0104 m
Then the distance covered during one period is
[tex]d=4A' = 4(0.0104)=0.0416 m[/tex]
The time period does not change, so it is still
T = 0.016 s
Therefore, the time needed to cover a distance of
d' = 8.40 m
this time is
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0416}=3.23 s[/tex]
So, the time taken has halved.
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The time taken for a wave to travel a distance of 8.40 m is approximately 0.120 seconds. The time taken for a point on the string to travel 8.40 m once the wave train has reached it is also approximately 0.120 seconds. Doubling the amplitude of the wave does not affect the time taken.
Explanation:To determine how long it takes for the wave to travel a distance of 8.40 m, we can use the formula:
Time = Distance / Speed
The speed of a wave is equal to the product of its frequency and wavelength:
Speed = Frequency x Wavelength
Substituting the given values into the formulas, we find that it takes approximately 0.120 seconds for the wave to travel 8.40 m along the string.
To calculate the time it takes for a point on the string to travel a distance of 8.40 m once the wave train has reached the point and set it into motion, we need to divide the distance by the speed of the wave pulse:
Time = Distance / Speed
Since the speed of the wave pulse is the same as the speed of the wave, the time it takes for a point on the string to travel 8.40 m is also approximately 0.120 seconds.
If the amplitude of the wave is doubled, it does not affect the time it takes for the wave to travel a distance of 8.40 m or for a point on the string to travel 8.40 m. The only thing that changes is the maximum displacement of the particles in the wave, but it does not impact the time taken.
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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s in 0.820 s. (a) What is the magnitude of the linear impulse experienced by a 66.0-kg passenger in the car during the time the car accelerates?
Answer:
[tex]I=336.6kgm/s[/tex]
Explanation:
The equation for the linear impulse is as follows:
[tex]I=F\Delta t[/tex]
where [tex]I[/tex] is impulse, [tex]F[/tex] is the force, and [tex]\Delta t[/tex] is the change in time.
The force, according to Newton's second law:
[tex]F=ma[/tex]
and since [tex]a=\frac{v_{f}-v_{i}}{\Delta t}[/tex]
the force will be:
[tex]F=m(\frac{v_{f}-v_{i}}{\Delta t})[/tex]
replacing in the equation for impulse:
[tex]I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)[/tex]
we see that [tex]\Delta t[/tex] is canceled, so
[tex]I=m(v_{f}-v_{i})[/tex]
And according to the problem [tex]v_{i}=0m/s[/tex], [tex]v_{f}=5.10m/s[/tex] and the mass of the passenger is [tex]m=66kg[/tex]. Thus:
[tex]I=(66kg)(5.10m/s-0m/s)[/tex]
[tex]I=(66kg)(5.10m/s)[/tex]
[tex]I=336.6kgm/s[/tex]
the magnitude of the linear impulse experienced the passenger is [tex]336.6kgm/s[/tex]
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N · m2/kg2, MEarth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km)
In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,
[tex]KE = \fract{1}{2}mv^2[/tex]
[tex]PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))[/tex]
Where,
M = Mass of Earth
m = Mass of Object
v = Velocity
r = Radius
G = Gravitational universal constant
Our values are given as,
[tex]m = 910 Kg[/tex]
[tex]r_1 = 1200 + 6371 km = 7571km[/tex]
[tex]r_2 = 6371 km,[/tex]
Replacing we have,
[tex]\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})[/tex]
[tex]v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})[/tex]
[tex]v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})[/tex]
[tex]v = 4456 m/s[/tex]
Therefore the speed of the object when striking the surface of earth is 4456 m/s
The speed of a 910-kg object striking the Earth after falling from 1200 km above the North Pole can be found using conservation of mechanical energy and the formula for gravitational potential energy.
Explanation:To determine the speed at which a 910-kg object strikes the surface of the Earth when released from rest at an altitude of 1200 km above the North Pole, we can use the conservation of mechanical energy. The total mechanical energy (kinetic plus potential) at the initial and final points must be equal since no external work is done on the system (we ignore atmospheric friction).
Initially, the object has potential energy due to its altitude above Earth and no kinetic energy since it's at rest. When it strikes the Earth, it has kinetic energy and minimal potential energy. Setting the initial and final total energies equal gives:
Ui + Ki = Uf + Kf,
where U is potential energy and K is kinetic energy. The potential energy can be calculated using the formula U = -GMEarthm/r, where m is the object's mass, r is the distance from the object to the center of Earth, and G is the gravitational constant. The kinetic energy is given by K = (1/2)mv2, where v is the velocity of the object.
