Answer:
D. The object will continue to move with a constant velocity
Explanation:
According to Newton's first law also known as law of inertia, states that an object at rest will remain at rest or, if in motion, will remain in motion at constant velocity unless acted on by a net external force.
Therefore, An object moving in the absence of a net force will continue to move at a constant velocity
In the absence of a net force, an object will continue to move with a constant velocity.
Explanation:The correct answer is D. The object will continue to move with a constant velocity. In the absence of a net force, an object will continue to move at a constant velocity. This means that the object will continue to move in a straight line at the same speed without slowing down or changing direction.
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A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. First briefly but precisely explain how this is possible, based on what you heard in the lecture. Then calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one tenth of a second.
Answer:
Explanation:
The ball was moving with velocity of 20 m /s earlier in horizontal direction . Due to kicking, additional V velocity was added to it at 40° because he kicked it at this angle but the ball travelled in the direction of resultant which was making an angle of 30° with the horizontal .
From the relation of inclination of resultant
Tan θ = V sinα / (u + V cosα) where α is angle between u and V , θ is inclination of resultant
Tan30 = [tex]\frac{Vsin40}{20+ Vcos40}[/tex]
[tex]\frac{1}{\sqrt{3 } } =\frac{V\times .64}{20+ V\times.766}[/tex]
20 + .766 V = 1.11 V
20 = .344 V
V = 58 m /s
To know the force , we shall apply concept of impulse
F x t = mv , F is force for time t creating a change of momentum mv
F x .1 = .4 x 58
F = 232 N
An electron is confined in a harmonic oscillator potential well. A photon is emitted when the electron undergoes a 3→1 quantum jump. What is the wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 3.6 N/m? (m el = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, ħ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s)
Answer:
4.74*10^-7 m
Explanation:
TO find the wavelength of the photon you calculate the energy of the photon emitted in a harmonic oscillator:
[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]
m: initial state = 3
n: final state = 1
k = 3.6N/m
By replacing the values of m for the electron, m,n and ħ you obtain:
[tex]E_{m-n}=(1.055*10^{-34}Js)\sqrt\frac{3.6N/m}{9.11*10^{-31}kg}}(3-1)=4.19*10^{-19}J[/tex]
Furthermore, this energy is equivalent to the expression:
[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]
By replacing you obtain:
[tex]\lambda=\frac{2\pi (1.055*10^{-34}Js)(3*10^8m/s)}{4.19*10^{-19}J}=4.74*10^{-7}m[/tex]
hence, the wavelength of the photon is 4.74*10^-7 m
The wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 3.6 N/m - [tex]4.74\times10^-7 m[/tex]
Formula:The energy of the photon emitted in a harmonic oscillator:
[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]
Given:m: initial state = 3
n: final state = 1
k = 3.6N/m
Solution:By replacing the values of m for the electron, m, n and ħ you obtain:
this energy is equivalent to the expression:[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]
By replacing you obtain:Thus, the wavelength of the photon is [tex]4.74\times10^-7 m[/tex]
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A bicyclist and a runner are waiting at a red light. When the light turns green they start to speed up and the bicyclist gets to a final speed of 20 mph in 5 seconds. The runner gets to a final speed of 11 mph in 3 seconds. Which one had the greater acceleration?
the bicyclist or the runner.
Answer:
The acceleration of bicyclist is greater than that of the runner.
Explanation:
It is given that,
Initial speed of both bicyclist and a runner is 0 as they both are waiting at a red light,
When the light turns green they start to speed up.
Final speed of the bicyclist is 20 mph in 5 seconds
The runner gets to a final speed of 11 mph in 3 seconds.
20 mph = 8.94 m/s
11 mph = 4.91 m/s
Acceleration of bicyclist is :
[tex]a_b=\dfrac{v}{t}\\\\a_b=\dfrac{8.94\ m/s}{5\ s}\\\\a_b=1.78\ m/s^2[/tex]
Acceleration of runner is :
[tex]a_r=\dfrac{v}{t}\\\\a_r=\dfrac{4.91\ m/s}{3\ s}\\\\a_r=1.63\ m/s^2[/tex]
It is clear that the acceleration of bicyclist is greater than that of the runner.
The acceleration of the bicyclist is 1.788m/s² and the acceleration of the runner is 1.639m/s².
Hence, the bicyclist has the greater acceleration.
