Answer:
a) x = 660 m, b) λ = 0.330 m, c) Δλ = 0.1 cm, d) Δf = 104
Explanation:
a) the distance in which the train of waves extends can be obtained from the uniform movement
v = x / t
x = v t
x = 330 2
x = 660 m
b) the speed of sound is related to the wavelength and frequency
v = λ f
λ = v / f
λ = 330/1000
λ = 0.330 m
c) The precision in the measurement of the wavelength refers to the error or uncertainty in the measurement, if the measurement is direct with a tape measure the precision is the appreciation of the tape measure, in general it is 0.1 cm
d) the accuracy of the frequency in general the frequency is calculated from the measurements of period T
f = 1 / T
The precision of the period is data by the chronometer used, in general time a press of 0.01s, by the response time of the people
Δf = df / dT ΔT
Δf = 1 / T² ΔT
Δf = 1 / (0.001)² 0.01
Δf = 104
As we can see, a much more precise system is needed to reduce the error
Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 5.43 m below the point of release and she hears the splash 0.85 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)
Answer:
Explanation:
Total time taken = 0.85 s .
Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85
5.43 / 343 + time taken by coin to fall by 5.43 m = .85
time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s
Let the initial velocity of throw of coin = u
displacement of coin s = 5.43 m
time take to fall t = .834 s
s = ut + 1/2 gt²
5.43 = u x .834 + .5 x 9.8 x .834²
5.43 = u x .834 + 3.41
u x .834 = 2.02
u = 2.42 m /s .
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Rick places the blue lens of one pair of 3D glasses over the red lens of another pair. He then looks through both lenses at the same time. What color will he see?
A. blue
B. black
C. red
D. white
Answer:
a
Explanation:
Concept Simulation 4.1 reviews the central idea in this problem. A boat has a mass of 4490 kg. Its engines generate a drive force of 4520 N due west, while the wind exerts a force of 890 N due east and the water exerts a resistive force of 1210 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign
Answer:
-0.54m/s²
Explanation:
According to Newton's second law of motion
F = ma
Force = mass * acceleration
Given
Mass m = 4490kg
Take the sum of forces
Sum of force along the east = 890+1210 = 2100N
Sum of forces along the west = -4520N
Net force = -4520+2100
Net force = -2420N
Acceleration = Net force/Mass
Acceleration = -2420/4490
Acceleration = -0.54m/s²
Hence the boat acceleration is -0.54m/s²
What is the weight of a 44.5 kg object?
Answer:
98.11 I think
Explanation:
I really hope this helps have a wonderful day
The electric field from two charges in the plane of the paper is represented by the dashed lines and arrows below.
Select a response for each statement below. (Use 'North' towards top of page, and 'East' to the right)
The magnitude of the E-field at Ris .... than at M.
The force on a (+) test charge at P is zero.
The magnitude of the charge on the left is .... that on the right.
The force on a (+) test charge at L is directed ....
The force on a (-) test charge at J is directed
The force on a (-) test charge at N is directed ....
The sign of the charge on the right is negative.
Answer:
a) electric field at point P must be zero
b) harged must be positive
c) force ais in the direction of the electric field
d) force is in the opposite direction to the electric field
e) force is in the opposite direction to the field
Explanation:
After reading your exercise, it is unfortunate that the diagram did not come out, but we are going to answer the questions in general.
a) force on a charge (+) is zero
this implies that the electric field at point P must be zero
F = q E
b) the magnitude of the charge on the left is on the right
this indicates that the charged must be positive since the lines must exit the charge
c) force on load directed towards (direction not indicated)
since the charge is positive the force at point L is in the direction of the electric field at this point
d) force on test load (-) does not indicate direction
The force on a negative charge is in the opposite direction to the electric field at point J
e) Force on a test load (-) at point N
the force is in the opposite direction to the field at point N
How does altitude from the surface of earth affect the time period of a simple pendulum
Answer:
because the strength of Earth's gravitational field is not uniform everywhere, a given pendulum swings faster, and thus has a shorter period, at low altitudes and at Earth's poles than it does at high altitudes and at the Equator.
Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 kg electric car be able to supply to climb a 3.00 x 102 m high hill in 2.00 min at a constant 22.0 m/s speed while exerting 7.00 x 102 N of force to overcome air resistance and friction
Answer:
[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]
Explanation:
Given that:
The air resistance and friction = 700 N
The gravity caused force = 716 × 9.8 = 7016.8
Total force = (7016.8 + 700) N
Total force = 7716.8 N
∴
[tex]13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}[/tex]
[tex]current(I) \times 10.92= 19292[/tex]
[tex]current(I) = \dfrac{19292}{10.92}[/tex]
[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]
point.
