Answers
Answer:
75mL
...............
The question is full of inconsistencies and conceptual errors. Students should not waste their time trying to answer it ... it will only mislead and confuse them.
The question is poor. It was written by someone unclear on the concepts. It should be ignored.
Related Questions
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.
What value of g will you report back to headquarters?
Answers
Answer:
The value of g is [tex]g =76.2 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the weight is [tex]m = 1.30 kg[/tex]
The spring constant [tex]k = 1.73 g/m = 1.73 *10^{-3} \ kg/m[/tex]
The second harmonic frequency is [tex]f = 100 \ Hz[/tex]
The number of oscillation is [tex]N = 200[/tex]
The time taken is [tex]t = 315 \ s[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated
[tex]l = \lambda[/tex]
Noe from the question the vibrating string is just half of the length of the main string so
Let assume the length of the main string is [tex]L[/tex]
So [tex]l = \frac{L}{2}[/tex]
The velocity of the vibrating string is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension on the string which can be mathematically represented as
[tex]T = mg[/tex]
So
[tex]v = \sqrt{\frac{mg}{k} }[/tex]
Then
[tex]f = \frac{v}{\frac{L}{2} }[/tex]
=> [tex]v = \frac{fL }{2}[/tex]
=> [tex]\sqrt{\frac{mg}{k} } = \frac{fL}{2}[/tex]
=> [tex]g = \frac{f^2 L^2 \mu}{4m}[/tex]
substituting values
[tex]g = \frac{(100) * (1.73 *10^{-3} )}{(4 * 1.30)} L^2[/tex]
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
Generally the period of oscillation is mathematically represented as
[tex]T_p = 2 \pi \sqrt{\frac{L}{g} }[/tex]
=> [tex]L = \frac{T^2 g}{4 \pi ^2}[/tex]
The period can be mathematically evaluated as
[tex]T_p = \frac{t}{N}[/tex]
substituting values
[tex]T_p = \frac{315}{200}[/tex]
[tex]T_p = 1.575 \ s[/tex]
Therefore
[tex]L = \frac{1.575^2 * g }{4 \pi ^2}[/tex]
[tex]L = 0.0628 ^2 g[/tex]
so
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
substituting for L
[tex]g = 3.326 ((0.0628) g)^2[/tex]
=> [tex]g = \frac{1}{(3.326)* (0.0628)^2}[/tex]
[tex]g =76.2 m/s^2[/tex]
A microstate is a state of a physical system described at the finest level of detail. A macrostate is a state of a physical system that is described in terms of the systems overall or average properties at a macroscopic level. A macrostate will generally consist of many different microstates. In defining a macrostate we ignore what is going on at the microscopic (atomic/molecular) level.
Suppose 24 students are enrolled in a class. For simplicity let’s assume that any student who is present is sitting completely still at his or her assigned seat, facing forward (so we don’t have to worry about students being in different locations, having different motions, etc.).
1. Which of the following would constitute a macrostate description of the class attendance and which would constitute a microstate description.
a. A list of the names of each student present today.
b. The number of students in attendance.
Answers
Answer:
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
Explanation:
You can fins the answer to this question by comparing the situation of the problem with a system of molecules with discrete energy.
Without importance of which molecules have a specific energy, but rather, what is the total amount of energy, you can get for different configurations of energy the same amount of the total energy. If different configurations of the energies of the molecules give you the same total energy of the system, you say that the macrostate is the same. In the case of the classroom, it does not matter how are distributed the students in the class, the total number of students is always the same. The macrostate is the same for what ever organization of the students in the class.
If you would interested in the energy of each molecules, you will obtain different configurations. In the case of the classroom. The names of the student will define a microstate because in this case there are many configurations.