Given the data in the question;
Since the runner and the bicyclist where initially at rest;
Initial velocity of bicyclist; [tex]u_b = 0[/tex]Final velocity of bicyclist; [tex]v_b = 20mph = 8.9408m/s[/tex]Time taken by the bicyclist; [tex]t_b = 5s[/tex]Acceleration is simply the rate at which velocity changes with respect to time. Formula for acceleration can be derived from the First Equation of Motion;
[tex]v = u + at\\\\at = v - u\\\\a = \frac{v - u}{t}[/tex]
Where a is acceleration, v is final velocity, u is initial velocity and t is time elapsed.
Now, to determine who has the greater acceleration, we substitute our given values into the expression above.
For the bicyclist;
[tex]a_b = \frac{v -u}{t} \\\\a_b = \frac{ 8.9408m/s - 0}{5s}\\ \\a_b = \frac{8.9408m/s}{5s}\\ \\a_b = 1.788m/s^2[/tex]
For the runner;
[tex]a_r = \frac{v -u}{t} \\\\a_r = \frac{ 4.91744m/s - 0}{3s}\\ \\a_r = \frac{4.91744m/s}{3s}\\ \\a_r = 1.639m/s^2[/tex]
The acceleration of the bicyclist is 1.788m/s² and the acceleration of the runner is 1.639m/s².
Therefore, the bicyclist has the greater acceleration.
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A point source of light is submerged 3.3 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have? Take the refraction index of water to be 1.333.
Answer:
Maximum Radius = 2.89m
Explanation:
The maximum radius will be determined by the angle of incidence which is equal to the critical angle. Now, any angle larger than that will make the light to be totally internally reflected. Hence, we can figure out that angle from Snell’s law where the refracted angle is 90°, and then use the tangent function.
From Snell's law;
n_air*sin90° = n_water*sin(θ _c)
Where;
θ_c is the critical angle
Refractive index of water; n_water = 1.333
Refractive index of air;n_air = 1
Thus;
1*1 = 1.33sinθ_c
sinθ_c = 1/1.33
θ_c = sin^(-1)0.7519
θ_c = 48.76°
Like I said earlier, we'll use tangent to find the radius.
Thus;
tanθ_c = d/R
From the question, d = 3.3m
Thus;
3.3/tan48.76 = R
So, R = 2.89m
After a switch is thrown to replace the battery in a DC LR circuit with a conducting wire (so that the circuit is still left complete), the time constant represents:
a. the time rate of change of the current in the circuit.b. the time rate of change of the induced emf in the circuit.c. the magnitude of the ratio of the current to the time rate of change of the current.d. all of the above.e. only (a) and (b) above
Answer:
c. the magnitude of the ratio of the current to the time rate of change of the current
Explanation:
In a LR circuit where battery is connected , the expression for decay of current is given by
[tex]i = i_0e^{-\frac{t}{\lambda}[/tex] where i is instantaneous current at time t , i₀ is maximum current , λ is constant .
differentiating on both sides with respect to t
di / dt = [tex]- \frac{i_0}{\lambda} e^{-\frac{t}{\lambda}[/tex]
[tex]\lambda = - \frac{i}{\frac{di}{dt} }[/tex]
So time constant is equal to magnitude of ratio of current to time rate of change of current .
The Smithsonain used to have a huge pendulum that was 21.0 m long. What was its period?
Answer:
T = 9.19 seconds
Explanation:
It is given that,
Length of the pendulum, l = 21 m
We need to find its time period. We know that the time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{L}{g}}[/tex]
g is acceleration due to gravity
So,
[tex]T=2\pi \sqrt{\dfrac{21}{9.8}}\\\\T=9.19\ s[/tex]
So, the period of the pendulum is 9.19 seconds.
Final answer:
The period of the Smithsonian's 21.0 m long pendulum is approximately 9.2 seconds, which is calculated using the formula for a simple pendulum's period T = 2π√(L/g).
Explanation:
The period of a pendulum primarily depends on its length and the acceleration due to gravity. Using the formula for the period of a simple pendulum T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), we can calculate the period of the Smithsonian's pendulum.
Given that the length L is 21.0 m and by taking the standard value for g, 9.81 m/s², we plug these values into the formula to find the period:
T = 2π√(21.0 m / 9.81 m/s²)
Calculating this we get:
T ≈ 2π√(2.14 s²)
T ≈ 2π√(2.14 s²)
T ≈ 2π(1.46 s)
T ≈ 9.2 s
This means the Smithsonian's pendulum would have taken approximately 9.2 seconds to complete one full swing, back and forth. This is the time it would take to go from one side to the other and back again, corresponding to one complete oscillation.
a star is observed to have a temperature of 3000 K and then luminosity of 105.