The initial distance ri to the center of Earth is 6357 km + 1200 km (release altitude), and when the object strikes Earth, rf = 6357 km (polar radius of Earth). We know Ki = 0 and Uf is small enough to be negligible at the Earth's surface.
By plugging in the values, solving the equation for v, and taking the square root, we find the speed of the object when it strikes the ground.
A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much work does the cannon do on the cannonball? (b) The cannon is aimed at a 25.0° angle above the horizontal. Assume gravity is the only other force acting on the cannonball as it moves through the cannon barrel. (That is, ignore all frictional forces.) What is the net work done by these 2 forces on the cannonball while it is in the cannon barrel?
To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,
[tex]W = F*d[/tex]
Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is
[tex]PE = -mgh[/tex]
The total work done would be given by the conservation and sum of these energies, that is to say
[tex]W_{net} = W+PE[/tex]
PART A) Applying the work formula,
[tex]W = F*d\\W = 2750*1.25\\W = 3437.5J[/tex]
PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then
[tex]PE = -mgh*sin25[/tex]
[tex]PE = -16*9.81*1.15sin25[/tex]
[tex]PE = -82.9J[/tex]
The net work would then be given by
[tex]W_{net} = W+PE[/tex]
[tex]W_{net} = 3437.5J-82.9J[/tex]
[tex]W_{net} = 3354.6J[/tex]
Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 11.8 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
To solve this problem it is necessary to apply the concepts based on Newton's second law and the Centripetal Force.
That is to say,
[tex]F_c = F_w[/tex]
Where,
[tex]F_c =[/tex]Centripetal Force
[tex]F_w =[/tex]Weight Force
Expanding the terms we have to,
[tex]mg = \frac{mv^2}{r}[/tex]
[tex]gr = v^2[/tex]
[tex]v = \sqrt{gr}[/tex]
Where,
r = Radius
g = Gravity
v = Velocity
Replacing with our values we have
[tex]v = \sqrt{(9.8)(11.8)}[/tex]
[tex]v = 10.75m/s[/tex]
Therefore the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top is 10.75m/s
How does the magnitude of the electrical force compare between a pair of charged particles when they are brought to half their original distance of separation? To one-quarter their original distance? To four times their original distance? (What law guides your answers?)
Answer:
a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law
Explanation:
Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.
F= k q1q2 / (r12)2
a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:
F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)
b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)
c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.
The empirical law that allows to find out easily these values, is the Coulomb´s Law.
Explanation:
In a double-star system, two stars of mass 4.6 x 1030 kg each rotate about the system's center of mass at radius 1.9 x 1011 m.
(a) What is their common angular speed?
(b) If a meteoroid passes through the system's center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system?
To carry out this exercise, it is necessary to use the equations made to Centripetal Force and Gravitational Energy Conservation.
By definition we know that the Centripetal Force is estimated as
[tex]F_c = M\omega^2R[/tex]
Where,
M = mass
[tex]\omega =[/tex] Angular velocity
R = Radius
From the 'linear' point of view the centripetal force can also be defined as
[tex]F_c = \frac{GM^2}{R^2}[/tex]
PART A ) Equating both equations we have,
[tex]\frac{M}{\omega^2R}=\frac{GM^2}{R^2}[/tex]
Re-arrange to find \omega
[tex]\omega = \sqrt{\frac{Gm}{r^3}}[/tex]
Replacing with our values
[tex]\omega = \sqrt{\frac{(6.67*10^{-11})(4.6*10^{30})}{(1.9*10^{11})^3}}[/tex]
[tex]\omega = 2.115*!0^{-7}rad/s[/tex]
Therefore the angular speed is [tex]\omega = 2.115*!0^{-7}rad/s[/tex]
PART B) For energy conservation we have to
[tex]KE_{min} = PE_{cm}[/tex]
Where,
[tex]KE_{min} =[/tex] Minimus Kinetic Energy
[tex]PE_{cm} =[/tex] Gravitational potential energy at the center of mass
Then,
[tex]\frac{1}{2} mv^2_{min} = \frac{2GMm}{R}[/tex]
Re-arrange to find v,
[tex]v_min = \sqrt{\frac{4GM}{R}}[/tex]
[tex]v_min = \sqrt{\frac{4(6.67*10^{-11})(2.2*10^{30})}{(1.9*10^{11})}}[/tex]
[tex]v_min = 5.55*10^4m/s[/tex]
Therefore the minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system is [tex]v_min = 5.55*10^4m/s[/tex]
You’re in a mall and you need some money to buy a nose ring, but you’re broke. You decide to scoop some quarters out of the fountain. The water in the fountain is one foot deep. How far below the water do the quarters appear to be? The index of refraction of water is 4/3.
