Answer:
A) 100
B) 10
C) 0.01
Explanation:
Let
Fs = force on slave cylinder
Fm = force on master cylinder
Given that the designed to exert a force 100 times as large as the one put into it. That is
Fs = 100 Fm
Using Pascal's law
Fs/As = Fm/Am
100Fm/As = Fm/Am
100/As = 1/Am
Cross multiply
As= 100Am
As/Am = 100
b) What must be the ratio of their diameters
Using their areas ratio, area of a cylinder is πr^2
A = πr^2 = π(D/2)^2
π(Ds/2)^2 ÷ π(Dm/2)^2 = 100
(Ds/2)^2 ÷ (Dm/2)^2 = 100
(Ds/Dm)^2 = 100
Ds/Dm = 10
(c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction
Let us consider the workdone at the input and at the output.
Work done = force × distance
Fs × Hs = Fm × Hm
100Fm × Hs = Fm × Hm
100Hs = Hm
100Hs/Hm = 1
Hs/Hm = 1/100
Hs/Hm = 0.01
In the United States, the average person eats about 3000 food Calories per day. What is the average power of this energy intake? 14.5 watts 14.5 watts 145 watts 145 watts 1450 watts 1450 watts 14 , 500 watts
Answer:
Explanation:
3000 Calorie = 3000 x 1000 calorie
1 calorie = 4.2 Joule
3000 x 1000 calorie = 4.2 x 3000 x 1000 J
= 12.6 x 10⁶ J
It is taken in one day
time period of one day = 24 x 60 x 60 second
= 86400 s
energy given by this intake in 86400s is 12.6 x 10⁶ J
power = energy / time
= 12.6 x 10⁶ / 86400 J/s
= 145.8 J/s or W
= 145 W.
Final answer:
The average power of energy intake for a person consuming 1,300 to 3,000 kcals per day translates to approximately 63 to 145 watts. This calculation is done by converting calories to joules and then dividing by the total number of seconds in a day.
Explanation:
In order to calculate the average power of the energy intake from food, we can convert the amount of energy consumed per day into watts. For a range of 1,300 to 3,000 kcal per day, we need to convert these values to joules, since 1 kcal = 4.184 kJ. Therefore, 1,300 kcal converts to 1,300 x 4.184 kJ = 5,439.2 kJ and 3,000 kcal to 3,000 x 4.184 kJ = 12,552 kJ.
Next, we divide each amount by the number of seconds in one day to find the average power in watts. There are 86,400 seconds in a day. Thus, the power for 1,300 kcal/day is 5,439.2 kJ / 86,400 s = approximately 63 watts, and for 3,000 kcal/day is 12,552 kJ / 86,400 s = approximately 145 watts.
The power range for 1,300 to 3,000 kcal/day is thus approximately 63 watts to 145 watts.
Ms. Baudino wants to challenge her AP Physics students to use high order thinking and problem solving skills while engaged in a meaningful, real-world learning activity. What software can provide her students with this type of learning environmentA. Presentation
B. Word processing
C. Gaming and simulation
D. Presentation
Answer:
C. Gaming and simulation
Explanation:
In a presentation, you are just going to read from something and resume that to present. So there is not much higher order thinking or problem solving.
Word processing you will just be writing a test, so not much to learn in these aspects.
However
In gaming, there are different strategies your team may assume(which means that the students have to use high order thinking to choose the best strategy), you have to work well as a group, depending on the game(which is a needed skill on the real-world), and other things.
So the correct answer is:
C. Gaming and simulation
Answer:
C. Gaming and Simulation
Explanation:
Problem solving and high order thinking refers to the ability to use knowledge, facts, and data to effectively solve problems. This means the ability to assess problems and find suitable solutions. It is the ability to develop a well thought out solution within a reasonable time frame.
Integrating games and simulation into the learning process has a positive effect in developing the cognitive and behavioral abilities of students. Simulations promote the use of critical and evaluative thinking. Because they are ambiguous or open-ended, they encourage students to contemplate the implications of a scenario. It allows the teacher to properly engage the students while assuming a real world situation.
Due to constant practice, it allows the student to grasp the concept of a subject matter. The students are placed in a world defined by the teacher and they have enough motivation to learn because they now catch fun with activities that involve critical thinking.
Other options stated such as presentation and word processing only helps the students to give reports of activities and researches carried out. Gaming and simulation are the only ones that develop the critical thinking and problem solving skills of the students.
In Problem 1, the rake angle was changed to a neutral rake angle. If the friction angle remains the same, determine: (a) the shear plane angle (b) the chip thickness (c) the shear strain for the operation.
Answer: seen in the explanations
Explanation:
Since no values for this problem is given, I will just give the formulas to solve the problem in mechanics of orthogonal plane cutting.
a) shear plane angle will be calculated using:
tan ϕ = r Cos α / 1- r sin α
Where :
r= chip thickness ratio
α = rake angle
ϕ = shear angle.