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
In a certain part of the North America Nebula, the amount of interstellar extinction in the visual wavelength band is 1.1 magnitudes. The thickness of the nebula is estimated to be 20 pc and it is located 700 pc from Earth. Suppose that a B spectral class main-sequence star is observed in the direction of the nebula and that the absolute visual magnitude of the star is known to be M(V) = -1.1 from spectroscopic data. Neglect any other sources of extinction between the observer and the nebula. Show all your work, assumptions, equations, and units.1. Find the apparent visual magnitude of the star if it is lying just in front of the nebula.2. Find the apparent visual magnitude of the star if it is lying just behind the nebula.
Answers
Answer:
Explanation:
The apparent magnitude of a star is related to the distance modulus as follows
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5+A_{\lambda}[/tex]
[tex]m_{\lambda}= \text {absolute visual magnitude}[/tex]
d = distance in parsec
[tex]A_{\lambda}=\text {interstellar extinction}[/tex]
Substitute
absolute visual magnitude = -1.1
distance =700pc
interstellar extinction = 0
to determine the apparent visual magnitude of the star lying in front of nebula
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5+A_{\lambda}[/tex]
[tex]=-1.1+5\log_{10}(700)-5+0\\\\=8.12[/tex]
Thus, the apparent visual magnitude of the star lying in front of nebula is 8.12
b) Substitute
absolute visual magnitude = -1.1
distance =700pc
interstellar extinction = 1.1
to determine the apparent visual magnitude of the star lying behind nebula
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5+A_{\lambda}[/tex]
[tex]=-1.1+5\log_{10}(700)-5+1-1\\\\=9.22[/tex]
the apparent visual magnitude of the star lying behind nebula is 9.22
c)
without taking extinction i.e 0, the distance of the star lying just behind nebula is calculated as follows
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5[/tex]
[tex]d=10^{(m_\lambda-M_{\lambda_5)/5}[/tex]
[tex]d=10^{(9.22+1.1+5)/5}\\\\=158.79pc[/tex]
Thus, without taking extinction , the distance of the star lying just behind nebula is 158.79pc
Compare the distance of nebula measured from earth with consideration of extinction to the distance of nebula without consideration of extinction
[tex]\frac{d_e}{d} =\frac{700pc}{1158.8pc}[/tex]
= 60.4%
thus, the percentage error in determining the distance if the interstellar extinction neglected is 60.4%
The 0.100 kg sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m from the center of the 5.00 kg mass. Assume that the only forces on the 0.100 kg sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg and 10.0 kg spheres are held in place at their initial positions.
Answers
Answer:
the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]3.3337*10^{-5} \ m/s[/tex]
Explanation:
Using the expression of the Change in Gravitational Potential Energy:
[tex]U= -(\frac{ Gm_1m_2 }{r_2} - \frac{ Gm_1m_2 }{r_1}) \\ \\ U=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :
[tex]U_1=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
[tex]U_1=6.67*10^{-11}*0.1 \ kg*10 \ kg(\frac{ 1 }{0.6 \ m} - \frac{ 1 }{0.8 \ m})[/tex]
[tex]U_1 = 2.778*10^{-11} J[/tex]
the change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;
[tex]U_2=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
[tex]U_2=6.67*10^{-11}*0.1 \ kg*5 \ kg(\frac{ 1 }{0.4 \ m} - \frac{ 1 }{0.2 \ m})[/tex]
[tex]U_2 = -8.335*10^{-11} J[/tex]
The net total change is:
[tex]U_{total } = U_1 +U_2[/tex]
[tex]U_{total} = 2.778*10^{-11} + (-8.335*10^{-11})[/tex]
[tex]U_{total} = -5.557*10^{-11}[/tex]
We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .
SO;
K.E = [tex]5.557*10^{-11}[/tex]
[tex]\frac{1}{2}mv^2 = 5.557*10^{-11}[/tex]
[tex]v^2 = \frac{2*5.557*10^{-11} \ J}{m_1}[/tex]
[tex]v=\sqrt{ \frac{2*5.557*10^{-11} \ J}{0. 1 \kg}[/tex]
v = [tex]3.3337*10^{-5} \ m/s[/tex]
Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]3.3337*10^{-5} \ m/s[/tex]
The speed of 0.10 sphere when it is moved to 0.20 kg to the left is :3.3337 *10^-5 m/s
What is speed ?The speed of any object is defined as the movement of any object with respect to the time.
