Answers
To help us solve this problem let's plot the points given in the table:
From the graph we notice that this the position can be modeled by a sine function, we also notice that the period of this function is 8. We know that a sine function can be modeled by:
[tex]A\sin(B(x+C))+D[/tex]where A is the amplitude, C is the horizontal shift, D is the vertical shift and
[tex]\frac{2\pi}{B}[/tex]is the period.
From the graph we have we notice that we don't have any horizontal or vertical shift, then C=0 and D=0. We also notice that the amplitude is 15, then A=15. Finally, as we said, the period is 8, then:
[tex]\begin{gathered} 8=\frac{2\pi}{B} \\ B=\frac{2\pi}{8} \\ B=\frac{\pi}{4} \end{gathered}[/tex]Plugging these values in the sine function we have:
[tex]x(t)=15\sin(\frac{\pi}{4}t)[/tex]If we graph this function along the points on the table we get the following graph:
We notice that we don't get an exact fit but we get a close one.
Now, that we have a function that describes the position we can find the velocity by taking the derivative:
[tex]\begin{gathered} x^{\prime}(t)=\frac{d}{dt}\lbrack15\sin(\frac{\pi}{4}t)\rbrack \\ =\frac{15\pi}{4}\cos(\frac{\pi}{4}t) \end{gathered}[/tex]Therefore, the velocity is:
[tex]x^{\prime}(t)=\frac{15\pi}{4}\cos(\frac{\pi}{4}t)[/tex]Once we have the expression for the velocity we can find values for the times we need, they are shown in the table below:
From the table we have that:
[tex]x^{\prime}(0.5)=10.884199\text{ cm/s}[/tex]And that:
• The earliest time when the velocity is zero is 2 s.
,• The second time when the velocity is zero is 6 s.
,• The minimum velocity happens at 4 s.
,• The minimum velocity is -11.780972 cm/s
Related Questions
Which is increased when the string of a stringed instrument is tightened?1) timbre2) pitch3) wavelength4) loudness
Answers
The pitch of a note is the same as its frequency.
Remember that te timbre depends on the shape and other properties of the vibrating object. The loudness is related to the amplitude of the wave, and the wavelength of sound is inversely proportional to the frequency.
On the other hand, the frequency of the sound of a string is proportional to the square root of the tension in the string. Then, if the tension increases (if the string is tightened), the frequency will also increase (and if the frequency increases, the wavelength decreases).
Therefore, the parameter that increases when a string of a stringed instrument is tightened, is the pitch.
Therefore, the answer is:
[tex]2)\text{ Pitch}[/tex]Please help me with this! (Sadly my previous tutor couldn't help me with this)
Answers
a.
The free body diagram (not at scale) for each crate is shown below:
b.
In this case we know that the force is just sufficient to keep the crates from sliding, this means that the acceleration of the system is zero.
From the free body diagram and Newton's second law we have that for the 45 kg crate that:
[tex]\begin{gathered} T-W=0 \\ T=W \\ T=(45)(9.8) \\ T=441 \end{gathered}[/tex]For the 35 kg crate the equations of motion would be:
[tex]\begin{gathered} F+f_f-T=0 \\ N-W^{\prime}=0 \end{gathered}[/tex]but we know that the force of friction is given by:
[tex]F_f=\mu N[/tex]and from the second equation of motion we have that:
[tex]\begin{gathered} N=W^{\prime} \\ N=(35)(9.8) \\ N=343 \end{gathered}[/tex]Then we have that:
[tex]\begin{gathered} F+F_f-T=0 \\ F+343\mu-441=0 \end{gathered}[/tex]Since the crates are not moving we need to use the static coefficient of friction, then:
[tex]\begin{gathered} F+343\mu-441=0 \\ F+343(0.5)-441=0 \\ F+171.5-441=0 \\ F-269.5=0 \\ F=269.5 \end{gathered}[/tex]Therefore the force applied is 269.5 N
c.
The diagram in this case is:
d.
