A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.
a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided
Answers
Answer:
a) p₂ = 1.88 kg*m/s
θ = 273.4 º
b) Kf = 37% of Ko
Explanation:
a)
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:\(p_{o1x} = 2.00 kg*m/s (1)\)
\(p_{o1y} = 0 (2)\)
We can do the same for the particle moving along the positive y-axis:\(p_{o2x} = 0 (3)\)
\(p_{o2y} = 4.00 kg*m/s (4)\)
Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:\(p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)\)
\(p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)\)
Now, the total initial momentum, along these directions, must be equal to the total final momentum.We can write the equation for the x- axis as follows:\(p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)\)
We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:\(p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)\)
Now, we can repeat exactly the same process for the y- axis, as follows:\(p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)\)
We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:\(p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)\)
Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:\(p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)\)
We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:\(tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)\)
The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.b)
Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:\(\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)\)
So, the final kinetic energy has lost a 37% of the initial one.Related Questions
Describe the image with a object distance of 65cm and a focal length of 20cm
Answers
Answer:
Imag distance = 28.9 cm
Explanation:
Given that,
Object distance, u = -65 cm
The focal length of the convex les, f = +20 cm
We need to find the image distance. The lens formula gives the relation between focal length, object distance and image distance is given by :
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-65}\\\\v=28.9\ cm\)
So, the image distance is 28.9 cm
Usain Bolt requires five seconds to step after he runs the 100 meter dash . When he wins the race he's running at 10 meters per second . What is his declaration? Please answer as soon as possible!
Answers
Answer:
d= -2
Explanation:
t=5s
S=100 meters
V= 10 m/ s
d=?
deceleration = (final velocity - initial velocity) / time. d = (vf - vi)/t.
The formula for acceleration can be used, recognizing that the final result must have a negative signe.
d= (0-10)/5
d= -2
8. Molecular gastronomy was created by two scientists from Oxford University.
True or false
Answers
True,
molecular gastronomy was created by two scientists from Oxford University. Molecular gastronomy is a sub-discipline of food science that investigates the physical and chemical transformations of ingredients during cooking. It was first established by physicist Nicholas Kurti and chemist Hervé This, both from Oxford University. The term "molecular gastronomy" was coined in 1988 during a workshop on the science of cooking, which aimed to bridge the gap between scientific research and culinary arts.
Molecular gastronomy explores various aspects of food, such as flavor, texture, and presentation, and combines scientific knowledge with culinary techniques. This innovative approach to cooking has led to the creation of many unique and modern dishes, often featuring unexpected combinations of ingredients and visually appealing presentations.
The objective of molecular gastronomy is not only to create innovative dishes but also to understand and improve traditional cooking methods. This has led to advancements in food technology, as well as a deeper understanding of the science behind cooking processes. As a result, molecular gastronomy has become a popular and influential trend in the culinary world, inspiring chefs and home cooks alike to experiment with new techniques and ingredients.
In summary, molecular gastronomy was created by two scientists from Oxford University, Nicholas Kurti and Hervé This. Their groundbreaking work has had a significant impact on the culinary world and has led to many exciting innovations in food preparation and presentation.
Answer ASAP and only if you know its correct
this is science btw
Answers
Answer:
jupiter
Explanation:
from the list, Jupiter is closest to the sun.
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.685 T, the strip was 0.107 mm thick, the current along the strip was 2.25 A, and the Hall voltage between the strip's edges was 2.59 mV.Find the density nof mobile electrons in the material. The elementary charge is 1.602×10−19 C.
Answers
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Which factor affects the color of the star?
Luminosity
Temperature
Apparent Magnitude
None of the aobve
Answers
Answer:
temperature
Explanation:
if you look at a hertzsprung-russel diagram. you can understand how I got that answer
Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x = 40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge? *
Answers
Answer:
Magnitude of the electrostatic force on the +32 µC charge, \(F_{net} = 12 N\)
Explanation:
Let q₁ = +32 µC, x₁ = 0
q₂ = +20 µC, x₂ = 40 cm = 0.4 m
q₃ = -60 µC, x₃ = 60 cm = 0.6 m
Let magnitude of the electrostatic force on the +32 µC charge due to the + 20 µC charge = F₁ (i.e force on q₁ due to q₂)
\(F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }\)
\(F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N\)
Let magnitude of the electrostatic force on the +32 µC charge due to the -60 µC charge = F₂ (i.e force on q₁ due to q₃)
\(F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }\)
\(F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N\)
The electrostatic force on the 32 µC charge, \(F_{net} = |F_{2} + F_{3}|\)
\(F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N\)
2. A man has a mass of 62 kilograms on Earth. What is his weight?
Answers
Answer:
about 608 newtons
Explanation:
The force the man exerts on the surface of the Earth is ...
F = ma
F = (62 kg)(9.8 m/s²) = 607.6 N
The man weighs about 608 newtons.
_____
In the formula, m = mass. In the fill-in, m = meters.
Answer:
620 N
Explanation:
You can use this formula to find thr weight of the man.
F = m a
F = Force (weight)
m = mass
a = accelaration
Let's find now,
F = m a
= 62 kg × 10ms₋²
≈ 620 N
Hope this helps you.
Let me know if you have any other questions :-)
In a certain two-slit interference pattern, eight bright fringes lie
within the second side peak of the diffraction envelope and diffraction minima coincide with two-slit interference maxima.
(a) What is the ratio of the slit separation to the slit width?