B) chip thickness
To get the chip thickness, we must calculate the chip thickness ratio using the formula:
Rt = Lc/ L
Where:
Lc = length of chip formed
L= uncut chip length.
Use the answer to solve for chip thickness with the formula;
Tc = t/ Rt
Where :
t = depth of cut
Rt = chip thickness ratio
C) shear strain for the operation will be solved using;
ε = cot β• + tan β•
Where:
ε = shear strain
β• = orthogonal shear angle.
A sphere moves in simple harmonic motion with a frequency of 4.80 Hz and an amplitude of 3.40 cm. (a) Through what total distance (in cm) does the sphere move during one cycle of its motion? cm (b) What is its maximum speed (in cm/s)? cm/s Where does this maximum speed occur? as the sphere passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these (c) What is the maximum magnitude of acceleration (in m/s2) of the sphere? m/s2 Where in the motion does the maximum acceleration occur? as the sphere passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these
Answer:
a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b) The maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
The maximum acceleration occurs at maximum excursion from equilibrium.
Explanation:
Given that :
Frequency (f) = 4.80 Hz
Amplitude (A) = 3.40 cm
a)
The total distance traveled by the sphere during one cycle of simple harmonic motion is:
d = 4A (where A is the Amplitude)
d = 4(3.40 cm)
d = 13.60 cm
Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b)
As we all know that:
[tex]x = Asin \omega t[/tex]
Differentiating the above expression with respect to x ; we have :
[tex]\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)[/tex]
[tex]v = A \omega cos \omega t[/tex]
Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;
Then:
[tex]v_{max} = A \omega[/tex]
We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium
substituting [tex]2 \pi f[/tex] for [tex]\omega[/tex] in the above expression;
[tex]v_{max} = A(2 \pi f)[/tex]
[tex]v_{max} = 3.40 cm (2 \pi *4.80)[/tex]
[tex]v_{max} = 102.54 \ cm/s[/tex]
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
[tex]v = A \omega cos \omega t[/tex]
By differentiation with respect to t;
[tex]\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)[/tex]
[tex]a =- A \omega^2 sin \omega t[/tex]
The maximum acceleration of the sphere is;
[tex]a_{max} =A \omega^2[/tex]
where;
[tex]w = 2 \pi f[/tex]
[tex]a_{max} = A(2 \pi f)^2[/tex]
where A= 3.40 cm = 0.034 m
[tex]a_{max} = 0.034*(2 \pi *4.80)^2[/tex]
[tex]a_{max} = 30.93 \ m/s^2[/tex]
The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of [tex]x = \pm A[/tex]
The sphere moves a total distance of 6.80 cm in one cycle. Its maximum speed is 102.5 cm/s as it passes through equilibrium and the maximum acceleration is at maximum excursion from equilibrium, with a magnitude of 18.5 m/s².
Explanation:The sphere's total distance moved in one cycle is twice the amplitude, as it moves to the maximum amplitude and back again. Therefore, in the case of this sphere, it moves through a total distance of 6.80 cm during one cycle. The maximum speed of an object in simple harmonic motion occurs as it passes through equilibrium. You can calculate this speed using the formula v = ωA, where ω is the angular frequency and A is the amplitude.
The angular frequency is ω = 2πf, where f is the frequency. Thus, the maximum speed of the sphere is approximately 102.5 cm/s. Lastly, the maximum magnitude of acceleration occurs at maximum excursion from equilibrium, and can be calculated with the formula derived from Newton's second law a = ω²A, thus the acceleration is 18.5 m/s².
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How could you increase the width (in centimeters) of the central maximum of the diffraction pattern on the screen? There may be more than one correct method, so select all those that would work. In each case, assume you only do the one named thing and make no other changes.
Answer: 1) use of light with longer wavelenght will increase the width of the central maxima
2) increasing the number of slit will increase the width of the central maxima
3) narrowing the slit will increase the central maxima.
Hola comastas muy bein
Answer:
v:
xddddddddddddddddd
Answer: hola
Explanation:спасибо за очки
It is proposed that future space stations create an artificial gravity by rotating. Suppose a space station is constructed as a 1000-m-diameter cylinder that rotates about its axis. The inside surface is the deck of the space station. What rotation period will provide "normal" gravity?
To solve this problem, apply the concepts related to the centripetal acceleration as the equivalent of gravity, and the kinematic equations of linear motion that will relate the speed, the distance traveled and the period of the body to meet the needs given in the problem. Centripetal acceleration is defined as,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Tangential Velocity
r = Radius
If we rearrange the equation to get the velocity we have,
[tex]v = \sqrt{a_c r}[/tex]
But at this case the centripetal acceleration must be equal to the gravitational at the Earth, then
[tex]v = \sqrt{gr}[/tex]
[tex]v = \sqrt{(9.8)(\frac{1000}{2})}[/tex]
[tex]v = 70m/s[/tex]
The perimeter of the cylinder would be given by,
[tex]\phi = 2\pi r[/tex]
[tex]\phi = 2\pi (500m)[/tex]
[tex]\phi = 3141.6m[/tex]
Therefore now related by kinematic equations of linear motion the speed with the distance traveled and the time we will have to
[tex]v = \frac{d}{t} \rightarrow \text{ But here } d = \phi[/tex]
[tex]v = \frac{\phi}{t} \rightarrow t = \frac{\phi}{t}[/tex]
[tex]t = \frac{3141.6}{70m/s}[/tex]
[tex]t = 44.9s[/tex]
Therefore the period will be 44.9s
Answer:
[tex]T \approx 44.88\,s[/tex]
Explanation:
"Normal" gravity is equal to 9.807 meters per squared second and cylinder must rotate at constant speed in order to simplify the equation of acceleration, which is in the radial direction. The centrifugal acceleration experimented by people allow them to be on the inside surface.