By using the expression of the Change in Gravitational Potential Energy:
[tex]U=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex][tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :
[tex]U_1=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
[tex]U_1=- 6.67\times 10^{-11}\times 0.1\times 10(\dfrac{1}{0.6}-\dfrac{1}{0.8})[/tex]
[tex]U_1=2.778\times 10^{-11][/tex]
The change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;
[tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
[tex]U_2=- 6.67\times 10^{-11}\times 0.1\times 5(\dfrac{1}{0.4}-\dfrac{1}{0.2})[/tex]
[tex]U_2=-8.335\times 10^{-11}\ J[/tex]
The net total change is:
[tex]U_{total}=U_1+U_2[/tex]
[tex]U_{total}=2.778\times 10^{-11}+(-8.335\times 10^{-11})[/tex]
[tex]U_{tot\leq al}=-5.557\times 10^{-11}[/tex]
We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .
SO;
[tex]\rm KE=5.557\times 10^{-11}[/tex]
[tex]\dfrac{1}{2}mv^2=5.557\times 10^{-11}[/tex]
[tex]v^2=\dfrac{2\times 5.557\times 10^{-11}}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times 5.557\times 10^{-11}}{m}}[/tex]
[tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]
Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]
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A 6,000-km undersea glass fiber phone line crosses the Atlantic ocean connecting the US and France. (a) How long does it take for a message to transverse this link? (b) How long does it take for a message to travel from US to France by using a satellite link? The satellite is stationed about 22,000 miles above the earth between the US and France. (c) Will two people having a conversation across these two different links notice the travel delays?
Answers
Answer:
Explanation:
Message will travel through the fibre at the speed of 3 x 10⁸ m /s which is also the speed of light .
a ) Time for a message to travel from US to France by using glass fibre link
time = distance / speed
= 6000 x 1000 / 3 x 10⁸
= .02 s .
b )
22000 miles = 1.6 x 22000
= 35200 km
Time for a message to travel from US to France by using a satellite link
time = distance / velocity
distance = 2 x √ ( 35200² + 3000² )
= 2 x 35327 km
= .7 x 10⁸ m .
time = .7 x 10⁸ / 3 x 10⁸
= .233 s
c )
Difference of time = .233 - .02
= .213 which is more than 1/6 s for which effect of hearing stays in the brain , which is called persistence of vision . So they will be heard as different sounds.
Puck A, of inertia mm, is attached to one end of a string of length ℓℓ , and the other end of the string is attached to a pivot so that the puck is free to revolve on a smooth horizontal surface. Puck B, of inertia 20m, is attached to one end of a string of length ℓ/4, and the other end of the string is attached to a second pivot so that BB is also free to revolve. In each case, the puck is held as far as possible from the pivot so that the string is taut and then given an initial velocity v perpendicular to the string.
How does the magnitude of the angular momentum of puck A about its pivot compare with that of puck B about its pivot?
Answers
Answer:
[tex]L_{B}[/tex] / [tex]L_{A}[/tex] = [tex]\frac{5mvl}{mvl}[/tex]= 5
Explanation:
Find the given attachment
The definitions of angular momentum allow to find the result for the relationship between the two angular moments is:
[tex]\frac{L_B}{L_A} = 5[/tex]
The angular momentum is the vector product of momentum and vector position.
L = r x p
Where the bold letters indicate vectors, r is the position vector and p the moment. In the body that is in a rotation can be written as a function of the moment of inertia and the angular velocity.
L = I w
Linear and angular variables are related.
v = w L
We substitute
L = I v L
The moment of inertia of a disk with respect to its center is:
I = ½ m r²
Let's write the angular momentum for each disk.
Disc A
[tex]L_A[/tex] = ½ [tex]m_A r^2 v_A L_A[/tex]
Disc B
[tex]L_B[/tex] = ½ [tex]m_B r^2 v_B L_B[/tex]
They indicate that the mass of disk A is m and that of disk b is 20m, the length of the string is l for disk A and l / 4 for disk B.