Answer: 0.75 ft
Explanation:
This situation is due to Refraction, a phenomenon in which a wave (the light in this case) bends or changes its direction when passing through a medium with an index of refraction different from the other medium. In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.
In addition, we have the following equation that states a relationship between the apparent depth [tex]{d}^{*}[/tex] and the actual depth [tex]d[/tex]:
[tex]{d}^{*}=d\frac{{n}_{1}}{{n}_{2}}[/tex] (1)
Where:
[tex]n_{1}=1[/tex] is the air's index of refraction
[tex]n_{2}=\frac{4}{3}=1.33[/tex] water's index of refraction.
[tex]d=1 ft[/tex] is the actual depth of the quarters
Now. when [tex]n_{1}[/tex] is smaller than [tex]n_{2}[/tex] the apparent depth is smaller than the actual depth. And, when [tex]n_{1}[/tex] is greater than [tex]n_{2}[/tex] the apparent depth is greater than the actual depth.
Let's prove it:
[tex]{d}^{*}=1 ft\frac{1}{1.33}[/tex] (2)
Finally we find the aparent depth of the quarters, which is smaller than the actual depth:
[tex]{d}^{*}=0.75 ft[/tex]
The average density of the body of a fish is 1080 kg/m3. To keep from sinking, the fish increases its volumeby inflating an internal air bladder. By what percentage must the fish increase its volume in order to beneutrally buoyant in fresh water? The density of air at room temperature is 1.2 kg/m3.
Answer:
f = 8 %
Explanation:
given,
density of body of fish = 1080 kg/m³
density of air = 1.2 Kg/m³
density of water = 1000 kg/m²
to protect the fish from sinking volume should increased by the factor f
density of fish + density of water x increase factor = volume changes in water
1080 +f x 1.2 =(1 + f ) x 1000
1080 + f x 1.2 = 1000 + 1000 f
998.8 f = 80
f = 0.0800
f = 8 %
the volume increase factor of fish will be equal to f = 8 %
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation? Give your answer in seconds with 3 significant figures. A = 13 B = 427
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So [tex]mg=Kx[/tex]
[tex]K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m[/tex]
Now we know that [tex]\omega =\sqrt{\frac{K}{m}}[/tex]
So [tex]\frac{2\pi }{T} =\sqrt{\frac{K}{m}}[/tex]
[tex]\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}[/tex]
[tex]T=2.1371sec[/tex]
The period of the resulting oscillation, named T, given a mass of (B + 25.0) g (with B = 427 g) and a spring displacement of (8.50 A) cm (with A = 13), when the object is pulled an additional 3.0 cm downward and released, is 2.345 seconds, when rounded to three significant figures.
Explanation:The problem given can be solved using the concept of simple harmonic motion, which is encountered in Physics. The object attached to the spring can be visualized as a simple harmonic oscillator. The period of the resulting oscillation can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass and k is the spring constant.
First, we compute the mass in grams, which is given by (B + 25.0) g, where B = 427 g. Therefore, the mass is (427 g + 25.0 g) = 452 g or 0.452 kg (since 1 kg = 1000 g).
Next, the spring constant k is calculated using Hooke’s law (F=kx), where F is the force exerted by the mass on the spring (F=mg), m is the mass and g is the acceleration due to gravity, and x is the displacement of the spring. From the problem, m = 0.452 kg, g = 9.8 m/s², and x = (8.50 A) cm = (8.50 x 13) cm = 110.5 cm or 1.105 m (since 1 m = 100 cm). Substituting these values into Hooke’s law gives: k = mg/x = (0.452 kg x 9.8 m/s²) / 1.105 m = 4.00 N/m.
Finally, the period T of the resulting oscillation is found using the formula T = 2π√(m/k) = 2π√(0.452 kg/4.00 N/m) = 2.345 s, rounded to three significant figures.
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A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp, measured from the horizontal, at a speed of 40.0 m/s (144 km/h). The top of the ramp is at the same height as the roofs of the buses and each bus is 20.0 m long.