[tex]g = \omega^{2}\cdot R[/tex]
The angular speed required to provide "normal" gravity is:
[tex]\omega = \sqrt{\frac{g}{R} }[/tex]
[tex]\omega = \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{500\,m} }[/tex]
[tex]\omega \approx 0.14\,\frac{rad}{s}[/tex]
The rotation period is:
[tex]T = \frac{2\pi}{\omega}[/tex]
[tex]T = \frac{2\pi}{0.14\,\frac{rad}{s} }[/tex]
[tex]T \approx 44.88\,s[/tex]
A bowling ball (which we can regard as a uniform sphere) has a mass of 3.63 kg and a diameter of 0.216 m. A baseball has a mass of 0.145 kg. If you connect these two balls with a lightweight rod, what must be the distance between the center of the bowling ball and the center of the baseball so that the system of the two balls and the rod will balance at the point where the rod touches the surface of the bowling ball?
Answer:
Distance will be 2.81 m
Explanation:
Detailed explanation and calculation is shown in the image below.
Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.
To move a 32 kg crate with a coefficient of static friction of 0.57 with the floor, the force necessary to start the crate moving is 180 N after rounding to two significant figures.
Explanation:To find the force necessary to start the crate moving, we need to use the concept of static friction. The force of static friction is given by the equation fs=μsN, where fs is the static friction, μs is the coefficient of static friction and N is the normal force. The normal force is equal to the product of the mass and gravitational acceleration (N = mg), where m is the mass and g is the acceleration due to gravity. Given that the mass of the crate is 32 kg and the coefficient of static friction is 0.57, we first calculate N = (32kg)(9.8m/s²) = 313.6 N.
Then, by substituting these values into the equation, we have fs = (0.57)(313.6 N) =178.85 N. Therefore, the force necessary to start the crate moving is 178.85 N, but we express the answer using two significant figures, so it is 180 N.
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A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Mercury's orbital period is 0.24 years, and Venus's is 0.62 years.
The average orbital radius of this planet is:
a) less than Mercury's.
b) between Mercury's and Venus's.
c) greater than Venus's.
Answer:
a) less than Mercury's.
Explanation:
For orbital time period of a planet , the expression is
T² = 4π² R³ / GM
T is time period , R is radius of orbit , G is universal gravitational constant , M is mass of the star or sun
T² ∝ R³
As radius of orbit increases , time period increases . The given planet is making around a star whose mass is equal that of sun so M is same as sun .
The given planet has time period equal to .01 years which is less than that of Mercury and Venus , hence its R will be less than orbit of both of them or less than mercury's .
If a 200 turn, 10-3 m2 cross-sectional area coil is immersed in a magnetic field such that the plane of the coil is perpendicular to the field. If the magnetic field increases by 6 T/s, then how much voltage is induced
Answer:
1.2 volt induced in coil.Explanation:
Given:
Number of turns [tex]N = 200[/tex]
Cross sectional area [tex]A = 10^{-3}[/tex] [tex]m^{2}[/tex]
Rate of increasing magnetic field [tex]\frac{dB}{dt} = 6[/tex] [tex]\frac{T}{s}[/tex]
From the faraday's law,
Induced emf is given by,
[tex]\epsilon = -N\frac{d \phi}{dt }[/tex]
Where [tex]\phi =[/tex] magnetic flux
[tex]\phi = AdB \cos(0)[/tex] ( because angle between normal coil and field is zero)
Where [tex]A =[/tex] area of coil
Put the value of [tex]\phi[/tex] in above equation,
Here we neglect minus sign
[tex]\epsilon = NA\frac{dB}{dt}[/tex]
[tex]\epsilon = 200 \times 10^{-3} \times 6[/tex]
[tex]\epsilon = 1.2[/tex] V
Therefore, 1.2 volt induced in coil
Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a current of 4.36 A .Is the force per meter exerted on the 4.36-A wire greater than, less than, or the same as the force per meter exerted on the 2.79-A wire
Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula
[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
[tex]I_1[/tex] is the current in the 1st wire
[tex]I_2[/tex] is the current in the 2nd wire
r is the separation between the wires
In this problem
[tex]I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m[/tex]
Substituting, we find the force per unit length on the two wires:
[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N[/tex]
However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).