4.
i. Explain why a particle moving with a constant speed along a circular
path has a radial acceleration.
ii. Show that the acceleration of a body moving in a circular path of
radius r with uniform speed v is → and draw a diagram to show the
direction of the acceleration.
2
iii. Show that the expression à = † is dimensionally correct.
Explanation:
Explanation: When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero. Explanation: When a particle moves with a uniform velocity, then dv/dt will be zero.
Canon launch is a 4.0 kg bowling ball with 50 J of kinetic energy what is the bowling ball speed
Answer:
5 m/s
Explanation:
50=1/2*4v^2
4*1/2=2
25*2=50
so...
square rood of 25 is 5
answer 5 m/s
sorry if that didn't make since
The name of the SI unit for magnetic field strength, such as that created around a current-carrying wire, is the
.
The name of the SI unit for magnetic field strength is Tesla. Magnetic fields are formed by moving electric charges.
What is Magnetic field strength?Magnetic field strength alludes to an actual amount that is utilized as one of the essential proportions of the power of the attractive field.
The SI unit of attractive field is tesla (T). 1 Tesla is defined as the magnetic field that carries 1C charge at the speed of 1m/s which is perpendicular to the force of 1 N.
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How much force is needed to accelerate a Kia Soul with a
mass of 1200 kg to 5 m/s2?
Answer:
[tex]\boxed {\boxed {\sf 6,000 \ Newtons}}[/tex]
Explanation:
Force is the product of mass and acceleration.
[tex]F=ma[/tex]
The mass of the Kia Soul is 1200 kilograms and its acceleration is 5 meters per square second.
[tex]m= 1200 \ kg \\a= 5 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 1200 \ kg * 5 \ m/s^2[/tex]
Multiply.
[tex]F= 6000 \ kg*m/s^2[/tex]
1 kilgram meter per square second is equal to 1 Newton. Our answer of 6000 kg*m/s² equals 6000 N[tex]F= 6000 \ N[/tex]
Answer:
Given :-Mass = 1200 kgAcceleration = 5 m/s²To Find :-Force
Solution :-We know that
F = ma
F = Force
m = mass
a = acceleration
F = 1200 × 5
F = 6000 N
[tex] \\ [/tex]
Fill in the question
help me pls it’s a usa test prep pretty easy
Answer:
Im 99.99999% sure its c
Explanation:
i cant see the pictures too well
The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the
Equator bring cooler waters to some regions, while currents flowing away from the Equator bring warmer waters to other regions.
North
British
Isles
Askan
North Atlantic
Azor
U.S.A
California
Gulf Stream
Loop
n
Canbean
North Equatorial
North Equatorial CC
North Fuatorial
Equator
South Equatorial
Not
South Equatorial
Image courtesy of NOAA
Judging from the map, which region probably has cooler summers than it would without the effect of a nearby ocean current?
A the Central U.S.
B. the British Isles
C. the U.S. East Coast
D. the US West Coast
Answer:
d
Explanation:
the US West Coast region probably has cooler summers than it would without the effect of a nearby ocean current.
what is ocean current ?
ocean current can be defined as the horizontal movement of seawater which is produced by gravity, wind, and water density, it play an major role in the determination of climates of coastal regions.
The movement of ocean water is continuous which can be up three types such as Waves, Tides, Currents
The streams of water which flow continuously on the ocean surface in specific directions are called ocean currents, it affect the temperature of ocean water as Warm ocean currents increase the temperature whereas Cold ocean currents decrease the temperature.
The magnitude of the ocean currents is about few centimeters per second to as much as 4 metres per second and the intensity of the ocean currents generally decreases with increasing depth.
There are two types of ocean currents such as Warm Ocean Currents
and Cold Ocean Currents
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Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangment of the four particles is defined as having zero electrical potential energy No work is required to bring one electron from infitinty to a location defined as the origin, while the other three particles remain at infiniuty. This is because no voltage exists near the origin until the first electron arrives. (a) Now, with the first electron remaining fixed at the origin, how much work is required to bring one of the remaining electrons from infinity to the coordinate (0 m, 2.00 m)? The other three particles remain at infinity. If this second electron was subsequently released, how fast would it be traveling once it returned to infinity? (b) Nļw, considering the two electrons fixed 2.00 m apart, how much work is required to bring the third electron from infinity to the coordinate (3.00 m, 0 m)? The other two particles remain at infinity. If this third electron was subsequently released, how fast would it be traveling once it returned to infinity? (c) Now considering the three fixed electrons at the coordinates described above. How much work is required to bring the last electron from infinity to the coordinate (3.00 m, 4.00 m)? If this forth electron was subsequently released, how fast would it be traveling once it returned to infinity? (d) Now considering the three fixed electrons at the coordinates described above. Finally, how much work is required to bring the proton from infinity to a coordinate of (1.00 m, 1.00 m)? If the proton is subsequently released and we assume that minimum separation distance between a proton and an electron is 1.00 pm, then how fast will the proton be traveling once it crashes into an electron?