We substitute
[tex]L_A[/tex] = ½ m r² v l
[tex]L_B[/tex] = ½ (20m) r² v ( [tex]\frac{l}{4}[/tex] )
The relationship between the angular moments is
[tex]\frac{L_B}{L_A} = \frac{20m \frac{l}{4} }{m l}\\ \frac{L_B}{L_A} = 5[/tex]
Consequently using the definitions of angular momentum we can find the result for the relationship between the two angular moments is:
[tex]\frac{L_B}{L_A} = 5[/tex]
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floating airboat conditions
Answers
Answer:
They are used in icy conditions.
Explanation:
They can also go over grassy plains and mucky swamps.
A wire of length L is used in an electric heater. When the potential
difference across the wire is 200 V, the power dissipated in the wire is
1000 W. The same potential difference is applied across a second similar wire of
length 2L. What is the power dissipated in the second wire?
Answers
Answer:500 W
Explanation:
Given
When length is L power dissipated is [tex]100\ W[/tex]
Potential difference applied is [tex]200\ V[/tex]
Resistance of wire is directly proportional to length of wire
Suppose initially Resistance is R
When L changes to [tex]2L[/tex]
Resistance changes to [tex]2R[/tex]
Initial Power dissipated is [tex]P_o=\frac{V^2}{R}[/tex]
For 2 R resistance and same voltage, Power dissipated is
[tex]P_1=\frac{V^2}{(2R)}[/tex]
So [tex]P_1=\frac{1}{2}\times \frac{V^2}{R}[/tex]
[tex]P_1=\frac{P_o}{2}[/tex]
So [tex]P_1=\frac{1000}{2}[/tex]
[tex]P_1=500\ W[/tex]
Two tiny conducting spheres are identical and carry charges of - 20.0 µC and +50.0 µC. They are separated by a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive of repulsive.
Answers
Answer:
a
The force experience by the two spheres is [tex]F_1 = 1.44*10^4 N[/tex]
This force is attractive cause the charge are unlike charges
b
The force experienced by the two spheres is [tex]F_2 = 3.24*10^{3} \ N[/tex]
The force is repulsive because the two charges are like charges
Explanation:
From the question we are told that
The charge on the first sphere is [tex]q_1 = -20.0 \mu C = 20 *10^{-6} C[/tex]
The charge on the second sphere is [tex]q_2 = 50 \mu C = 50*10^{-6} C[/tex]
The distance of separation is [tex]d = 2.50 \ cm = \frac{2.50}{100} = 0.025 \ m[/tex]
The electrostatic force experienced by the two spheres is mathematically represented as
[tex]F_1 = \frac{k q_1 q_2}{d^2}[/tex]
Now k is the coulomb’s constant with a value of [tex]k = 9*10^9 N \cdot m^2 /C^2[/tex]
So
[tex]F_1 = \frac{9*10^9 * 20 *10^{-6} * 50*10^{-6}}{0.025}[/tex]
[tex]F_1 = 1.44*10^4 N[/tex]
When the sphere are brought together the charge on each sphere would be the average of the total charge and this can be mathematically evaluated as
[tex]q = \frac{q_1 + q_2 }{2}[/tex]
[tex]q = \frac{(-20 + 50)*10^{-6} }{2}[/tex]
[tex]q = 15 \mu C[/tex]
So when they are seperated the electrostatic force experienced is
[tex]F_2 = \frac{kq^2}{d^2}[/tex]
[tex]F_2 = \frac{ 9*10^9 * (15 *10^{-6})}{0.025}[/tex]
[tex]F_2 = 3.24*10^{3} \ N[/tex]
You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 150 kPa , and the depth of the water will be 13.6 m . The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 kPa .
A) Find the net downward force on the tank's flat bottom, of area 2.15 m^2 , exerted by the water and air inside the tank and the air outside the tank.
Answers
Answer:
630.93 kN of force.