Answer:
Dare devil can cross 7 buses.
Explanation:
given,
angle of inclination of ramp = 32°
speed of the motorcycle = 40 m/s
length of bus = 20 m
how many buses daredevil can clear =?
to solve this we need to calculate the range of the daredevil
considering it as projectile
the range of motorcyclist
[tex]R = \dfrac{V^2 sin (2\theta)}{g}[/tex]
[tex]R = \dfrac{40^2 sin (2\times 32^0)}{9.8}[/tex]
[tex]R =163.26 \times sin(2\times 32^0)[/tex]
[tex]R =146.74\ m[/tex]
length of bus is given as 20 m
Number of bus daredevil can cross
[tex]N = \dfrac{147.74}{20}[/tex]
[tex]N =7.34[/tex]
Dare devil can cross 7 buses.
Answer:
The number of buses are 7.
Explanation:
Given that,
Angle =32°
Speed = 40.0 m/s
Length of bus = 20.0
We need to calculate the range of bus
Using formula of range
[tex]R=\dfrac{v^2\sin2\theta}{g}[/tex]
Where, g = acceleration due to gravity
v = initial velocity
Put the value into the formula
[tex]R=\dfrac{(40.0)^2\times\sin(2\times32)}{9.8}[/tex]
[tex]R=150.20\ m[/tex]
We need to calculate the number of buses
Using formula of number of buses
[tex]N=\dfrac{R}{L}[/tex]
Where, R = range
L = length of bus
[tex]N=\dfrac{150.20}{20.0}[/tex]
[tex]N=7[/tex]
Hence, The number of buses are 7.
The heart produces a weak magnetic field that can be used to diagnose certain heart problems. It is a dipole field produced by a current loop in the outer layers of the heart.
a. It is estimated that the field at the center of the heart is 90 pT. What current must circulate around an 8.0-cm-diameter loop, about the size of a human heart, to produce this field?
To produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, a current of approximately 0.045 Amps must circulate around the loop.
Explanation:To calculate the current needed to produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, we can use the formula for the magnetic field strength at the center of a circular loop. The formula is given by B = (µ₀I)/(2R), where B is the magnetic field strength, µ₀ is the permeability of free space, I is the current, and R is the radius of the loop.
First, let's convert 90 pT to Tesla by dividing by 10^12. This gives us a value of 9 x 10^-11 T. We can now substitute this value into the formula along with the given radius of 4.0 cm (half of the diameter) to solve for the current.
B = (µ₀I)/(2R)
9 x 10^-11 T = (4π x 10^-7 T m/A)(I)/(2(0.04 m))
Simplifying the equation, we find:
I = (2π(0.04 m)(9 x 10^-11 T))/(4π x 10^-7 T m/A) = 0.045 A
Therefore, a current of approximately 0.045 Amps must circulate around the 8.0-cm-diameter loop to produce a magnetic field of 90 pT at its center.
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A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?
Answer:2
Explanation:
Given
Velocity of bicycle is 5 m/s towards north
radius of rim [tex]r=20 cm[/tex]
mass of rim [tex]m=2 kg[/tex]
Angular momentum [tex]\vec{L}=I\cdot \vec{\omega }[/tex]
[tex]I=mr^2=2\times 0.2^2=0.08 kg-m^2[/tex]
[tex]\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s[/tex]
[tex]L=0.08\times 25=2kg-m^2/s[/tex]
direction of Angular momentum will be towards west
A Texas cockroach of mass 0.117 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 17.3 cm, rotational inertia 5.20 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 1.91 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 2.87 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer
given,
mass of cockroach = 0.117 Kg
radius = 17.3 cm
rotational inertia = 5.20 x 10⁻³ Kg.m²
speed of cockroach = 1.91 m/s
angular velocity of Susan (ω₀)= 2.87 rad/s
final speed of cockroach = 0 m/s
Initial angular velocity of Susan
L_s = I ω₀
L_s = 5.20 x 10⁻³ x 2.87
L_s=0.015 kg.m²/s
initial angular momentum of the cockroach
L_c = - mvr
L_c = - 0.117 x 1.91 x 0.173
L_c = - 0.0387 kg.m²/s
total angular momentum of Both
L = 0.015 - 0.0387
L = - 0.0237 kg.m²/s
after cockroach stop inertia becomes
I_f = I + mr^2
I_f = 5.20 x 10⁻³+ 0.117 x 0.173^2
I_f = 8.7 x 10⁻³ kg.m²/s
final angular momentum of the disk
L_f = I_f ω_f
L_f = 8.7 x 10⁻³ x ω_f
using conservation of momentum
L_i = L_f
-0.0237 =8.7 x 10⁻³ x ω_f
[tex]\omega_f = \dfrac{-0.0237}{8.7 \times 10^{-3}}[/tex]
[tex]\omega_f = -2.72\ rad/s[/tex]
angular speed of Susan is [tex]\omega_f = -2.72\ rad/s[/tex]
The value is negative because it is in the opposite direction of cockroach.