Answer:
a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,
d) W = - 4.93 10⁻²⁸ J
Explanation:
a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m
If we use the law of conservation of energy, work is the change in energy of the system
W = ΔU = U_∞ -U
the potential energy for point charges is
U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]
in this case we only have two particles
U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]
the distance is
r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]
r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)
r₁₂ = √2= 1.4142 m
we substitute
W = k \sum \frac{q_i q_j}{r_{ij} }
let's calculate
W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142
W = 1.63 10⁻²⁸ J
b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0
in this case we have two fixed electrons
U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]
in this case all charges are electrons
q₁ = q₂ = q₃ = q
W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]
the distances are
r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0
r₁₃ = 3
r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2
r₂₃ = √13
r₂₃ = 3.606 m
let's look for the job
W = U
let's calculate
W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]
W = 1.407 10⁻²⁷ J
c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,
y₄ = 4.00 m
W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]
all charges are equal q₁ = q₂ = q₃ = q₄ = q
W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]
let's look for the distances
r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]
r₁₄ = 5 m
r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]
r₂₄ = √13 = 3.606 m
r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]
r₃₄ = 4 m
we calculate
W = 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]
W = 1.68 10⁻²⁸ J
d) we take the proton to the location x5 = 1m y5 = 1m
W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]
in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value
W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]
we look for distances
r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2
r₁₅ = √ 2 = 1.4142 m
r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]
r₂₅ = √2 = 1.4142 m
r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]
r₃₅ = √5 = 2.236 m
r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]
r₄₅ = √13 = 3.606 m
we calculate
W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]
W = - 4.93 10⁻²⁸ J
Scientists have investigated how quickly hoverflies start beating their wings when dropped both in complete darkness and in a lighted environment. Starting from rest, the insects were dropped from the top of a 50 - cm tall box. In the light, those flies that began flying 200 m s after being dropped avoided hitting the bottom of the box 87 % of the time, while those in the dark avoided hitting only 25 % of the time.
Required:
a. How far would a fly have fallen in the 200 ms before it began to beat its wings?
b. How long would it take for a fly to hit the bottom if it never began to fly? In seconds.
Answer:
Explanation:
a )
Hoverfly will fall with acceleration equal to g .
Initial velocity of fall of hoverflies u = 0
displacement ( vertical ) h = ?
time t = 0.2 s
acceleration due to gravity g = 9.8 m / s²
h = ut + 1/2 g t²
= 0 + .5 x 9.8 x .2²
= .196 m
= 19.6 cm
b )
Time taken to fall by 50 cm or 0.5 m under free fall from initial position .
.5 = 0 + .5 x 9.8 t²
t² = .1020
t = .319 s = 319 ms .
If a total 50 J of work are done on an object, it's energy...
Answer:
0.0119502868 kilocalorie
Explanation:
Answer:
increases by 50
Explanation:
What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?
Answer:
speed = 10.1215 m/s
Explanation:
speed = distance / time
speed = 100 / 9.88 = 10.1215 m/s
what is a vector quantity?
Answer:
A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude
For the questions below, include units if applicable. If necessary, use a separate sheet of paper for 1, 6c and 7c. Tire pressure is in part a function of the temperature of the tire.
1. Based on everyday experience, state (in words) the relationship between tire pressure and temperature. Look at the data below and see if the numbers support your statement.
2. Prepare a hand-drawn plot of the two variables on the reverse side of this worksheet. Include a title, axis labels (with units), and a trendline. Estimate the tire pressure when the temperature is 18.6°C: Estimate the temperature of the air in the tire when the pressure is 37.0 psi: 3.
a. Prepare a plot using graphing software. Include a title, axis labels (with units), the equation of the best-fit Line and the R? value on the graph.
b. Re-write the equation of the best-fit line substituting "Temperature" for x and "Pressure" for y directly on the graph.
c. Attach the fully labeled graph to this worksheet.