What is the color of the star?
Answer: the first box is 'red' and the second box is 'supergiants'
Explanation: just did it edg. And it was correct
A star is observed to have a temperature of 3000 K and then luminosity of 105. The color of the star is red.
What is spectra of star?A spectrum is a collection of all visible light. The region of the electromagnetic spectrum that is visible to the human eye is the light that we see, which includes the rainbow's hues.
All electromagnetic energy emits some kind of radiation, whether it is in the form of visible light or another type, and it also radiates heat. Other stars emit heat and light, just like our sun does. Numerous star measurements have revealed a strong correlation between star temperature and star hue.
Given parameters:
Temperature of the star: T = 3000 K.
luminosity of the star = 105.
For this temperature of the star, the color of the star is red.
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A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center of the sphere and at a distance 6.00 cm from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume. Express your answers in newtons per coulomb separated by a comma. Repeat part A if the sphere is a solid conductor.
Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C
You attach a 1.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.400 m and release it from rest. Assume the mass slides on a horizontal surface with negligible friction. The mass reaches a speed of zero again 0.600 s after release (for the first time after release). What is the maximum speed of the mass (in m/s)
Answer:
2.09 m/s
Explanation:
As the spring is stretched initially , and the mass released from rest i.e v=0. Also, The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.600 s. This illustrates half oscillation of the system.
Therefore, for the period of a full oscillation of the system
T= 2t => 2(0.6)=> 1.2 s
As the frequency is the reciprocal of the period, we have
f= 1/T => 1/1.2
f= 0.833 Hz
The angular frequency'ω' is given by,
ω= 2πf => 2π x 0.833=>5.23 rad/s
For the maximum velocity of the object in a spring-mass system:
V[tex](_{max} )[/tex]= Aω
where A is the amplitude of the oscillation. As here, the amplitude of the motion corresponds to the initial displacement of the object (A=0.400 m)
V[tex](_{max} )[/tex]= 0.4 x 5.23 =>2.09 m/s
Unpolarized light of intensity I0 is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45o to that of the first filter, while the axis of the third filter is oriented at 90o to that of the first filter. What is the intensity of the light transmitted through the third filter
Answer:
The intensity of the light transmitted through the third filter is [tex]I_3 = \frac{I_o}{8}[/tex]
Explanation:
From the question we are told
The intensity of the unpolarised light [tex]I_o[/tex]
The angle between the first and second polarizer is [tex]\theta _1 = 45^o[/tex]
The angle between the first and third polarizer is [tex]\theta _2 = 90^o[/tex]
Generally the intensity of light emerging from the first polarizer is mathematically represented as
[tex]I_1 = \frac{I_o}{2}[/tex]
According to Malus law the intensity of light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_1 cos^2 (\theta_1)[/tex]
Substituting for [tex]I_1[/tex] and [tex]\theta _1[/tex]
[tex]I_2 = \frac{I_o}{2} cos^2 (45)[/tex]
[tex]I_2 = \frac{I_o}{4 }[/tex]
According to Malus law the intensity of light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_2 cos ^2 (\theta_2 - \theta_1)[/tex]
Substituting for [tex]I_2[/tex] and [tex]\theta _1 \ and \ \theta _2[/tex]
[tex]I_3 = \frac{I_o}{4} cos ^2 (90 - 45)[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Describe what ballistic stretching is and why it’s harmful. Then, provide at least two examples of how one should properly stretch? (Site 1)
Answer:
When doing ballistic stretching, it is using motion to bounce and stretch your body past its natural range of motion. In doing so you can harm yourself if you don't have a professional to help you as you might tear, damage, or pop your tendons, ligaments, or joints.
Explanation:
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The ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.
The stretching activity that utilizes the momentum of body to achieve greater range of motion and flexibility, is known as Ballistic stretching. It is one of the intense stretching method that involves the bouncing movements to force the body beyond the normal range of motion.
This can be harmful if an athlete do not have a professional trainer to train for the cause. This may cause tear, damage of tendons, ligaments, or joints.
Following are the ways to perform a perform a proper stretch:
One should balance its body weight by standing on its feet together.The bending of knees should be done in a steady manner, and before this proper warm ups are needed to be done.It is not required to start with higher intensity, one can go with 5-10 repetitions for initial days, after that the repetitions can be increased gradually.Thus, we can conclude that the ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.