[tex]|\omega_f |= 2.72\ rad/s[/tex]
b) the mechanical energy is not conserved because cockroach stopped in between.
a kg object and a kg object are separated by m. find the magnitude of the net gravitational force exerted by these objects on a kg object placed midway between them. at what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
Answer:
F₂ = 0 N
x = 0.5*d = 0.5 m
Explanation:
Given
m₁ = m₂ = m₃ = m = 1 Kg
d = distance between m₁ and m₃ = d₁₃ = 1 m
d₁₂ = 0.5*d = 0.5*(1 m) = 0.5 m = d₃₂
G = 6.673*10⁻¹¹ N*m²/ Kg²
a) Find the magnitude of the net gravitational force exerted by these objects on a kg object (F₃) placed midway between them.
we can apply
F₂ = F₁₂ + F₃₂
then we use the formula
F₁₂ = G*m₁*m₂ / d₁₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₁₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₁₂ = 2.6692*10⁻¹⁰ N (←)
and
F₃₂ = G*m₃*m₂ / d₃₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₃₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₃₂ = 2.6692*10⁻¹⁰ N (→)
we get
F₂ = F₁₂ + F₃₂ = (- 2.6692*10⁻¹⁰ N) + (2.6692*10⁻¹⁰ N) = 0 N
b) At what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
From a) It is known that x = 0.5*d = 0.5*(1 m) = 0.5 m
Nevertheless, we can find the distance as follows
If
F₂ = 0 ⇒ F₁₂ + F₃₂ = 0 ⇒ F₁₂ = - F₃₂
If
x = distance between m₁ and m₂
d - x = distance between m₂ and m₃
we get
F₁₂ = G*m² / x²
F₃₂ = G*m² / (d - x)²
If F₁₂ = - F₃₂
⇒ G*m² / x² = - G*m² / (d - x)²
⇒ 1 / x² = - 1 / (d - x)²
⇒ (d - x)² = - x²
⇒ d²- 2*d*x + x² = - x²
⇒ 2*x² - 2*d*x + d² = 0
⇒ 2*x² - 2*1*x + 1² = 0
⇒ 2*x² - 2*x + 1 = 0
Solving the equation we obtain
x = 0.5 m = 0.5*d
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at rest. After he lands on the raft, the raft, with him on it, moves away from the pier at 1.6 m/s. What was Tyrone's speed as he ran off the end of the pier?
Answer:u=4.04 m/s
Explanation:
Given
Mass m=85 kg
mass of Raft M=130 kg
velocity of raft and man v=1.6 m/s
Let initial speed of Tyrone is u
Conserving Momentum as there is no external Force
[tex]mu=(M+m)v[/tex]
[tex]85\times u=(85+130)\cdot 1.6[/tex]
[tex]u=\frac{215}{85}\cdot 1.6[/tex]
[tex]u=2.529\cdot 1.6=4.04 m/s[/tex]
Final answer:
Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.
Explanation:
To calculate Tyrone's speed as he ran off the end of the pier, we can use the principle of conservation of momentum. According to this principle, the initial momentum of the system (Tyrone and the raft) is equal to the final momentum. Tyrone's initial momentum is given by his mass (85 kg) multiplied by his speed (which we need to find). The mass of the raft is 130 kg and its initial velocity is 0 m/s since it was at rest. The final momentum of the system is given by the mass of Tyrone and the raft (215 kg) multiplied by their final velocity (1.6 m/s).
Using the equation for conservation of momentum, we have:
(85 kg)(v) = (215 kg)(1.6 m/s)
Solving for the velocity v:
v = (215 kg)(1.6 m/s) / 85 kg
v ≈ 4.04 m/s
Therefore, Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.
The driver of a car, which is initially at rest at the top A of the grade, releases the brakes and coasts down the grade with an acceleration in feet per second squared given by a = 3.22-0.004v2 , where v is the velocity in feet per second. Determine the velocity vB at the bottom B of the grade.