In the formula 4h²o how many total hydrogen atoms are there
A bubble of air has a volume of 16 cm3 when at a depth of 6.5 m. As the bubble rises, it expands. What is the bubble’s volume just below the surface of the water? (Note: Assume the temperature of the air in the bubble doesn’t change, in which case pressure times volume is constant.)
Final answer:
The volume of the bubble just below the surface of the water is 16 cm3.
Explanation:
To solve this problem, we can use Boyle's law, which states that the pressure times the volume of a gas is constant if the temperature remains constant. In this case, the volume of the air bubble at the bottom of the water is 16 cm3 and the depth is 6.5 m. As the bubble rises, the pressure decreases, causing the volume to increase. We can set up a proportion to solve for the volume just below the surface of the water.
Using the formula P1V1 = P2V2, where P1 is the pressure at the bottom of the water, V1 is the initial volume, P2 is the pressure just below the surface, and V2 is the volume just below the surface, we can plug in the values and solve for V2.
Since the temperature of the air in the bubble doesn't change, we can assume that the pressure times the volume is constant. Therefore, we have:
P1V1 = P2V2
Using the given values, we have:
(1 atm)(16 cm3) = (?, just below the surface) V2
To find V2, we can rearrange the equation:
V2 = (P1V1) / P2
Since the pressure just below the surface is atmospheric (1 atm), we can substitute that value in:
V2 = (1 atm)(16 cm3) / (1 atm)
V2 = 16 cm3
Therefore, the volume of the bubble just below the surface of the water is 16 cm3.
20 POINTS! TRUE OR FALSE:
If a stationary box does not move when you push against it is because of kinetic friction.
A) TRUE
B) FALSE
Answer:
False.
Explanation:
Kinetic energy means it must move
Answer: I believe that it is A) True
Disclaimer: I'm not quite sure. We learned about this recently though so possibly I'm right. Good luck though!
Explanation:
My Notes You push a box of mass 24 kg with your car up to an icy hill slope of irregular shape to a height 5.7 m. The box has a speed 12.1 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) with a horizontal velocity before immediately falling off a sheer cliff to the ground (with no drag). (a) What is the speed of the box at the top of the hill?
Answer:
The speed of the box at the top of the hill will be 5.693m/s.
Explanation:
The kinetic energy of the box at the bottom of the hill is
[tex]K.E = \dfrac{1}{2}mv^2[/tex]
putting in [tex]m =24kg[/tex] and [tex]v = 12.1m/s[/tex] we get
[tex]K.E = \dfrac{1}{2}(24kg)(12.1)^2\\\\K.E = 1756.92J[/tex]
Now, the potential energy this box gains as it rises [tex]h =5.7m[/tex] up the hill is
[tex]P.E = mgh[/tex]
[tex]P.E = (24kg)(10ms/s^2)(5.7m)\\\\P.E = 1368[/tex]
Therefore, the energy left [tex]E_{left}[/tex] in the box at the top if the hill will be
[tex]E_{left} =K.E - P.E = 1756.92J-1368J\\[/tex]
[tex]\boxed{E_{left} = 388.92J}[/tex]
This left-over energy must appear as the kinetic energy of the box at the top of the hill (where else could it go? ); therefore,
[tex]\dfrac{1}{2}mv_t^2= 388.92J[/tex]
putting in numbers and solving for [tex]v_t[/tex] we get:
[tex]\boxed{v_t = 5.693m/s.}[/tex]
Thus, the speed of the box at the top of the hill is 5.693m/s.
A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop is:
Answer:
To oppose applied magnetic field current will flow in anticlockwise direction
Explanation:
Given:
Uniform magnetic field directed vertically upward.
Current will flow in clockwise direction
Here loop of wire is suddenly contracts to half so diameter of loop is reduced.
Hence less number of magnetic field line pass through the loop. This change in magnetic field lines lead to flow of current.
Now from lenz law flow of induced current will oppose the cause of its production
Therefore, to oppose applied magnetic field current will flow in anticlockwise direction.
Two objects moving in opposite directions with the same speed v undergo a totally inelastic collision, and half the initial kinetic energy is lost. Find the ratio of their masses.