In this case we know that the 35 kg is sliding to the right at constant velocity, this means that the acceleration for the system is zero. (Notice that the difference with the previous case is that the friction points to the left)
From the discussion in part b we know that for the 45 kg block:
[tex]T-W=0[/tex]and then:
[tex]T=441[/tex]For the 35 kg we have that:
[tex]\begin{gathered} F-F_f-T=0 \\ N-W^{\prime}=0 \end{gathered}[/tex]from the previos discussion we know that:
[tex]N=343[/tex]and since in this case the crates are moving we need to use the kinetic coefficient of friction, then we have:
[tex]\begin{gathered} F-F_f-T=0 \\ F-(0.3)(343)-441=0 \\ F-102.9-441=0 \\ F-543.9=0 \\ F=543.9 \end{gathered}[/tex]Therefore in this case the force applied is 543.9 N
e.
In this case the free body diagram is:
f.
Since the crates are moving with an accelearion of 0.5 m/s^2 we have for the 45 kg crate that:
[tex]T-W=ma[/tex]from where:
[tex]\begin{gathered} T-(9.8)(45)=(45)(0.5) \\ T-441=22.5 \\ T=441+22.5 \\ T=463.5 \end{gathered}[/tex]For the 35 kg crate we have that:
[tex]\begin{gathered} F-F_f-T=m^{\prime}a \\ N-W^{\prime}=0 \end{gathered}[/tex]from the previous discussion we know that N=343, plugging this in the first equation we have:
[tex]\begin{gathered} F-(0.3)(343)-463.5=(35)(0.5) \\ F-102.9-463.5=17.5 \\ F-566.4=17.5 \\ F=566.4+17.5 \\ F=583.9 \end{gathered}[/tex]Therefore the force applied in this case is 583.9 N
If a hammer is dropped from a height of 52 m and there is no air resistance, what is the acceleration the hammer experiences while it is falling towards the ground?
Answers
The hammer is fallingtowards the ground under the action of only one force which is gravity.
Thus, the acceleration of the hammer while falling towards the ground is the acceleration due to gravity.
Hence, the acceleration of the hammer in the given case is,
[tex]g=9.81ms^{-2\text{ }}[/tex]A 590 kg elevator accelerates upward at 1.1 m/s2 for the first 15 m of its motion. How much work is done during this part of its motion by the cable that lifts the elevator? Neglect any friction.
Answers
The work done by the elevator is - 86.730kj
It should be converted to energy in order to move an object. Energy can be transferred by the use of force. Work done refers to the amount of energy used by a force to move an object.
We are given that ,
The mass of the elevator = m = 590 kg
The acceleration of the elevator upward = 1.1m/s²
The height of the elevator = d = 15m
Therefore, T is the tension of the cable pulling the elevator upwards then the formulation of the work done of the elevator may be given as,
Work done = Force × distance
W = F × d
i.e. F = - mg
Above negative force due to gravity is acting opposite direction to the motion having an upward acceleration.
Thus , from above two equations we can write as,
W = - (mg)(d)
W = -(590kg)(9.8m/s²)(15m)
W = - 86730j
W = - 86.730kj
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An object has 25 J of kinetic energy and 35 J of potential energy. What isthe total energy possessed by the object? *
Answers
The total energy possessed by the object is 60 J
Given:
The kinetic energy of the object, K=25 J
The potential energy of the object, P=35 J
To find:
The total energy possessed by the object.
Explanation:
The kinetic energy of an object is the energy possessed by it due to its motion. The kinetic energy is directly proportional to the square of the velocity of the object.
The potential energy of an object is the energy possessed by it due to its position. The gravitational potential energy of an object is directly proportioinal to the mass ofthe
If the planet Mercury has a mass of planet 3.3×10²³ kg and a radius of 2400 km - calculate the magnitude of the gravitational field on its surface?