Answer:
28.3 ft/s
Explanation:
We are given that
Initial velocity of a car=[tex]u=0[/tex]
Acceleration of the car=[tex]a=3.22-0.004v^2[/tex]
We have to find the velocity [tex]v_B[/tex] at the bottom of the grade.
Distance covered by car=600 ft
We know that
Acceleration=a=[tex]\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}[/tex]
[tex]v=\frac{ds}{dt}[/tex]
[tex]ds=\frac{vdv}{a}[/tex]
Taking integration on both side and taking limit of s from 0 to 600 and limit of v from 0 to [tex]v_B[/tex]
[tex]\int_{0}^{600}ds=\int_{0}^{v_B}\frac{vdv}{3.22-0.004v^2}[/tex]
[tex][S]^{600}_{0}=-\frac{1}{0.008}[ln(3.22-0.004v^2)]^{v_B}_{0}[/tex]
Using substitution method and [tex]\int f(x)dx=F(b)-F(a)[/tex]
[tex]600-0=-\frac{1}{0.008}[ln(3.22-0.004v^2_B)-ln(3.22)][/tex]
[tex]600=-\frac{1}{0.008}(ln(\frac{3.22-0.04v_B}{3.22}))[/tex]
[tex]ln(m)-ln(n)=ln(\frac{m}{n})[/tex]
[tex]600\times 0.008=-ln(\frac{3.22-0.04v_B}{3.22})[/tex]
[tex]-4.8=ln(\frac{3.22-0.04v_B}{3.22})[/tex]
[tex]\frac{3.22-0.04v_B}{3.22}=e^{-4.8}[/tex]
[tex]ln x=y\implies x=e^y[/tex]
[tex]3.22-0.004v^2_B=3.22e^{-4.8}[/tex]
[tex]3.22-0.004v^2_B=0.0265[/tex]
[tex]3.22-0.0265=0.004v^2_B[/tex]
[tex]0.004v^2_B=3.22-0.0265=3.1935[/tex]
[tex]v_B=\sqrt{\frac{3.1935}{0.004}}=28.3 ft/s[/tex]
Hence, the velocity [tex]v_B[/tex] at the bottom of the grade=28.3 ft/s
The velocity of the car at the bottom of the grade is 28.37 ft/s.
What is acceleration?Acceleration is the rate of change of velocity with time.
Motion of the car from top to bottom of the gradeThe velocity of the car increases as the car moves downwards and it will be maximum when the car reaches equilibrium position. At equilibrium position (bottom of the grade) the acceleration will be zero but the velocity will be maximum.
a = 3.22 - 0.004v²
0 = 3.22 - 0.004v²
0.004v² = 3.22
v² = 3.22/0.004
v² = 805
v = √805
v = 28.37 ft/s
Thus, the velocity of the car at the bottom of the grade is 28.37 ft/s.
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Suppose that there are N=108 two-state systems, each with energy difference E=6 × 10-21 J between the two states. The environment temperature is 300 K. The number of systems in their high energy states is M. Therefore, the free energy is a function of M: F(M).What is dU/dM?
Answer:
The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].
Explanation:
Given that,
Number [tex]N=10^{8}[/tex]
Energy difference = 6\times10^{-21}\ J[/tex]
Temperature T =300 K
We need to calculate the value of [tex]\dfrac{dU}{dM}[/tex]
We know that,
[tex]\dfrac{dU}{dM}=\dfrac{energy\ difference}{Change\ in\ number\ of\ system}[/tex]
[tex]\dfrac{dU}{dM}=\dfrac{6\times10^{-21}}{10^{8}}[/tex]
[tex]\dfrac{dU}{dM}=6\times10^{-29}\ J/unit[/tex]
Hence, The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].
A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 3.10m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.210rad/s relative to the earth. The radius of the turntable is 3.10m , and its moment of inertia about the axis of rotation is 81.0kg?m2
Question: Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)
This problem can be resolved using the principle of conservation of angular momentum. Initially, the positive momentum of runner and negative of turntable cancel each other, making the total momentum zero. When the runner stops, the angular momentum of the turntable is expected to compensate for the total momentum of the system.
Explanation:In Physics, this is a classic problem involving the conservation of angular momentum. Angular momentum is given as L = Iω for an object rotating about an axis, where I is the moment of inertia and ω is the angular velocity. The total initial angular momentum of the system is zero because the positive runner's angular momentum cancels out the negative turntable's angular momentum.