Explanation:
Pressure inside the tank is 150 kPa
The acceleration due to gravity on Mars g is 3.71 m/s^2.
The depth of water h is 13.6 m.
Pressure due to air outside tank is 93 kPa
The density of water p is 1000 kg/m^3
Pressure of the water on the tank bottom will be equal to pgh
Pressure of water = pgh
= 1000 x 3.71 x 13.6 = 50456 Pa
= 50.456 kPa.
Total pressure at the bottom of the tank will be pressure within tank and pressure due to water and pressure outside tank.
Pt = (150 + 50.456 + 93) = 293.456 kPa
Force at the bottom of the tank will be pressure times area of tank bottom.
F = Pt x A
F = 293.456 x 2.15 m^2 = 630.93 kN
Identify three pollutants released into the air when fossil fuels are burned.
Answers
Answer: Carbon dioxide (CO2), Carbon monoxide (CO), Methane (CH4)
A framed picture hangs from two cords attached to the ceiling.
A picture of a picture frame hanging by two cables at the center of the frame at the same length and angle from the vertical.
Which shows the correct free body diagram of the hanging picture?
A free body diagram with two force vectors, the first pointing downward labeled F Subscript g Baseline, the second pointing upward labeled F Subscript N Baseline.
A free body diagram with three force vectors, the first pointing south labeled F Subscript p Baseline, the second pointing northeast labeled F Subscript T Baseline, and the third pointing northwest labeled F Subscript N.
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T.
Answers
Answer:
Answer is C
Explanation:
I just did it E2020
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T. Thus option C is correct.
What is free body diagram?Free body diagram is defined as a drawing of an interesting object with all of its surroundings removed and the forces at work on the subject's body clearly depicted.
It is also defined as a diagram that reduces the loads and moments acting on a component or system of components.
Free body diagrams are streamlined depictions of an object and the force vectors operating on it in a problem. Since the diagram will depict this body without its surroundings, it is "free" of its environment.
Thus, a free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T. Thus option C is correct.
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How will you determine the direction
of a torque? Explain.
Answers
When two forces on the same object or equal and opposite these forces are called
Answers
Answer:
These two forces are called action and reaction forces
Explanation:
Why does the gravitational force obey an inverse square law?
Answers
Explanation:
the gravity force obeys the inverse square law because according to the newtons law of gravitaion the force of the gravitation is directly proportional to the product of masses ... In this way it depends upon the masses directly.We donot notice any gravitational force because we have very small masses as compared to the plantery universe and the solar system ... due to there greater masses they exert a large force and thus by increasing the distance the force of attraction varies inversely to the force of gravitation...thus gravitional force is a variable quantity ..............varies inversely to the square of distance between the to bodies...
F ∝1/r2
A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
Answers
Answer:
Explanation:
For a person suffering from nearsightedness, far point is 3.5 m . That means , an object beyond 3.5 m will not be clearly visible by the person . Ray of light coming from 3.5 m will be focussed on retina . For an object to be visible from infinity , ray of light coming from infinity should appear to be coming from 3.5 m . In other words , the object placed at infinity should form a virtual image at 3.5 m . This can be done by concave lens.
u = ∝ ; v = 3.5
Using lens formula
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{-3.5} -0 = \frac{1}{f}[/tex]
f = - 3.5 m .
Which factor indicates the amount of charge on the source charge?
A. the number of field lines on the test charge
B. the number of field lines on the source charge
C. the direction of lines on the source charge
D. the direction of lines on the test charge
Answers
Answer:
B. the number of field lines on the source charge
Explanation:
As we know that electric flux is defined as the number of electric field lines passing through a given area.
So here electric flux due to a point charge "q" is given by
so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.
The factor indicates the amount of charge on the source charge is the number of field lines on the source charge.
What is electric flux?The electric flux is defined as the number of electric field lines passing through a given area.
The electric flux due to a point charge q is given by the number of filed lines through particular closed area.
We know that flux depends on the magnitude of charge and number of field lines starting from a point charge will depends on the magnitude of the charge.