Answers
The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
Explanation:The mass of mercury, m = 3.3×10²³ kg
The radius, r = 2400 km
r = 2400 x 1000m
r = 2.4 x 10⁶m
Note that the magnitude of the gravitational field on the surface of the planet is the acceleration due to gravity on that planet
It is given by the formula:
[tex]g=\frac{Gm}{r^2}[/tex]Substitute the values of G, m, and r into the formula above
[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{2.2\times10^{13}}{5.76\times10^{12}} \\ g=3.82m/s^2 \end{gathered}[/tex]The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
if you had only one telescope and wanted to take both visible-light and ultraviolet pictures of stars, where should you locate your telescope?
Answers
If we just had one telescope and wanted to photograph stars in both visible and ultraviolet light, we should put it in space.
While visible light is observable from Earth, ultraviolet light can only be seen from space. Indeed, Hubble's ability to observe ultraviolet light gives it a major advantage over larger ground-based observatories.
Rank the visible light hues from left to right according to the altitude in the atmosphere where they are totally absorbed. All visible light wavelengths reach the Earth's surface, which is why we can see all colors and why visible-light telescopes perform well on the ground.
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Scientists might make a computer model of volcanic eruptions. What is the
biggest benefit of this model?
Answers
Answer:
Computer Model May Help to More Accurately Predict Volcano Eruptions. Scientists at the GFZ German Research Center in Potsdam, Germany, have developed a computer model which they say boosts the accuracy of volcanic eruption prediction.
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A person takes a trip, driving with a constant speed of 93.5 km/h, except for a 22.0-min rest stop. If the person's average speed is 71.6 km/h, find the following.(a) How much time is spent on the trip? h(b) How far does the person travel? km
Answers
We will have the following:
First, we determine the time:
[tex]\begin{gathered} \frac{93.5km/h+0km/h}{H}=71.6km/h\Rightarrow H=\frac{935}{716}h \\ \\ \Rightarrow H\approx1.3h \end{gathered}[/tex]So, the time spent on the trip was 935/716 h; that is approximately 1.3 hours.
Now, we deteremine the distance traveled:
[tex]\begin{gathered} d=(93.5km/h)(\frac{935}{716}-\frac{11}{30}h)\Rightarrow d=87.81513035...km \\ \\ \Rightarrow d\approx87.82km \end{gathered}[/tex]So, the distane traveled was approximately 87.82 km.
A bungee jumper jumps off a bridge and bounces up and down several times.She finally comes to rest 30 m below the bridge from which she just jumped.If her mass is 50 kg and the spring constant of the bungee cord is 10 N/m,how much energy was lost due to air resistance while she was bouncing?(Recall that g = 9.8 m/s2)A. 9200 NB. 7330 Nc. 10,200 ND. 8605 N
Answers
Given data
*The given height is h = 30 m
*The given mass is m = 50 kg
*The spring constant of bungee cord is k = 10 N/m
*The value of the acceleration due to the gravity is g = 9.8 m/s^2
The net change in potential energy is calculated as
[tex]\begin{gathered} \Delta U_p=mgh \\ =(50)(9.8)(30) \\ =14700\text{ J} \end{gathered}[/tex]The spring stretch is calculated by using the relation as
[tex]\begin{gathered} F=mg \\ kx=mg \\ x=\frac{mg}{k} \\ =\frac{(50)(9.8)}{10} \\ =49\text{ m} \end{gathered}[/tex]The energy stored in spring is calculated as
[tex]\begin{gathered} U_s=\frac{1}{2}kx^2 \\ =\frac{1}{2}(10)(49)^2 \\ =12005 \end{gathered}[/tex]The energy was lost due to air resistance while she was bouncing is calculated as
[tex]\begin{gathered} \Delta E=\Delta U_p-U_s_{}_{} \\ =14700-12005 \\ =2695\text{ J} \end{gathered}[/tex]Hence, the energy was lost due to air resistance while she was bouncing is 2695 J
A man takes in 0.95 × 107 J of energy each day from consuming food, and maintains a constant weight. what power in watt supplied by the food?