When the runner comes to rest, he no longer contributes an angular momentum. The turntable has to account for all of the angular momentum. We therefore use the conservation of angular momentum to solve for the new angular momentum of the turntable after the runner stops:
Equating moment of inertia of the runner (considering the runner as a point mass, moment of inertia, I = m*r^2) and moment of inertia of the turntable and using the equation L_initial = L_final, we can find the final angular velocity of the system:
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The final angular velocity of the system is 6.853 rad/s.
The initial angular momentum [tex]\(L_{i}\)[/tex] of the turntable is:
[tex]\[ L_{i} = I\omega_{i} \][/tex]
where I is the moment of inertia of the turntable.
The initial angular momentum [tex]\(L_{runner,i}\)[/tex] of the runner is:
[tex]\[ L_{runner,i} = mvr \][/tex]
since the angular momentum of a particle moving in a circle is the product of its mass, velocity, and the radius of the circle.
The final angular momentum [tex]\(L_{f}\)[/tex] of the system is:
[tex]\[ L_{f} = I\omega_{f} + mvr_{f} \][/tex]
where [tex]\(r_{f}\)[/tex] is the final radius at which the runner is at rest relative to the turntable, which is the same as the initial radius r since the runner is still on the turntable.
By conservation of angular momentum:
[tex]\[ L_{i} + L_{runner,i} = L_{f} \] \[ I\omega_{i} + mvr = I\omega_{f} + mvr \][/tex]
Since the runner is at rest relative to the turntable in the final state, (vr = 0), and the equation simplifies to:
[tex]\[ I\omega_{i} + mvr = I\omega_{f} \][/tex]
Now we can solve for [tex]\(\omega_{f}\)[/tex]:
[tex]\[ \omega_{f} = \frac{I\omega_{i} + mvr}{I} \] \[ \omega_{f} = \omega_{i} + \frac{mvr}{I} \][/tex]
Given:
[tex]\(m = 56.0\) kg\\ \(v = 3.10\) m/s\\ \(r = 3.10\) m\\ \(I = 81.0\) kg\(\cdot\)m\(^2\)\\ \(\omega_{i} = 0.210\) rad/s[/tex]
Plugging in the values:
[tex]\[ \omega_{f} = 0.210 + \frac{56.0 \times 3.10 \times 3.10}{81.0} \] \[ \omega_{f} = 0.210 + \frac{56.0 \times 9.61}{81.0} \] \[ \omega_{f} = 0.210 + \frac{538.16}{81.0} \] \[ \omega_{f} = 0.210 + 6.643 \] \[ \omega_{f} = 6.853 \][/tex]
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanbag is moving exactly horizontally when it strikes and sticks to the inside wall of the crate, which is lined with velcro. As a result of this impact, the crate slides backwards 0.935 meters along the concrete, which has a coefficient of kinetic friction of \mu_k = 0.480μ k =0.480. The beanbag has a mass of 0.354 kg and the empty crate has a mass of 3.77 kg. Calculate the speed of the beanbag vv at the moment it strikes the crate. You may neglect the effect of air resistance.
Answer:
31.42383 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction = 0.48
s = Displacement = 0.935 m
[tex]m_1[/tex] = Mass of bean bag = 0.354 kg
[tex]m_2[/tex] = Mass of empty crate = 3.77 kg
[tex]v_1[/tex] = Speed of the bean bag
[tex]v_2[/tex] = Speed of the crate
Acceleration
[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]
[tex]a=--9.81\times 0.48=4.7088\ m/s^2[/tex]
From equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s[/tex]
In this system the momentum is conserved
[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s[/tex]
The speed of the bean bag is 31.42383 m/s
The speed of the beanbag when it strikes the crate is 34.46 m/s.
Inelastic collision:
In an inelastic collision, two bodies collide, stick together and move with the same velocity. The momentum is conserved while the kinetic energy is not.
As the beanbag sticks to the crate they move together with the same velocity until they stop after 0.935 m due to frictional force.
Let M be the combined mass of the crate and beanbag. the frictional force acting on the system is:
[tex]F=-\mu_k Mg\\\\Ma=-\mu_k Mg\\\\a=-\mu g\\\\a=-0.480\times9.8\;m/s^2\\\\a=4.7\;m/s^2[/tex]will be the acceleration of the system.