Thus, the correct option is B.
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Consider two cars, a 700kg Porsche and a 600kg Honda Civic. The Porsche is speeding along at 40 m/s (mph) and the Civic is going half the speed at 20 m/s. If the two cars brake to a stop with the same constant acceleration, lets look at whether the amount of time required to come to a stop or the distance traveled prior to stopping is influenced by their initial velocity.
1. A car traveling 5m/s slams on its brakes, creating an acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
2. The same car traveling for 10m/s applies the same acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
Answers
Answer:
Explanation:
To find the distance covered by the car after it applied brakes, we use 3rd equation of motion.
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
1.
We have:
Vi = Initial Velocity = 5 m/s
Vf = Final Velocity = 0 m/s (Since, car finally stops)
a = deceleration = - 2 m/s²
s = distance covered by the car = ?
Therefore,
s = [(0 m/s)² - (5 m/s)²]/2(- 2 m/s²)
s = 6.25 m
2.
We have:
Vi = Initial Velocity = 10 m/s
Vf = Final Velocity = 0 m/s (Since, car finally stops)
a = deceleration = - 2 m/s²
s = distance covered by the car = ?
Therefore,
s = [(0 m/s)² - (10 m/s)²]/2(- 2 m/s²)
s = 25 m
Hence, the distance traveled by the car is affected by the initial speed in accordance with a direct relationship.
If you have two spoons of the same
size, one silver and one stainless
steel, there is a quick test to tell
which is which. Hold the end of a
spoon in each hand, then lower them
both into a cup of very hot water.
One spoon will feel hot first. Is that
the silver spoon or the stainless steel
spoon? Explain.
Answers
Answer: Silver
Explanation:Silver has ahigh thermal conductivity of 429 watts/meter-k. Stainless steel on the other hand, has low thermal conductivity of 160 watts/meter-k.
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____is the distance traveled during a specific unit of time
Answers
Derive the expression of the mechanical energy for under damped system.
Answers
Answer: the expression of the mechanical energy for under damped system is;
x(t)=Ae−γ/2tcos(ωdt+ϕ), where ωd=ω02−γ2/4
γ = damping rate, and
ω0 = the angular frequency of the oscillator without damping.
Explanation:
The physical situation in mechanical energy defined through out the world has three possible results depending on the value of a (which is a constant value), which depends on the value of what is under our radical. This expression can either be positive, negative, or equal to zero which will result in overdamping, underdamping, and critical damping, as the case may be.
γ2 >4ω²0 This is the Over Damped case. Here, the system returns to equilibrium by exponentially decaying towards zero, and the system will not pass that equilibrium position more than once.
γ² < 4ω²0 this is the Under Damped case. Here, the system moves back and forth as it slowly returns to equilibrium and the amplitude of the system decreases over time.
Finally, γ² = 4ω²0
This is the Critically Damped case. Here, the system returns to equilibrium very fast without moving back and forth and without passing the equilibrium position at all.
Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This radiation is known as Cherenkov radiation. Cherenkov radiation is found in many interesting places such as particle detectors and nuclear reactors and can even be seen by astronauts when cosmic rays traverse their eyes. It should be stressed that the particle is never going faster than the speed of light in vacuum (or ccc), just faster than the speed of light in the material (which is always less than ccc). The creation of Cherenkov radiation occurs in much the same way that a sonic boom is created when a plane is moving faster than the speed of sound in the air. The various wavefronts that propagate in the material add coherently to create an effective shock wave. In this problem, you will become familiar with this type of radiation and learn how to use its properties to get information about the particles that created it. Part A What is the threshold velocity vthreshold(water)vthreshold(water)v_threshold (water) (i.e., the minimum velocity) for creating Cherenkov light from a charged particle as it travels through water (which has an index of refraction of n
Answers
Answer:
to create the particle the speed must be greater than 2.25 10⁸ m / s
Explanation:
In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium
n = c / v
where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio
in this case they give us that the medium matter water them that has a refractive index of
n = 1,333
we clear
v = c / n
let's calculate
v = 3 10⁸ / 1,333
v = 2.25 10⁸ m / s
to create the particle the speed must be greater than 2.25 10⁸ m / s
A typical machine tests the tensile strength of a sheet of material cut into a standard size of 5.00 centimeters wide by 10.0 centimeters long. The machine consists of one clamp that holds the entire width (5.00 centimeters) so that it hangs vertically. A second clamp is placed on the lower end of the object, to which a variable downward force is applied. The force is slowly increased until the object ruptures, and the breaking force is recorded.