Answer:
[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]
Explanation:
We name M the larger mas, and m the smaller mass, so base on this we write the following conservation of momentum equation for the collision:
[tex]P_i=P_f\\M\,v_i -m\,v_i=(M+m)\,v_f\\(M-m)\,v_i = (M+m)\,v_f\\\\\frac{v_i}{v_f} = \frac{(M+m)}{(M-m)}[/tex]
We can write this in terms of what we are looking for (the quotient of masses [tex]\frac{m}{M}[/tex]:
[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}[/tex]
We use now the information about Kinetic Energy of the system being reduced in half after the collision:
[tex]K_i=2\,K_f\\\frac{1}{2} (M+m)\,v_i^2= 2*\frac{1}{2} (M+m)v_f^2\\v_i^2=2\,v_f^2\\\frac{v_i^2}{v_f^2} =2[/tex]
We can combine this last equation with the previous one to obtain:
[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}\\(\frac{v_i}{v_f})^2 = \frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}\\2=\frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}[/tex]
where solving for the quotient m/M gives:
[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]
A falcon is soaring over a prairie, flying at a height of 49.0 m with a speed of 13.5 m/s. The falcon spots a mouse running along the ground and dives to catch its dinner. Ignoring air resistance, and assuming the falcon is only subject to the gravitational force as it dives, how fast will the falcon be moving the instant it is 5.00 m above the ground
Answer:
32.3 m/s
Explanation:
We can solve this problem by applying one of Newton's equations of motion:
[tex]v^2 = u^2 + 2gs[/tex]
where v = final velocity of falcon
u = initial velocity of falcon
g = acceleration due to gravity ([tex]9.8 m/s^2[/tex])
s = distance moved by falcon
From the question, we have that:
u = 13.5 m/s
s = initial height - new height = 49.0 - 5.0 = 44.0 m
Hence, to find the velocity when it has traveled 44 m towards the ground (5 m above the ground):
[tex]v^2 = 13.5^2 + (2 * 9.8 * 44)\\\\\\v^2 = 182.25 + 862.4\\\\\\v^2 = 1044.65\\\\\\v = \sqrt{1044.65} \\\\\\v = 32.3 m/s[/tex]
The velocity of the falcon at the instant when it is 5.0 m above the ground is 32.3 m/s
Light of wavelength 614 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?
Answer:
The least and second least thicknesses of the film are 0.115 um and 0.346 um respectively.
Explanation:
Optical path length ===> 2n * t = (m + 0.5) * λ
λ = 614 nm , n = 1.33
Substitute in the parameters in the equation.
∴ 2(1.33) * t = (m + 0.5) * 614
2.66 * t = 614m + 307
t = (614m + 307) / 2.66 .............(1)
(a) for m = 0
t = (614m + 307) / 2.66
t = (614(0) + 307) / 2.66
t = 307 / 2.66
t = 115 nm == 0.115 um
(b) for m = 1
t = (614(1) + 307) / 2.66
t = (614 + 307) / 2.66
t = 921 / 2.66
t = 346.24 nm = 0.346 um
why is the motion of simple pendulum not strictly a simple harmonic motion
Final answer:
The motion of a simple pendulum is not strictly a simple harmonic motion. While it exhibits some characteristics of simple harmonic motion, such as periodic motion and a restoring force, the restoring force is not linearly proportional to the displacement.
Explanation:
The motion of a simple pendulum is not strictly a simple harmonic motion because it does not follow Hooke's law. Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke's law. However, a simple pendulum is governed by the law of conservation of energy and the restoring force is not directly proportional to the displacement.
While a simple pendulum exhibits some characteristics of simple harmonic motion, such as periodic motion and a restoring force, it deviates from true simple harmonic motion because the restoring force is not linearly proportional to the displacement. Instead, the restoring force for a simple pendulum is proportional to the sine of the displacement angle.
For example, when a pendulum is displaced to one side, it experiences a restoring force in the opposite direction. But the magnitude of the restoring force does not increase linearly with the displacement. Instead, it follows the relationship F = -mg sin(theta), where F is the restoring force, m is the mass of the pendulum bob, g is the acceleration due to gravity, and theta is the displacement angle.
A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the plane by an ideal spring that is aligned with the surface and attached to a wall above the brick. The spring has a spring constant (force constant) of 120 N/m. By how much does the spring stretch with the brick attached?
Answer: 0.137 m
Explanation:
Given
Mass of brick, m = 3 kg
Angle of inclination, Φ = 34°
Force constant of the spring, k = 120 N/m
The force of the brick, F can be gotten using the relation
F = mg
F = 3 * 9.8
F = 29.4 N
Now, the force parallel to the incline, F(p) can be gotten using the formula,
F(p) = F sinΦ, so that
F(p) = 29.4 * sin 34
F(p) = 29.4 * 0.559
F(p) = 16.4 N
The stretch distance then is,
d = F(p) / k * 1 m
d = 16.4 / 120
d = 0.137 m
Thus, the spring stretched by a distance of 0.137 m
Suppose you stand with one foot on ceramic flooring and one on a wool carpet, making contact over a 77.2 cm2 area with each foot. Both the ceramic and the carpet are 2.60 cm thick and are 10.0°C on their bottoms. At what rate must each foot supply heat to keep the top of the ceramic and carpet at 33.0°C? The thermal conductivity of ceramic is 0.84 J/(s · m · °C) and that of wool is
Answer:
[tex]P_{c[/tex]= 5.74W
[tex]P_{w}[/tex]=0.27 W
Explanation:
d= 2.6cm =>0.026m (for both the ceramic and the carpet )
Thermal conductivity of wool '[tex]k_{w}[/tex]'= [tex]k_{carpet}[/tex] = [tex]k_{wool}[/tex]= 0.04J/(sm °C)
Thermal conducticity of carpet '[tex]k_c}[/tex]' = 0.84 J/(sm °C)
Area 'A'= [tex]A_{carpet}[/tex]= [tex]A_{ceramic}[/tex]= 77.2cm²=> 77.2 x [tex]10^{-4}[/tex]m²
[tex]T_h}[/tex]=33.0°C
[tex]T_c}[/tex]=10.0°C
Average Power [tex]P_{avg}[/tex] is determined by dividing amount of energy'Q' by time taken for the transfer't':
[tex]P_{avg}[/tex] = Q/Δt
Due to conductivity, heat of flow rate will be P= dQ/dt
P=[tex]\frac{dQ}{dt}[/tex] = [tex]\frac{kA[T_{h}-T_{c} ]}{d}[/tex]
For CERAMIC:
[tex]P_{c[/tex]=[tex]\frac{k_{c} A[T_{h}-T_{c} ]}{d}[/tex] => [0.84 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026
[tex]P_{c[/tex]= 5.74W
For WOOL CARPET:
[tex]P_{w}[/tex]= [tex]\frac{k_{w} A[T_{h}-T_{c} ]}{d}[/tex]=> [0.04 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026
[tex]P_{w}[/tex]=0.27 W
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.25 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.540 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.51 m and t = 0.150 s .