4. What is the value of the slope for the relationship between temperature and pressure?
5. Determine the percent error using the definition of percent error: Use 0.145 psi/" for the "Actual" value of the slope. %error = Actual-Experimental % Error Actual
6. Based on your computer-generated graph,
a. visually estimate the tire pressure when the temperature is 18.6°C:
b. calculate the tire pressure at this temperature using the equation of the best fit line: the graph to ensure that this value is reasonable.
c. compare the calculated pressure to the two visually interpolated values (Steps 2 and 6a). Comment on any discrepancies.
7. Based on your computer-generated graph,
a. visually estimate the temperature of the air in the tire when the pressure is 37.0 psi:
b. calculate the temperature of the air in the tire at this pressure: Use the graph to ensure that this value is reasonable.
c. compare the calculated temperature to the two visually interpolated values (Steps 2 and 7a). Comment on any discrepancies.
Data:
Temperature (x) Tire Pressure, psi (y)
12.9°C 3.39 x 10
15.4C 34.25
-2.10 F 2.68 x 10
19.5 °C 3.50 x 10
29.6 'F 36.53
Answer:
All answer are explained below in the explanation section.
Explanation:
1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.
As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.
Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.
2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.
Tire pressure at temperature 18.6 degree C is ~ 35 psi.
Temperature at air pressure of 37 psi is ~26.1 degree C
3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.
3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32
3. c) It is already done in part a of this question.
4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.
5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]
Plugging in the values, we get:
Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:
% Error = 21.33%
6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi
6. b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.
6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.
7. a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.
7. b.) Estimation of temperature from best fit value of line is = 26.64 degree C
7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.
Can someone please help, ty!!
Will mark brainliest.
Answer:
4. unbalanced and Accelerating
5. balance and rest
What is the mass of 9.11 moles of
ozone, 03?
The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
What is molecular mass?Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule. The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.
To compute the Molecular Mass of [tex]$O_{3}[/tex]
Atomic mass of oxygen(O) = 16
As [tex]$O_{3}[/tex] contains 3 atoms,
The molecular mass of [tex]$O_{3}[/tex]
= (16 x 3) = 48g/mol
Hence the mass of 9.11 moles O3
= 9.11 mol x 48g/mol
= 437.28g
= 0.43728kg
Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
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Atoms of which pair of elements will form covalent bonds in a compound? ASAP PLZ
A. Li and Al
B. C and O
C. Co and Fe
D. Na and F
Answer:B
Explanation:
C and O
Atoms of C and O pair of elements will form covalent bonds in a compound.Therefore the correct option is B.
What is a Chemical compound?The chemical compound is a combination of two or more either similar or dissimilar chemical elements
for example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.
These chemical compounds are formed because of different types of bonds between the constituents elements ,the chemical bonds are mainly ionic bonds, covalent bonds,s, and hydrogen bonds.
Ionic bonds are formed due to the transfer of electrons between two bond forming pairs differentiated by their electronegativity.
Covalent bonds are formed by the sharing of electrons.Generally organic compound are formed as the reason of covalent bonds.
The carbon and oxygen atoms share their valence electrons to form a covalent bond, therefore the correct option is B.
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If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow
Answer:fastest,same,slow down,opposite,slow
Explanation:
A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.
Velocity of a satellite around the planet.If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).
While moving towards the Earth (along the path from D to A) there is a component of force in the same direction as the motion; this causes the satellite to speed up.
While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.
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What current is needed in the solenoid's wires?
A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.8 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 μT local value of the earth's field.
What current is needed in the solenoid's wires? Express your answer with the appropriate units.
PLZZZZ HELPPPPPPPPPppppp
Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits
Answer:
a) A' = 0.345 m, b) f = 2,800 Hz
Explanation:
b) The angular velocity of a simple harmonic motion is
w =[tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and frequency are related
w = 2π f
we substitute
f = 1 /2π √k/m
indicates that the initial frequency value f = 3.96 Hz
in this case the mass is reduced by half
m ’= m / 2
we substitute
f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]
f = √1/2 (2π √k/m)
f = 1 /√2 3.96
f = 2,800 Hz
a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.
When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment
initial instant. Right before the division
p₀ = (m₁ + m₁) v
final instant. Right after the split
p_f = m₁ v '
p₀ = p_f
(2 m₁) v = m₁ v ’
v ’= 2v
At this point we can use conservation of energy for the system with only half the block.