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If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t − 0.83t2. (a) Find the velocity of the arrow after two seconds. m/s (b) Find the velocity of the arrow when t = a. m/s (c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s (d) With what velocity will the arrow hit the surface? m/s
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
A group of campers travels to a cabin which has no electrical power. In order to provide for a heater and lights, which device would be appropriate? a resistor, an insulator, a generator, or a voltmeter
Answer:A generator is the appropriate device to be used.
Explanation:
A generator is a device which transforms mechanical energy to electrical energy.
The different parts of a generator includes:
- Engine, alternator, fuel system, voltage regulator, lubricating system, cooling and exhaust system, Battery charger and control panel.
The generator works on the principle of electromagnetic induction which states that
that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field. This movement creates a voltage difference between the two ends of the wire or electrical conductor, which in turn causes the electric charges to flow, thus generating electric current.
Therefore, the use of a generator is the best device for the campers to provide electrical power source for their heaters and light in the cabin.
A generator would be the appropriate device to power a heater and lights in a cabin with no electrical power, as it can convert mechanical energy into electrical energy required to operate these devices.
To provide power for a heater and lights in a cabin with no electrical power, the appropriate device among the options provided is a generator. A generator converts mechanical energy into electrical energy, which can then be used to power electrical devices such as heaters and lights. It does not rely on external power sources, making it suitable for remote locations like a cabin.
Other devices mentioned, like a resistor, are components typically used within electronic circuits to control current flow or convert electrical energy into heat. An insulator is a material that prevents the transfer of electricity, and a voltmeter is a tool used to measure voltage across points in a circuit, rather than to provide power. None of these would be suitable to power a heater and lights in the absence of an existing power source.
Ehren is trying to increase his mile run from eight to six minutes. He has decided to run some sandy hills near his home. Which principle of overload is at work? progression time frequency intensity
Answer:
The correct answer is intensity.
Explanation:
The principle of overload is a basic sports fitness training concept which means that in order to improve, athletes must continually work harder as they their bodies adjust to existing workouts. Thus, overloading also plays a role in skill learning.
Now, since erhen wants to increase his mile run from eight to six minutes by run some sandy hills near his home, it means the principle at work is making his training runs more intense for better speed results. Thus the principle at work is intensity because he is trying to do something that makes him run faster.
Final answer:
Ehren is employing the Overload Principle, focusing on the component of intensity to improve his mile run time by running up sandy hills. This method increases resistance and is aligned with the Progression Principle to safely enhance his fitness and performance.
Explanation:
Ehren is working on improving his mile run time and has decided to include running up sandy hills as part of his training. This implementation of intensity in his workouts is a component of the Overload Principle. The principle of overload necessitates a "greater than normal workload or exertion" for an individual to improve in aspects such as aerobic endurance, muscular strength, endurance, and flexibility. By running on sandy hills, Ehren increases the resistance and difficulty compared to running on a flat surface, thus intensifying his training sessions to drive physiological adaptations.
The Progression Principle is also at play here, which entails the gradual increase in stress placed on the body to safely enhance fitness without risking overuse or injury. This principle aligns with the Overload Principle, as it supports the idea of incrementally adding stress to the body through exercises to foster continual improvements. Overall, Ehren's choice to run sandy hills is applying both the intensity and progression components of the overload to achieve his goal of increasing his running pace.
A favorite physics demonstration at the University of Texas at Austin is a giant skateboard about 6 feet long, with about the same mass as a physics professor. Suppose the skateboard rolls with negligible friction on the level classroom floor. The professor is standing at rest on the skateboard, of length L, and the end of the board opposite to the professor is a distance d from the wall. d L If the professor and board have the same mass, and if the professor slowly walks towards the wall, how far is he from the wall when he stops at the opposite end of the skateboard from where he started? (Note his initial distance from the wall is d + L.)
Answer:[tex]d+\frac{L}{2}[/tex]
Explanation:
Given
Length of skateboard is L
distance of skateboard from the wall is d
Suppose mass of skateboard is M
so mass of Professor is M
When Professor moves towards wall skateboard started moving away from wall.
If the professor moves L distance on the skateboard
Therefore relative displacement of the skateboard is
[tex]=\frac{ML}{M+M}[/tex]
[tex]=\frac{L}{2}[/tex]
Therefore professor is at a distance of [tex]d+\frac{L}{2}[/tex] from wall
Final answer:
The professor is 0 meters from the wall when he stops at the opposite end of the skateboard from where he started.