Answers
Given data
*The given energy is E = 0.95 × 10^7 J
*The given time is t = 1 day = (24 × 60 × 60) s = 86400 s
The formula for the power in watt supplied by the food is given as
[tex]P=\frac{E}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} P=\frac{(0.95\times10^7)}{(86400)} \\ =1.09\times10^2\text{ W} \end{gathered}[/tex]Hence, the power in watt supplied by the food is P = 1.09 × 10^2 W
Jacob, who has a mass of 74.8 kg, is sitting in a chair at rest. Calculate his weight in newtons
and round the answer to one decimal place.
Answers
Jacob's weight when sitting in a chair with a mass of 74.8 kg is 733.0 N
What is weight?Weight can be defined as the gravitational pull of a body.
To calculate Jacob's weight, we use the formula below.
Formula:
W = mg............. Equation 1Where:
W = Jacob's weightm = Jacob's massg = Acceleration due to gravity.From the question,
Given:
m = 74.8 kgg = 9.8 m/s²Substitute these values into equation 1
W = (74.8×9.8)W = 733.0 NHence, the weight of Jacob is 733.0 N.
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In a chosen coordinate system, the position of an object in motion can have negative values.Question 2 options:TrueFalse
Answers
True
Explanations:The position of an object can take any sign, it can be negative, zero or positive. This depends on the coordinate system chosen.
The position of an object (whether moving or static) is specified with respect to a frame of reference, and can be positive, zero or negative depending on the coordinate system chosen
Therefore, we can conclude that, in a chosen coordinate system, the position of an object in motion can have negative values.
What is the normal force of a 0.0037 kg tennis ball rolling at a constant speed of 3 m/s
across a desk?
Answers
The normal force of the rolling ball is 0.037 N
Mass of tennis ball= 0.0037 kg
Constant speed= 3m/s
we need to apply the concept of laws of motion
Since the ball is rolling at a constant speed, it is an example of uniform motion.
So,
Normal force=weight of the body
= 0.0037 x 10 ( acceleration of gravity= 10 m/s²)
Normal force= 0.037 N
Therefore the normal force on the ball is 0.037 N
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Four masses are arranged as shown. They are connected by rigid, massless rods of lengths 0.780 m and 0.500 m. What torque must be applied to cause an angular acceleration of 0.750 rad/s2 about the axis shown?
Answers
Given,
The length of the rods;
L=0.780 m
l=0.500 m
The angular acceleration, α=0.750 rad/s²
The masses;
m_A=4.00 kg
m_B=3.00 kg
m_C=5.00 kg
m_D=2.00 kg
The moment of inertia of the given system of masses is given by,
[tex]\begin{gathered} I=\Sigma mr^2 \\ =m_A(\frac{L}{2})^2+m_B(\frac{L}{2})^2+m_C(\frac{L}{2})^2+m_D(\frac{L}{2})^2 \\ =(\frac{L}{2})^2(m_A+m_B+m_C+m_D) \end{gathered}[/tex]Where r is the distance between each mass and the axis of rotation.
On substituting the known values,
[tex]\begin{gathered} I=(\frac{0.780}{2})^2(4.00+3.00+5.00+2.00) \\ =2.13\text{ kg}\cdot\text{m}^2 \end{gathered}[/tex]The torque required is given by,
[tex]\tau=I\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \tau=2.13\times0.750 \\ =1.6\text{ Nm} \end{gathered}[/tex]Thus the torque that must be applied to cause the required acceleration is 1.6 Nm
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision.. A)each car has half the impulse of before B)each car has double the impulse of before C)each car have the same impulse D)each car has an impulse in ratio to its mass.
Answers
We are given the following information.
The collision is elastic.
Car B has twice the mass of car A.
Recall that in an elastic collision, the momentum and the kinetic energy are conserved.
Impulse is basically the change in momentum.
Since car B has 2 times the mass of car A, the momentum of car B will be 4 times the momentum of car A.
This means that the impulse of car B will be greater than the impulse of car A.