Answer:
The transverse displacement is [tex]y(1.51 , 0.150) = 0.055 m[/tex]
Explanation:
From the question we are told that
The generally equation for the mechanical wave is
[tex]y(x,t) = Acos (kx -wt)[/tex]
The speed of the transverse wave is [tex]v = 8.25 \ m/s[/tex]
The amplitude of the transverse wave is [tex]A = 5.50 *10^{-2} m[/tex]
The wavelength of the transverse wave is [tex]\lambda = 0540 m[/tex]
At t= 0.150s , x = 1.51 m
The angular frequency of the wave is mathematically represented as
[tex]w = vk[/tex]
Substituting values
[tex]w = 8.25 * 11.64[/tex]
[tex]w = 96.03 \ rad/s[/tex]
The propagation constant k is mathematically represented as
[tex]k = \frac{2 \pi}{\lambda}[/tex]
Substituting values
[tex]k = \frac{2 * 3.142}{0. 540}[/tex]
[tex]k =11.64 m^{-1}[/tex]
Substituting values into the equation for mechanical waves
[tex]y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))[/tex]
[tex]y(1.51 , 0.150) = 0.055 m[/tex]
You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom floor. What work do you do on the child for each part of the trip and for the whole trip
To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations
[tex]\text{Mass of the child} = m = 25kg[/tex]
[tex]\text{Acceleration due to gravity} = g = 9.81m/s^2[/tex]
[tex]\text{Height lifted} = h = 0.80m (Upward)[/tex]
Work done to upward the object
[tex]W = mgh[/tex]
[tex]W = (25)(9.81)(0.8)[/tex]
[tex]W = 196.2J[/tex]
Horizontal Force applied while carrying 10m,
[tex]F = 0N[/tex]
[tex]W = 0J[/tex]
Height descended in setting the child down
[tex]h' = -0.8m (Downwoard)[/tex]
[tex]W = mgh'[/tex]
[tex]W = (25)(9.81)(-0.80)[/tex]
[tex]W = -196.2J[/tex]
For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.
To calculate the work done on the child during different parts of the trip and for the whole trip, we need to consider the weight of the child and the vertical distances involved. The work done in lifting the child up and setting the child back down is equal to the weight of the child multiplied by the vertical distance. The work done in carrying the child horizontally is zero.
Explanation:In this problem, we need to calculate the work done on the child during different parts of the trip and for the whole trip. Work, W, is defined as the product of force and displacement, W = Fd. Since the child is being lifted vertically, the work done in lifting the child up is equal to the weight of the child multiplied by the vertical distance lifted, Wup = mgh. The work done in carrying the child horizontally is zero since the displacement is perpendicular to the force. Finally, the work done in setting the child back down is also equal to the weight of the child multiplied by the vertical distance lowered, Wdown = mgh. Therefore, the total work done on the child for the whole trip is the sum of the work done in lifting the child up and setting the child back down, Wtotal = Wup + Wdown.
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Which one of the following substances can be separated into several different elements?
A: Air
B: Iron
C: hydrogen
D: nickel
The substance that can be separated into several different elements is air. The correct option is A.
Air, which is a mixture mainly composed of nitrogen and oxygen, among other gases, can thus be separated into its component elements. Substances like Iron (Fe), Hydrogen (H), and Nickel (Ni) are elements and cannot be broken down into simpler substances by ordinary chemical means.
he uniform 110-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position withθ= 0. Determine the initial angular accelerationα of the beam and the magnitudeFAof the force supported by the pin at A due to the application of the force P = 350 N on the attached cab
Answer:
The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]
The magnitude of the force at A is 832.56N
Explanation:
Here, m is the mass of the beam and l is the length of the beam.
[tex]I =\frac{1}{3} ml[/tex]
[tex]I=\frac{1}{3} \times110\times4^2\\I=586.67kgm^2[/tex]
Take the moment about point A by applying moment equilibrium equation.