Let the initial velocity of the system be u and the final velocity is 0, then from the third equation of motion:
[tex]0=u^2-2as\\\\u=\sqrt[]{2as}\\\\u=\sqrt{2\times4.7\times0.935}\;m/s\\\\u=2.96\;m/s[/tex]
So the momentum after collision is
Mu = (0.354 + 3.77)×2.96 = 12.2 kgm/s
Which must be equal to the initial momentum that is contributed only by the beanbag, since the crate is on rest. Let the speed of the bean bag be v then initial momentum should be:
12.2 = 0.354 × v
v = 34.46 m/s
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If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes? Express your answer in meters.
Answer:
f = 0.84 m
Explanation:
given,
near point of the distance from eye(di) = 0.60 m
focal length calculation = ?
distance he can read (d_o)= 0.350 m
now,
Using lens formula
[tex]\dfrac{1}{f} = \dfrac{1}{d_0} + \dfrac{1}{d_i}[/tex]
[tex]\dfrac{1}{f} = \dfrac{1}{0.35} + \dfrac{1}{-0.6}[/tex]
[tex]\dfrac{1}{f} = 1.19[/tex]
[tex] f = \dfrac{1}{1.19}[/tex]
f = 0.84 m
The focal length of the lens f = 0.84 m
5. A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with the velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.
Answer:
The recoil velocity of the raft is 1.205 m/s.
Explanation:
given that,
Mass of the swimmer, [tex]m_1=55\ kg[/tex]
Mass of the raft, [tex]m_2=210\ kg[/tex]
Velocity of the swimmer, v = +4.6 m/s
It is mentioned that the swimmer then runs off the raft, the total linear momentum of the swimmer/raft system is conserved. Let V is the recoil velocity of the raft.
[tex]m_1v+m_2V=0[/tex]
[tex]55\times 4.6+210V=0[/tex]
V = -1.205 m/s
So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.
Answer:
The recoil velocity of the raft would be [tex]v_{r}\approx 1.2\frac{m}{s}[/tex] (pointing to the left if the swimmer runs to the right)
Explanation:
The problem states that the swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.
To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.
Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:
[tex]p_{i}=0[/tex]
Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as
[tex]p_{f}=mv-Mv_{r}=0[/tex]
The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.
Then, from that previous relation we can clear
[tex]v_{r}=\frac{m}{M}v=\frac{55}{210}*4.6\frac{m}{s}=\frac{253}{210}\frac{m}{s}\approx1.2\frac{m}{s}[/tex]
wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).
An object moves in simple harmonic motion with amplitude 14cm and period 0.2 seconds. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a positive direction. Give the equation modeling the displacement d as a function of time t.
Answer:
d= 14 sin 31.41 t
Explanation:
Lets take the general equation of the motion
d= D sinωt
d=Displacement
ω=Natural angular frequency
t=Time
D=Amplitude
Given that D= 14 cm
We know that time period T
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Given that T= 0.2 s
[tex]0.2=\dfrac{2\pi}{\omega}[/tex]
ω=31.41 rad/s
D= 14 cm
Therefore
d= D sinωt
d= 14 sin 31.41 t
Final answer:
The equation modeling the displacement d of an object in simple harmonic motion with an amplitude of 14 cm and a period of 0.2 seconds, starting at 0 cm and moving in a positive direction at t = 0, is d(t) = 14cos(10πt + π/2) cm.
Explanation:
The displacement d of an object in simple harmonic motion can be modeled by the function d(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. Since the amplitude A is 14 cm and the period T is 0.2 seconds, and given that the initial displacement at time t = 0 is 0 cm and it initially moves in a positive direction, we can determine ω using the formula ω = 2π/T and φ using the fact that the cosine function must start at 0.
Therefore, the angular frequency ω = 2π/0.2 s = 10π rad/s, and since the function starts at 0 and moves in a positive direction, φ = +π/2. Hence, the equation that models the displacement d as a function of time t is d(t) = 14cos(10πt + π/2) cm.
The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitude of the acceleration due to gravity g at the planet surface depend on the planet radius R ? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2
Answer:
the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)
Explanation:
according to newton's law of universal gravitation ( we will neglect relativistic effects)
F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet
if we assume that the planet has a spherical shape, the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,
since M= ρ* V = ρ* 4/3πR³ , ρ= density
F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR
from Newton's second law
F= m*g = G*ρ*m* 4/3πR
thus
g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R
g ∝ R