Option D says that each car has an impulse in ratio to its mass meaning that a car with a larger mass will have a larger impulse and vice versa.
Therefore, we can conclude that after the collision, each car has an impulse in ratio to its mass.
A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?
Answers
Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:
[tex]\begin{gathered} d=vt \\ =(3.1\frac{m}{s})(314s) \\ =973.4m \end{gathered}[/tex]That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:
[tex]\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=\frac{d}{2\pi} \\ =\frac{973.4m}{2(3.14...)} \\ =154.9...m \end{gathered}[/tex]The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:
[tex]a_c=\frac{v^2}{r}[/tex]Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:
[tex]a_c=\frac{(3.1\frac{m}{s})^2}{(154.9m)}=0.062\frac{m}{s^2}[/tex]Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.
What is the displacement and distance of the car from t=0s to t=10s
Answers
We are given a graph of distance vs time. To determine the displacement we must add up the total distance covered by the car.
We notice from the graph that the distance from t = 0 and t =2 is:
[tex]d_{0-2}=2m[/tex]From t = 2 and t = 3 there is no change
If Hubble's constant had a value of 95 km/s/Mpc, what would be the age of the Universe?
Answers
The age of universe will be 0.0103 million years.
Hubble's law states that the rate at which any galaxy is receding from another galaxy is proportional to its distance from the galaxy. In simple form,
v = H₀ × d
where v is the velocity of the galaxy, d is its distance, and H₀ is the Hubble's constant.
Given: Hubble's constant, H₀ = 95 km/s/Mpc.
The term 1 / H₀ is called the Hubble's time and gives the age of the universe. That is, if 't' is the age of the universe, then
t = 1 / H₀.
Substitute the given value.
⇒ t = 1 / (95 km/s/Mpc)
(1 pc = 3.086 ×[tex]10^{13}[/tex] km)
⇒ t = 1 / (95/3.086 ×[tex]10^{13}[/tex] ) s
⇒ t = (1 / 30.784) ×[tex]10^{13}[/tex] s
⇒ t = 0.0325 ×[tex]10^{13}[/tex] s
(1 year = 3.154 ×[tex]10^{7}[/tex] s)
⇒ t = 10305.68 years
⇒ t = 0.0103 million years
Therefore, the age of universe will be 0.0103 million years.
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name the quatity whose SI unit is j/kg/°c
Answers
Answer:
Explanation:
Specific heat:
[ c ] = 1 J / (kg·°C)
The initial earthquake had a Richter scale reading of 3.24. The aftershock was 7470 times as strong as the initial earthquake. What was the Richter scale reading of the aftershock?(a) 7.11(b) 6.48(c) 4.32(d) 3.87
Answers
In order to determine the Richter scale reading of the aftershock, use the following formula:
[tex]M_2-M_1=\log (\frac{I_2}{I_1})[/tex]Where I1 and I2 are the intensity of both earthquake and its aftershock, M1 and M2 are the magnitude on the Richter Scale for earthquake and aftershock.
In this case, you have:
I1 = intensity earthquake
I2 = 7470*I1
M1 = 3.24
M2 = ?
solve the equation above for M2 and replace the values of the given parameters:
[tex]undefined[/tex]All matter is composed of quarks and leptons. Is this true or false?
Answers
All matters are made up of protons, neutrons and electrons. And protons and neutrons are made up of fundamental particles known as quarks. Leptons are fundamental particles with half-integer spin. An electron is a lepton.
Thus the given statement is true.
A 2.0 microF capacitor is connected across a 60 Hz voltage source, and a current of 2.0 mA is measured on a VOM. What is the capacitive reactance of the circuit?
Answers
Let's write down and name the variables we know.
C: capacitance; C = 2 μF = 2*10^-6 F
f: frequency of voltage source; f = 60 Hz
Xc: capacitive reactance of circuit (we are solving for this)
We also know that ω = 2πf = 120π.
From this information, we can use the following equation:
Xc = 1/(ωC)
And we can solve for Xc.