A strip of aluminum foil with a thickness of 15.0 micrometers and matching the size recommendations of the machine is placed in the machine and tested. The force needed to rupture the foil is found to be 233 newtons. What is the tensile strength of the aluminum foil sample?
Answers
Answer:
Explanation:
tensile strength is stress that is needed to break the wire made of the material .
Here force required to break the sheet of material = 233 N
cross sectional area of the foil = breadth x thickness
= 5 x 10⁻² x 15 x 10⁻⁶ m²
= 75 x 10⁻⁸ m²
breaking stress = force / cross sectional area
= 233 / 75 x 10⁻⁸
= 3.1 x 10⁸ Pa .
Tensile strength = 3.1 x 10⁸ Pa .
The transfer of energy from one organism to the next in an ecosystem begins with a producer. A producer is an organism that produces its own food. A consumer is an organism in a food chain that obtains energy from producers or other consumers; consumers may be herbivores or carnivores.
Which food chain correctly describes the flow of energy in an ecosystem?
A) grass → grasshopper → fish → human
B) grasshopper → fish → human
C) human → fish → grasshopper
D) human → grasshopper → fish → grass
Answers
Answer:
A.)
Explanation:
The answer is food chain A, as plants produce their own food using photosynthesis, therefore making them producers.
A grasshopper then eats this grass, making it a consumer as it is obtaining its energy from the producer. It is then eaten by a fish and a human. This chain accurately describe the flow of energy in an ecosystem, as it goes, producer, consumer, consumer, consumer
Hope this helps a little
Answer:
sorry
Explanation:
patulong
4
Select the correct answer.
Which of the following is evidence of the past existence of glaciers?
A.
rift valleys
B. fjords
C.
estuaries
D.
oceanic ridges
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Answer:
B. fjords
explanation :
Fjords were created by glaciers. In the Earth's last ice age, glaciers covered just about everything. Glaciers move very slowly over time, and can greatly alter the landscape once they have moved through an area. This process is called glaciation.The fjords are one of the glaciers that existed in the past. They are one of the many glacial relief forms that can give us a insight into the size and power of the glaciers.
Answer:
B. fjords
I hope this helps
Which planet is an inner planet, and which one is an
outer planet? Explain your answer. (this is a essay question and you have to type a answer)
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Answer:
Inner planets rotate slowly and so they take more time to complete a rotation. Whereas, outer planets rotate faster and so they take less time to complete a rotation.
Explanation:
The inner planets are the ones that are much closer to the sun and as such they have much smaller orbits around them and take much less time to circle around it. These planets are much smaller, which enables them rotate slowly, so they take more time to complete a rotation..
Whereas, the outer planets differ from the inner planets in that they are much further from the sun and so take much more time for them to be able to make a single circle around the su and also, because they are much bigger, it takes them much more time to spin around their own axis, thus Outer planets rotate faster, so they take less time to complete a rotation
Jake is a farmer who owns horses and cattle. How often should Jake deworm his animals to keep them healthy?
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Explanation:
After every 3 to 4 months
1. What is the potential energy of a 5.0-kg
object located 2.0 m above the ground?
A. 2.5 J
C. 98 J
B. 10 J
D. 196 J
Answers
Answer:
C. 98 J
Explanation:
The appropriate formula is ...