Explanation:
To solve this problem, we can apply the principle of conservation of momentum. The initial momentum of the system, consisting of the professor and the skateboard, is equal to the final momentum of the system. Initially, both the professor and the skateboard are at rest, so the initial momentum is zero.
When the professor walks towards the wall, he exerts a force on the ground, causing an equal and opposite force on him (according to Newton's third law). This force propels the skateboard forward, resulting in a change in momentum of the professor-skateboard system. The total momentum is conserved during this process.
When the professor reaches the opposite end of the skateboard, he comes to a stop. At this point, the skateboard would have moved a certain distance, which we'll call x. If the professor and the board have the same mass, the professor would have moved a distance equal to (d + x) from the wall, while the board would have moved a distance equal to x from the wall.
From the conservation of momentum, we can write:
(0) + (m)(v) = (m)(v) + (M)(V)
Here, m represents the mass of the professor, v represents his initial velocity, M represents the mass of the skateboard, and V represents its final velocity.
Since the professor starts from rest, his initial velocity v is zero. The final velocity V of the skateboard can be calculated using the equation:
(m)(v) = (M)(V)
From the given information, we know that the professor's mass is equal to the skateboard's mass, so m = M. Plugging this into the equation, we get:
(M)(0) = (M)(V)
This simplifies to:
0 = V
Since the final velocity V of the skateboard is zero, we can conclude that the professor comes to a stop at the opposite end of the skateboard. Therefore, he would be a distance x from the wall. To find the value of x, we need to analyze the forces and motion of the system.
When the professor exerts a force on the ground, causing an equal and opposite force on him, the system experiences a net force. According to Newton's second law, the net force on the system is equal to the product of the mass of the system and its acceleration.
Let's define the positive direction as the direction the professor is moving towards the wall. The net force on the system in the positive direction is:
F_net = F_applied - F_opposing
Where F_applied is the force exerted by the professor on the ground, and F_opposing is the sum of all the opposing forces, such as friction.
Since the skateboard rolls with negligible friction, the opposing forces can be considered negligible. Therefore, we have:
F_net = F_applied
From Newton's second law, we can write:
F_net = (M + m)a
Where a is the acceleration of the system.
Plugging in the given information, we have:
150 N = (2M)a
Solving for a gives:
a = 75 N / M
Now, let's consider the motion of the skateboard. Since there is no friction, the only horizontal force acting on the skateboard is the force exerted by the professor on the ground. This force causes the skateboard to accelerate in the positive direction.
The distance x is related to the acceleration a and the time taken t to reach the opposite end of the skateboard. We can use the kinematic equation:
x = (1/2)at^2
Since the professor is initially at rest and comes to a stop at the opposite end, his final velocity vf is zero. We can use the equation:
vf = vi + at
Where vi is the initial velocity of the professor, and a is the acceleration.
Since the professor is initially at rest, his initial velocity vi is zero. Plugging in the known values, we have:
0 = 0 + (75 N / M)t
This simplifies to:
t = 0
Since the time t is zero in this case, the professor reaches the opposite end of the skateboard instantaneously. Therefore, the distance x is also zero.
In conclusion, the professor is 0 meters from the wall when he stops at the opposite end of the skateboard from where he started.
Two equal masses m are constrained to move without friction, one on the positive x axis and one on the positive y axis. They are attached to two identical springs (force constant k) whose other ends are attached to the origin. In addition, the two masses are connected to each other by a third spring of force constant k', The springs are chosen so that the system is in equilibrium with all three springs relaxed (length equal to unstretched length). What are the normal frequencies? Find and describe the normal modes
Answer:
Check the explanation
Explanation:
The potential energy (is the energy by virtue of a particular object's location relative to that of other objects. This term is most of the time linked or associated with restoring forces such as a spring or the force of gravity.) seems to be U = mgy [tex](1/2)(k k')(x^2 y^2)[/tex]. In fact, the mgy term has disappeared from the development.
Kindly check the attached image below for the step by step explanation to the question above.
A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in the opposite directions at 0.45 m/s. Find the child's mass.