[tex]\sum M_A =I \alpha[/tex]
[tex]P \sin45^0 \times 3 =I \alpha[/tex]
Here, P is the force applied to the attached cable and [tex]\alpha[/tex] is the angular acceleration.
Substitute 350 for P and 586.67kg.m² for I
[tex]350 \sin 45^0 \times3=568.67 \alpha[/tex]
[tex]\alpha =1.3056rad/s^2[/tex]
The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]
Find the acceleration along x direction
[tex]a_x = r \alpha[/tex]
Here, r is the distance from center of mass of the beam to the pin joint A.
Substitute 2 m for r and 1.3056rad/s² for [tex]\alpha[/tex]
[tex]a_x = 2\times 1.3056 = 2.6112m/s^2[/tex]
Find the acceleration along the y direction.
[tex]a_y = r \omega ^2[/tex]
Here, ω is angular velocity.
Since beam is initially at rest,ω=0
Substitute 0 for ω
[tex]a_y = 0[/tex]
Apply force equilibrium equation along the horizontal direction.
[tex]\sum F_x =ma_x\\A_H+P \sin45^0=ma_x[/tex]
[tex]A_H + 350 \sin45^0=110\times2.6112\\\\A_H=39.75N[/tex]
Apply force equilibrium equation along the vertical direction.
[tex]\sum F_y =ma_y\\A_V-P \cos45^0-mg=ma_y[/tex]
[tex]A_v +350 \cos45^0-110\times9.81=0\\A_V = 831.61\\[/tex]
Calculate the resultant force,
[tex]F_A=\sqrt{A_H^2+A_V^2} \\\\F_A=\sqrt{39.75^2+691.61^2} \\\\= 832.56N[/tex]
The magnitude of the force at A is 832.56N
Answer:
a) Initial angular acceleration of the beam = 1.27 rad/s²
b) [tex]F_{A} = 851.11 N[/tex]
Explanation:
[tex]tan \theta = \frac{opposite}{Hypothenuse} \\tan \theta = \frac{3}{3} = 1\\\theta = tan^{-1} 1 = 45^{0}[/tex]
Force applied to the attached cable, P = 350 N
Mass of the beam, m = 110-kg
Mass moment of the inertia of the beam about point A = [tex]I_{A}[/tex]
Using the parallel axis theorem
[tex]I_{A} = I_{G} + m(\frac{l}{2} )^{2} \\I_{G} = \frac{ml^{2} }{12} \\I_{A} = \frac{ml^{2} }{12} + \frac{ml^{2} }{4} \\I_{A} = \frac{ml^{2} }{3}[/tex]
Moment = Force * Perpendicular distance
[tex]\sum m_{A} = I_{A} \alpha\\[/tex]
From the free body diagram drawn
[tex]\sum m_{A} = 3 Psin \theta\\ 3 Psin \theta = \frac{ml^{2} \alpha }{3}[/tex]
P = 350 N, l = 3+ 1 = 4 m, θ = 45°
Substitute these values into the equation above
[tex]3 * 350 * sin 45 = \frac{110 * 4^{2}* \alpha }{3} \\\alpha = 1.27 rad/s^{2}[/tex]
Linear acceleration along the x direction is given by the formula
[tex]a_{x} = r \alpha[/tex]
r = 2 m
[tex]a_{x} = 2 * 1.27\\a_{x} = 2.54 m/s^{2}[/tex]
the linear acceleration along the y-direction is given by the formula
[tex]a_{y} = r w^{2}[/tex]
Since the beam is initially at rest, w = 0
[tex]a_{y} = 0 m/s^{2}[/tex]
General equation of motion along x - direction
[tex]F_{x} + Psin \theta = ma_{x}[/tex]
[tex]F_{x} + 350sin45 = 110 * 2.54\\F_{x} = 31.913 N[/tex]
General equation of motion along y - direction
[tex]F_{x} + Pcos \theta - mg = ma_{y}[/tex]
[tex]F_{y} + 350cos45 - 110*9.8 = m* 0\\F_{y} = 830.513 N[/tex]
Magnitude [tex]F_{A}[/tex] of the force supported by the pin at A
[tex]F_{A} = \sqrt{F_{x} ^{2} + F_{y} ^{2} } \\F_{A} = \sqrt{31.913 ^{2} + 850.513 ^{2} } \\F_{A} = 851.11 N[/tex]
A disk 8.04 cm in radius rotates at a constant rate of 1 220 rev/min about its central axis. (a) Determine its angular speed. rad/s (b) Determine the tangential speed at a point 3.02 cm from its center. m/s (c) Determine the radial acceleration of a point on the rim. magnitude km/s2 direction (d) Determine the total distance a point on the rim moves in 2.02 s. m
Answer:
A) 128 rad/s
B) 3.87 m/s
C) 1317.17 m/s²
D) 20.79m
Explanation:
A) We are given the angular speed as 1220 rev/min. Now let's convert it to rad/s.