Starting point. Where the block divides
Em₀o = K = ½ m v'²
Final point. Point of maximum elongation
Em_f = Ke = ½ k A²
how energy is conserved
Em₀ = Em_f
½ m’ v’² = ½ k A’²
we substitute the previous expressions
½ m/2 (2v)² = ½ k A’²
A’² = 2 m v² / k (1)
Let's use the conservation of energy with the initial conditions, before dividing the block
½ m v2 = ½ k A2
A² = mv² / k = 5.95 10⁻² m²
we substitute in 1
A'² = 2 A²
A ’²= 2 5.95 10⁻²
A ’²= 11.9 10⁻² m
A' = 0.345 m
An astronaut named Sandra Bullock has drifted too far away from her spaceshuttle while attempting to repair the Hubble Space telescope. She realizes that theshuttle is moving away from her at 3 m/s. On her back is a 10 kg jetpack which consistsof an 8 kg holding tank filled with 2 kg of pressurized gas. Without the jetpack, sheand her space suit have a mass of 80 kg.
Required:
a. She is able to use the gas to propel herself in the same direction as the shuttle. The gas exits the tank at a uniform rate with a constant velocity of 100 m/s, relative to the tank (and her). After the gas in the tank has been released, what is her velocity?
b. After this, she throws her empty tank into space and relies on the conservation of momentum to increase her speed to match that of the shuttle. With what velocity (in her frame of reference!) will she have to throw the tank?
Answer:
a) v_f = 0.898 m / s, b) v₂ = -6.286 m / s
Explanation:
a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s
Initial instant. Before you start to pass gas
p₀ = (M + Δm) v
M is the mass of the astronaut M = 80Kg and Δm the masses of the gases
Final moment. When you expel the gases
p_f = M (v + Δv) + Δm (v-v_e)
where v_e is the gas velocity v_e = 100 m / s
momentum is conserved
p₀ = p_f
M v + Δm v = Mv + M Δv + Δm v -Δm ve
0 = M Δv - Δm v_e
if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM
M dv = -v_e dM
∫ dv = - v_e ∫ dM / M
We solve, between the lower limits v₀ = v with M = M₀ and the upper limit v = v_f for M = M_f
v_f - v₀ = - v_e (ln M_f - Ln M₀)
v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])
v_f = v₀ + v_e ln (\frac{M_o}{M_f})
let's calculate
v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)
v_f = -1.3 +2.20
v_f = 0.898 m / s
b) launch the jetpack to increase its speed up to the speed of the platform
initial instant. Before launching the tanks
p₀ = (M + m') v_f
final instnte. After launching the tanks
p_f = M v₁ + m' v₂
indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it
p₀ = p_f
(M+ m) v_f = M v₁ + m v₂2
v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]
v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f
let's calculate
v₂ = 80/10 (0.898 - 0) + 0.898
v₂ = -7.1874 + 0.898
v₂ = -6.286 m / s
Acceleration is the rate ot change of the velacity a -dejdt so it is the slope of the Velocity vs. Time graph Because it is dficult to drag the person in a consistent and reproducible way use the Expression Evakaator under the Special Features menu for this question lick Reset A and type in the hr on z t * t * t " t in the Expression Evaluator Click the Play button and let the simulation run roughly 5 sin ulation seconds before ressing the Pause but use the zoom buttons to a 쪄 the p s they the screen You should see 8 p at s ar l what you got in the previous question, but much smoother Look at the Postion vs Time. Velocity vs Time and Acceleration vs. Time piets h
a) the velocity is zero but the acceleration is negative
When the person is 8 to to the tight of the origin
b) the velocity is zero but the acceleration is positive
c) both the velocity and the acceleration are zero
d) both the velocity and the acceleraton are nonzero
Answer:
a) the body is changing direction,
b)the body must go to the left and the acceleration to the right
c) the movement has not started.
d) all points of the motion
Explanation:
In this exercise you are asked to find in which position you have the following characteristics of the movement
a) The velocity is zero and the acceleration is negative
This is when the body reaches the end of the travel and turns around, in this case the speed is zero and the acceleration has the opposite direction to the movement.
In this case the body moves to the right and the acceleration is to the left, therefore the speed decreases
b) The velocity is zero, but the acceleration is positive
This occurs at the points where the speed is changing direction, specifically for this case the body must go to the left and the acceleration to the right
c) Both are zero
This only occurs where the body is stopped and the movement has not started.
d) both the velocity and the relation are nonzero.
This is at all points of the motion since the velocity is constantly changing as long as there is an acceleration
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride
Answer:
Answer is explained in the explanation section below.
Explanation:
In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.
a) You pass through the highest point:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.
And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.
b) You pass through the lowest point of the ride:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.
And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.