Answer:
The child's mass is 14.133 kg
Explanation:
From the principle of conservation of linear momentum, we have;
(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂) × v₃ - m₃ × v₄
We include the negative sign as the velocities were given as moving in the opposite directions
Since the child and the ball are at rest, we have;
v₁ = 0 m/s and v₂= 0 m/s
Hence;
0 = m₁ × v₃ - m₂ × v₄
(m₁ + m₂)× v₃ = m₃ × v₄
Where:
m₁ = Mass of the child
m₂ = Mass of the scooter = 2.4 kg
v₃ = Final velocity of the child and scooter = 0.45 m/s
m₃ = Mass of the ball = 2.4 kg
v₄ = Final velocity of the ball = 3.1 m/s
Plugging the values gives;
(m₁ + 2.4)× 0.45 = 2.4 × 3.1
(m₁ + 2.4) = 16.533
∴ m₁ + 2.4 = 16.533
m₁ = 16.533 - 2.4 = 14.133 kg
The child's mass = 14.133 kg.
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.85 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2840 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.38 V/m, (b) in the negative z direction and has a magnitude of 5.38 V/m, and (c) in the positive x direction and has a magnitude of 5.38 V/m
Answer:
Explanation:
a ) Magnetic force on proton
= B q v , B is magnetic field , q is charge with velocity v
= 2.85 x 10⁻³ x 1.6 x 10⁻¹⁹ x 2840
= 12.95 x 10⁻¹⁹ N
Its direction will be towards positive z - direction according to Fleming's left hand rule.
force on proton due to electric field = charge x electric field.
= 1.6 x 10⁻¹⁹ x 5.38
= 8.608 x 10⁻¹⁹ N
this force will be along the field ie in positive z direction so both the forces are acting in the same direction so they will add up.
total force = (12.95 + 8.608)x 10⁻¹⁹
= 21.558 x 10⁻¹⁹ N .
b ) in this case , both the forces are acting in opposite direction . net force
= (12.95 - 8.608)x 10⁻¹⁹
= 4.342 x 10⁻¹⁹ N
c ) In this case both the forces are acting perpendicular to each other
resultant = √(12.95² + 8.608²) x 10⁻¹⁹ N
= 15.54 x 10⁻¹⁹ N .
A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wheel (nose wheel) is 0.800 m behind the nose, and the main wheels are 3.02 m behind the nose. What percentage of the airplane's weight is supported by the nose wheel?
Answer:
The percentage of the weight supported by the front wheel is A= 19.82 %
Explanation:
From the question we are told that
The center of gravity of the plane to its nose is [tex]z = 2.58 m[/tex]
The distance of the front wheel of the plane to its nose is [tex]l = 0.800\ m[/tex]
The distance of the main wheel of the plane to its nose is [tex]e = 3.02 \ m[/tex]
At equilibrium the Torque about the nose of the airplane is mathematically represented as
[tex]mg (z- l) - G_B *(e - l) = 0[/tex]
Where m is the mass of the airplane
[tex]G_B[/tex] is the weight of the airplane supported by the main wheel
So
[tex]G_B =\frac{mg (z-l)}{(e - l)}[/tex]
Substituting values
[tex]G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}[/tex]
[tex]G_B = 0.8018 mg[/tex]
Now the weight supported at the frontal wheel is mathematically evaluated as
[tex]G_F = mg - G_B[/tex]
Substituting values
[tex]G_F = mg - 0.8018mg[/tex]
[tex]G_F = (1 - 0.8018) mg[/tex]
[tex]G_F = 0.1982 mg[/tex]
Now the weight of the airplane is = mg
Thus percentage of this weight supported by the front wheel is [tex]A = 0. 1982 *100 =[/tex] 19.82 %
A small spotlight mounted in the bottom of a swimming pool that is 4.5 m deep emits light in all directions. On the surface of the pool, directly above the light, the area illuminated is a circle. Determine the maximum radius of this circle. The index of refraction of water is 1.33.
Answer:
Maximum radius = 5.1m
Explanation:
For us to get the radius of the circle of light, we have to first calculate the critical angle which is the angle of incidence above which total internal reflection occurs, i.e. the angle of incidence when θ2 = 90° .
At the point where the total internal reflection occurs, the light ray doesn't pass through the interface, thus, this point is on the edge of the circle of light.
From Snell's law, we have:
n_water * sin (θ1) = n_air * sin(θ2)
Thus;
sin(θ1) = n_air/(n_water * sin(θ2))
Since the critical angle is the value of θ1 when θ2 = 90° and that sin(90°) =1, thus;
sin(θ1) = (n_air/n_water) * sin(90°) = n_air/n_water
We are told the Refractive index of water is 1.33. meanwhile the Refractive index of air is not given but it has a constant value of 1.