1220 (rev/min) x (2πrad/1rev) x (1min/60sec) = (1220 x 2π)/60 rad/s = 127.76 rad/s ≈ 128 rad/s
B) Formula for tangential speed is given as;
v = ωr
We know that ω = 128 rad/s
Also, r = 3.02cm = 0.0302m
Thus, v = 128 x 0.0302 = 3.87 m/s
C) Formula for radial acceleration is given as;
a_c = v²/r
From earlier, v = ωr
Thus, a_c = v²/r = (ωr)²/r = ω²r
On the rim, r = 8.04cm = 0.0804
a_c = 128² x 0.0804 = 1317.17 m/s²
D) We know that; distance/time = speed
Thus, distance = speed x time
D = vt
From earlier, v = ωr
Thus, D = ωrt
Plugging in the relevant values ;
D = 128 x 0.0804 x 2.02 = 20.79m
A mass m at the end of a spring of spring constant k is undergoing simple harmonic oscillations with amplitude A. show answer No Attempt 33% Part (a) At what positive value of displacement x in terms of A is the potential energy 1/9 of the total mechanical energy? x = | α β π θ A B d g h j k m P S t ( ) 7 8 9 HOME ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. No Attempt No Attempt 33% Part (b) What fraction of the total mechanical energy is kinetic if the displacement is 1/2 the amplitude? No Attempt No Attempt 33% Part (c) By what factor does the maximum kinetic energy change if the amplitude is increased by a factor of 3? All content © 2020 Expert TA, LLC
Answer:
a) x = 0.33 A , b) K = ¾ Em , c) the kinetic energy increases 9 times
Explanation:
a) In a simple harmonic motion the mechanical energy is conserved and is expressed by the relation
Em = ½ k A²
At all points of movement the mechanical energy and
Em = K + U
remember that potential energy is
U = ½ k x²
they ask us for the point where
U = 1/9 Em
we substitute
½ k x² = 1/9 (½ k A²)
x = √1/9 A
x = 0.33 A
b) At all points the equation for mechanical energy is
Em = K + U
K = Em - U
K = ½ k A² - ½ k x²
K = ½ k (A² -x²)
at point x = ½ A
K = ½ k (A² - ¼ A²)
K = (½ k A²) ¾
K = ¾ Em
U = ½ k (½A) 2
U = ¼ Em
the fraction of energy e
U / K = 1/3
c) Kinetic energy is
K = ½ k v²
the system is described by the expression
x = A cos (wt + Ф)
speed is defined
v = dx / dt
v = - A w sin (wt + Ф)
we substitute
K = ½ k (- A w sin ( wt + Ф))²
We write this equation for the initial amplitude A
K₀ = ½ k (w sin (wt + Ф))² A²
we write it for the new amplitude A´ = 3 A
K = ½ k (w sin (wt +Ф))² (3A)²
the relationship between this energies is
K / K₀ = 9
whereby the kinetic energy increases 9 times
The displacement for 1/9 of mechanical energy is A/3. When the displacement is half amplitude, 3/4 of total mechanical energy becomes kinetic energy. If the amplitude increases by a factor of 3, the maximum kinetic energy increases by a factor of 9.
Explanation:In this context, we know that the total mechanical energy (E) of the system is constant and is the sum of kinetic and potential energy. The potential energy (U) of a mass-spring system is given by the formula U = 1/2 kx².
Part (a)
We want to find the displacement x such that the potential energy is 1/9 total mechanical energy, thus we set U = E/9. From substituting U = 1/2kx² and E = 1/2kA² in the equation,
we find that x²/A² = 1/9, which gives x = A/3.
Part (b)
We know that when the particle is at x = A/2 the potential energy is U = 1/2 k(A/2)² = 1/8 kA².
The total energy E = 1/2 kA², so the kinetic energy (K) is given by E - U = 1/2kA² - 1/8kA² = 3/8 kA².
The fraction of the total energy which is kinetic is (K/E) = (3/8)/(1/2) = 3/4.
Part (c)
The maximum kinetic energy is the total mechanical energy (since potential energy is zero at maximum kinetic energy), which is E = 1/2*kA².
If we triple the amplitude, the new energy E' = 1/2*k(3A)² = 9/2*kA².
The factor by which the maximum kinetic energy changes is (E'/E) = (9/2)/(1/2) = 9.
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The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
Answer:
B.
Explanation:
I'm taking the test right now.
The final momentum of the system is 0.4kgm/s
According to the law conservation of momentum, the momentum of the system before the collision and after the collision remains consevred.
if [tex]m_{1}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{1}[/tex] and [tex]v_{1f}[/tex] respectively, and
if [tex]m_{2}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{2}[/tex] and [tex]v_{2f}[/tex] respectively
then, [tex]m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{1f} + m_{2}v_{2f}[/tex]
Momentum after collision = [tex]m_{1}v_{1} + m_{2}v_{2}[/tex][tex]=0.6*0.5+0.5*0.2[/tex]
Momentum after collision = 0.4 kgm/s
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