PE = mgh . . . . . m is mass; below, m is meters
PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2
PE = 98 J
You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-cm-long piece of the wire is left over from its installation. Using an analytical balance, you determine that the mass of the spare piece is 14.5 μg . You then hang a 0.400-kg mass from the lower end of the long, suspended wire. When a small-amplitude transverse wave pulse is sent up that wire, sensors at both ends measure that it takes the wave pulse 24.7 ms to travel the length of the wire.
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Answer:
Explanation:
Let L be the length of the wire.
velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s
mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m
m = 7.25 x 10⁻⁷ kg / m
Tension in the wire = Mg , M is mass hanged from lower end.
= .4 x 9.8
= 3.92 N
expression for velocity of wave in the wire
[tex]v = \sqrt{\frac{T}{m} }[/tex] , T is tension in the wire , m is mass per unit length of wire .
40.48 L = [tex]\sqrt{\frac{3.92}{7.25\times10^{-7}} }[/tex]
1638.63 L² = 3.92 / (7.25 x 10⁻⁷)
L² = 3.92 x 10⁷ / (7.25 x 1638.63 )
L² = 3299.64
L = 57.44 m /s
A lion and a pig participate in a race over a 2.20 km long course. The lion travels at a speed of 18.0 m/s and the pig can do 2.70 m/s. The lion runs for 1.760 km and then stops to tease the slow-moving pig, which eventually passes by. The lion waits for a while after the pig passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.
(a) How far (in m) is the pig from the finish line when the lion resumes the race? (b) For how long in time (in s) was the lion stationary?
Answers
The lion covers a distance of [tex]\left(18.0\frac{\rm m}{\rm s}\right)t[/tex] after [tex]t[/tex] seconds, so it reaches the 1.760 km mark at time
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=1760\,\mathrm m\implies t\approx97.8\,\mathrm s[/tex]
The pig travels a distance of [tex]\left(2.70\frac{\rm m}{\rm s}\right)t[/tex], so that it has moved
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(97.8\,\mathrm s)=264\,\mathrm m[/tex]
in the time it takes for the lion to move 1.760 km.
(a) The lion has 0.44 km left in the race, which would take it
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=440\,\mathrm m\implies t\approx24.4\,\mathrm s[/tex]
to finish.
In order for the lion and pig to cross the finish line at the same time, the lion needs to resume running once the pig has 24.4 s remaining to the finish line; this happens when it is
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(24.4\,\mathrm s)\approx\boxed{65.9\,\mathrm m}[/tex]
away from the end.
(b) The lion is stationary for as long as it takes the pig to cover the distance between [65.9 m away from the finish line] and [264 m from the starting line], or (2.20 km - 65.9 m) - 264 m = 1.87 km, which takes it
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)t=1870\,\mathrm m\implies t=\boxed{935\,\mathrm s}[/tex]
(a) The distance of the pig from the finish line is 428.3 m
(b) The lion was stationary for 4.32 s
The given parameters;
total distance covered by both animals, d = 2.20 km = 2,200 mspeed of the lion, [tex]v_l[/tex] = 18 m/sspeed of the pig, [tex]v_p[/tex] = 2.7 m/sinitial distance traveled by the lion, = 1.760 km = 1,760 mWhen the pig meets the lion, both have covered a total distance of 1,760 m
The remaining distance to be covered = 2,200 m - 1,760 m = 440 m
Let the time both animals finished the remaining distance = tApply relative velocity concept; as the pig is moving, the lion is closing the gap between them until the finish line.
[tex](V_l - V_p)t = 440\\\\(18- 2.7)t = 440\\15.3 t = 440\\\\t = \frac{440}{15.3} \\\\t = 28.76 \ s[/tex]
If the lion had maintained a constant motion without stopping, the time it would have finished the remaining race is calculated as;
[tex]t_l = \frac{440}{18} = 24.44 \ s[/tex]
(b) This show that the lion was stationary for (28.76 - 24.44 )s = 4.32 s
(a) The distance traveled by the pig during the 4.32 s that the lion was stationary is calculated as;
d = 4.32 x 2.7 = 11.7 m
Thus, the distance of the pig from the finish line is (440 - 11.7)m = 428.3 m.
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