Thus, we can determine θ1:
sin(θ1) = (nair/nwater) = 1/1.33
sin(θ1) = 1/1.33
(θ1) = sin^(-1)(1/1.33)
(θ1) = 48.75°
The question when looked at critically, depicts a right triangle with vertices including the light and the extremity of the circle, and we know one of its angles(θ1 = 48.75°) and one of its sides(4.5 m).
Thus, from trigonometric ratio, we can determine the radius as;
r/4.5 = tan(θ1)
r = 4.5tan(48.75°) = 5.1 m
Derive the equation relating the total charge QQ that flows through a search coil (Conceptual Example 29.3) to the magnetic-field magnitude BB. The search coil has NN turns, each with area AA, and the flux through the coil is decreased from its initial maximum value to zero in a time ΔtΔt. The resistance of the coil is RR, and the total charge is Q=IΔtQ=IΔt, where II is the average current induced by the change in flux.
Answer:
Explanation:
flux through the coil = NBA
Change in flux = NBA - 0 = NBA
rate of change of flux = NBA / Δt
emf induced = NBA / Δt
current i = emf / resistance
= NBA / (RΔt)
Charge flowing through the search coil
= NBA Δt/ (RΔt)
Q = NBA/R
calcular la longitud de un péndulo que oscila a 10 Hz en santa fe de bogota, sabiendo que en esta ciudad la aceleracion de la gravedad es de 978 cm/s2.
Answer:
[tex]L=2.48*10^{-3} m[/tex]
Explanation:
The period equation for a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
and we know that T = 1/f, where f is the frequency, so we will have:
[tex]\frac{1}{f}=2\pi \sqrt{\frac{L}{g}}[/tex]
Now, we just need to solve this equation for L.
[tex]\frac{1}{2\pi f}=\sqrt{\frac{L}{g}}[/tex]
[tex]L=\frac{g}{(2\pi f)^{2}}[/tex]
g is the gravity in Bogota (g=9.78 m/s^{2})f is 10 HzL is the lenght of the pendulum[tex]L=\frac{9.78}{(2\pi*10)^{2}}[/tex]
[tex]L=2.48*10^{-3} m[/tex]
I hope it helps you!
Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon, which has been named Dactyl. Assume the mass of Ida is 4.4 x 1016 kg, the mass of Dactyl is 2.6 x 1012 kg, and the distance between the center of Dactyl and Ida is 95 km. G = 6.672x10-11 N-m2/kg2 Part Description Answer Save Status A. Assuming a circular orbit, what would be the orbital speed of Dactyl? (include units with
The orbital speed of Dactyl around Ida is approximately 161.6 m/s.
Explanation:The orbital speed of an object can be calculated using the formula v = √ (G * (M + m) / r), where v is the orbital speed, G is the gravitational constant (6.672x10-11 N-m2/kg2), M is the mass of the primary object (Ida), m is the mass of the secondary object (Dactyl), and r is the distance between the center of the two objects.
Plugging in the given values, we have:
v = √ ((6.672 x 10-11 N-m2/kg2) * ((4.4 x 1016 kg) + (2.6 x 1012 kg)) / (95,000 m))
Simplifying the equation, we get:
v = 161.6 m/s
Therefore, the orbital speed of Dactyl around Ida is approximately 161.6 m/s.
Final answer:
To calculate the orbital speed of Dactyl in a circular orbit around Ida, you can use the formula for orbital speed.
Explanation:
Orbital speed of Dactyl:
The orbital speed of a satellite in a circular orbit can be calculated using the formula: v = √(G * M / r), where v is the orbital speed, G is the gravitational constant, M is the mass of the central body, and r is the distance from the center of the central body.
Substitute the given values: G = 6.67²x10-11 N-m²/kg², M = 4.4 x 10¹⁶ kg, and r = 95 km (converted to meters).
A helicopter, which starts directly above you, lands at a point that is 4.50 km from your present location and in a direction that is 25° north of east. You want to meet the helicopter at it's landing site, however, you must travel along streets that are oriented either east-west or north-south. What is the minimum distance you must travel to reach the helicopter?
Answer:
5.98 km
Explanation:
This question can be easily solved by using the trigonometric properties of a right angled triangle.