Xc = 1/(120π*2*10^-6)
Xc = 1326.291 Ω
Do you think we can see the planets without telescope? explain your answer
Answers
Answer:
No, how do you tell that?
Telescope is an example of technology that use on seeing the object on far place try to do that, you acn see a planet like small pieceof like an tungaw
Accomplishment of SPTA Officers and all parents this 1st Quarter.. Enjoy watching..Mabbalo
Explanation:
HOPE IT HELP TO YOUA duck is paddling due East at 1.1 m/s across the river while it flows due South at 0.55 m/s What is her resultant velocity?
Answers
The diagram representing this scenario is shown below
A right angle triangle is formed. R represents the resultant velocity. To find R, we would apply the pythagorean theorem which is expressed as
hypotenuse^2 = one leg^2 + other leg^2
From the diagram,
hypotenuse = R
one leg = 1.1
other leg = 0.55
Thus,
R^2 = 1.1^2 + 0.55^2 = 1.5125
Taking square root of both sides,
R = square root of 1.5125
R = 1.23
The resultant velocity is 1.23 m/s
Determine the maximum static friction when the normal force is 890 N and the coefficient of static friction is 0.55.
Answers
Answer:
489.5 N
Explanation:
The maximum static friction can be calculated as
[tex]F_f=\mu_sF_N[/tex]Where μ is the static friction and Fn is the normal force.
So, replacing the values, we get:
[tex]F_f=0.55(890N)=489.5N[/tex]Therefore, the maximum static friction is 489.5 N
Sound is an example of a longitudinal wave because the wave travels by compressions and rarefactions of air particles?True or false
Answers
To find
The given statement is true or false.
Explanation
The sound waves propag5ate th6roug5h6 a medium by the compression and rarefraction of the m
The figure shows a 100 W light bulb 1 meter away from my finger. If my finger tip has an area of 1 cm2 and if the wavelength of the light from the bulb is λ = 588 nm = 588 × 10−9 m, then show that the number of photons hitting my finger per second is about 1015γ/second.1 Watt of power is 1 Joule/second.Number of photons per second?
Answers
We are given the following information
Energy of bulb = 100 W = 100 Joules/second
Area of fingertip = 1 cm² = 0.0001 m²
Wavelength of light = 588×10⁻⁹ m
Number of photons per second = ?
Let us first convert the wavelength into energy
[tex]E=\frac{h\cdot c}{\lambda}[/tex]Where h is the plank's constant (6.626x10⁻³⁴J.s), c is the speed of light (3×10⁸ m/s) and λ is the wavelength.
[tex]\begin{gathered} E=\frac{6.626\times10^{-34}\cdot3\times10^8}{588\times10^{-9}} \\ E=3.3806\times10^{-19}\; \frac{J}{s} \end{gathered}[/tex]I want to know if I’m doing this correctly, and if not, then what would be the best process to do solve it?
Answers
Given
[tex]3\times10^{-10}m^{}[/tex]
Factor Name Symbol
10-1 deci d
10-2 centi c
10-3 milli m
10-6 micro µ
10-9 nano n
10-12 pico p
10-15 femto f
10-18 atto a
10-21 zepto z
10-24 yocto y
a. nanometers
0.3 nm
[tex]3\times10^{-1}nm^{}[/tex]b. picometers
300 pm
[tex]3\times10^2pm[/tex]upon leaving her club, the golf ball moved upward to a height above the surrounding trees. is the ke and pe increasing, decreasing, or staying the same?
Answers
ANSWER
PE increases and KE decreases
EXPLANATION
As described, the golf ball is moving and changing its height, like in the following diagram,
By the law of conservation of energy, the total energy when the ball starts moving and during the whole motion until it stops, must be the same. This total energy is the sum of the potential energy and the kinetic energy.
When the club hits the ball, it gives it a certain amount of kinetic energy but no potential energy. As the ball starts going uphill, the potential energy starts to increase, since it depends on the height of the object. Therefore, to maintain the total energy constant, the kinetic energy must decrease.