See attachment for pictorial explanation
To get x we have
Sinθ = opp / hyp
Sin25 = x / 4.5
x = 4.5 sin 25
x = 4.5 * 0.423
x = 1.9 km
To get y, we have
Cosθ = adj / hyp
Cos25 = y / 4.5
y = 4.5 cos 25
y = 4.5 * 0.906
y = 4.08 km
x + y = 1.9 + 4.09 = 5.98 km
Thus, the minimum distance required is 5.98 km
Final answer:
To find the minimum distance to the helicopter, calculate the eastward and northward components using trigonometry and sum them.
Explanation:
The minimum distance you must travel to reach the helicopter is determined by decomposing the direct diagonal path into two perpendicular paths that correspond to the grid of streets running east-west and north-south. Since the direction to the helicopter is 25° north of east, you can use trigonometry to find the lengths of the east and north legs of your journey. Using the cosine function for the eastward distance (cos(25°) × 4.50 km) and the sine function for the northward distance (sin(25°) × 4.50 km), you can calculate the exact distances you need to travel east and north:
Eastward distance = cos(25°) × 4.50 km
Northward distance = sin(25°) × 4.50 km
Sum these two distances to get the total minimum distance you need to travel.
Two forces act on a block as shown in the picture.
Right
20 N
Left
10 N
What is the net force on the block?
Answer:
10n to the right
Explanation:
(a) The roof of a large arena, with a weight of 410 kN, is lifted by 34 cm so that it can be centered. How much work is done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised 4000 N (about 1/4 of the car's weight) by 5 cm, how much work did her force do on the car?
Answer:
a) W = 139.4 kJ
b) W = 200J
Explanation:
a) given;
Force F = 410kN
distance d = 34cm = 0.34m
Workdone = force × distance = Fd
Substituting the values;
Workdone = 410kN × 0.34m = 139.4 kJ
b) given;
Force F = 4000N
distance d = 5cm = 0.05m
Workdone = force × distance = Fd
Substituting the values;
Workdone = 4000N × 0.05m = 200J
In a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was the same as the track was 200 m. (i) What is the total distance to be covered by the athletes? (ii) What is the displacement of the athletes when they touch the finish line? (iii) Is the motion of the athletes uniform or non uniform? (iv) Is the distance moved by and displacement of athletes at the end of the race equal?
Answer:
1) The track is 200 meters long, the athletes do 4 rounds, so the total distance ran is:
4*200 meters = 800 meters
2) The displacement is defined as the difference between the final position and the initial position.
When you are in a track, and you run a full round, you end in the same position where you started, so the total displacement is 0 meters.
3) As they are running in a closed track (for example, a circular track), they must change the direction of motion at some point, so the motion is not uniform (uniform motion happens when the movement is always in the same direction and always at the same speed)
4) no, is not equal because in the end, the total distance is 800 meters and the displacement is 0 meters.
A glass lens that has an index of refraction equal to 1.57 is coated with a thin layer of transparent material that has an index of refraction equal to 2.10. If white light strikes the lens at near-normal incidence, light of wavelengths 495 nm in air and 660 nm in air are absent from the reflected light. What is the thinnest possible layer of material for which this can be accomplished?
Answer:
Find the given attachment
Answer:
471.4 nm
Explanation:
Please kindly see attachment for the step by step and very detailed solution to the given problem.
The attached file gave an explicit solution
Suppose a 2.0×10−62.0×10−6-kgkg dust particle with charge −1.0×10−9C−1.0×10−9C is moving vertically up a chimney at speed 6.0 m/sm/s when it enters the +2000-N/CN/C E⃗ E→ field pointing away from a metal collection plate of an electrostatic precipitator. The particle is 4.0 cmcm from the plate at that instant. Find the time needed for the particle to hit the plate. Express your answer with the
Answer:
Explanation:
mass of particle m = 2 x 10⁻⁶ kg
charge q = 1 x 10⁻⁹ C
electric field E = 2000 N/C
force on charge = E q
= 2000 x 1 x 10⁻⁹
acceleration = force / mass
= 2000x10⁻⁹ / 2 x 10⁻⁶
= 1 m /s²
initil velocity u = 6 m /s
distance s = 4 x 10⁻²
time = t
s = ut + .5 t²
4 x 10⁻² = 6t + .5 x 1 x t²
t² + 12t - .08 = 0
= .0067 s .
= 6.7 ms .
force= mass x acceleration
Which object would have the greatest force?
Answer:the object with the highest mass and with the highest acceleration
Explanation:
Force is directly proportional to mass
The higher the mass, the higher the force