(b) How many bright fringes lie within the first side peak?
Answers
a) The ratio of the slit separation to the slit width is 2.
b) There are four bright fringes within the first side peak.
How to determine ratio and brightness?a) The ratio of the slit separation to the slit width is 2. This is because the second side peak of the diffraction envelope is located at an angle of
2λ/d, where λ = wavelength of light and d = slit width.
The diffraction minima coincide with the two-slit interference maxima, which are located at angles of λ/d.
Therefore, the ratio of the slit separation to the slit width is 2.
(b) There are four bright fringes within the first side peak. This is because the first side peak of the diffraction envelope is located at an angle of
λ/d, where λ = wavelength of light and d = slit width.
The diffraction minima coincide with the two-slit interference maxima, which are located at angles of λ/d.
Therefore, there are 4 bright fringes within the first side peak.
Find out more on interference pattern here: https://brainly.com/question/17112134
#SPJ1
Which of the following is not a true statement?
A
B
C
D
Answers
Answer:
I think A
Explanation:
because I don't think a unknowned number can be a division problem
3) Two particles are travelling along a straight line AB of length 20m.
At the same instant one particle starts from rest at A and travels
towards B with a constant acceleration of 2 ms72 and the other
particle starts from rest at B and travels towards A with a constant
acceleration of 5 ms2, find how far from A the particles collide.
Answers
At distance of 40/7 m from A the particles will collide.
Let particles collide at C
And let distance AC = x and BC = y
then, x+y = 20
As the equation used for distance required here is: s= ut +1/2 a\(t^{2}\) and here u=0
Then, 1/2 (2)\(t^{2}\) +1/2 (5)\(t^{2}\) =20
7/2\(t^{2}\) =20
\(t^{2}\) = 40/7
Now
x= 1/2 (2)\(t^{2}\) = 40/7 m
Therefore, at distance of 40/7 m from A the particles will collide.
Learn more about distance here;
https://brainly.com/question/15172156
#SPJ9
14 An oil tank has a base of area 2.5 m2 and is filled with oil to a depth of 1.2m.
The density of the oil is 800 kg / m
What is the force exerted on the base of the tank due to the oil?
Answers
Answer:
F = 23544 [N]
Explanation:
To solve this problem we must use the following expression for static pressure in fluids.
\(P=Ro*g*h\)
where:
P = pressure [Pa]
Ro = density = 800 [kg/m³]
g = gravity acceleration = 9.81 [m/s²]
h = depth = 1.2 [m]
Now replacing:
\(P = 800*9.81*1.2\\P=9417.6 [Pa]\)
And the pressure is defined as the relationship of force between the area.
\(P=F/A\\\)
where:
F = force [N]
a = area = 2.5 [m²]
\(F = P*A\\F = 9417.6*2.5\\F = 23544 [N]\)
The potential energy of a catapult was completely converted into kinetic energy by releasing a small stone with a mass of 20 grams. The velocity of the stone when it reached the target was 15.0 meters/second. What was the value of the kinetic energy of the stone when it reached the target?
Answers
Subsequently, the value of the kinetic energy of the stone when it reached the target is 2.25 joules.
Kinetic energy calculation .
The kinetic energy (KE) of an question is given by the equation:
KE= 1/2 * mass * velocity²
In this case, the mass of the stone is 20 grams, which is rise to 0.02kg and the speed of the stone is 15m/s
Plugging these values into the equation, we are able calculate the active vitality of the stone:
KE = 1/2 * 0.02kg * (15.0mls)²
KE = 1/2 * 0.02kg * 225 m²/ s²
KE = 2.25J
Subsequently, the value of the kinetic energy of the stone when it reached the target is 2.25 joules.
Learn more about kinetic energy below.
https://brainly.com/question/25959744
#SPJ1
6.
The minimum value of coefficient of friction (u)
such that block of mass '5 kg' remains at rest is
5 kg
μ
ng
3 kg
(1) 0.3
(3) 0.6
(2) 0.5
(4) 0.4
Answers
Answer:
2
Explanation:
It is simple coman sense
When a skater pulls her arms in, it
reduces her moment of inertia from
2.12 kg m² to 0.699 kg-m². If she was
initially spinning 3.25 rad/s, what is
her final angular velocity?
Answers
The skater's final angular velocity is approximately 9.86 rad/s.
The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, the skater has an angular momentum of:
L_initial = I_initial * ω_initial
Substituting the given values:
L_initial = 2.12 kg m² * 3.25 rad/s
The skater's final angular momentum remains the same, as angular momentum is conserved:
L_final = L_initial
The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:
L_final = I_final * ω_final
0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s
Solving for ω_final:
ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²
Hence, the skater's final angular velocity is approximately 9.86 rad/s.
For more such questions on angular velocity, click on:
https://brainly.com/question/29566139
#SPJ8
I NEED THIS FILLED ASAP
Answers
Frequencies of sound waves higher than those we can hear are called __?__.
Answers
Answer:
Ultrasound
Explanation:
ultra = above ultrasound = above hearing sound
The image shows a molecular model of a compound using balls and sticks. Each ball is an atom. If you were to use balls and sticks to model a molecule of an element how must your model differ from what’s shown
Answers
Element molecules consist of a single type of atom and are held together by covalent bonds. The model would have uniform-colored balls connected by sticks representing these bonds.
The image represents a molecular model of a compound. Balls represent atoms, while sticks show chemical bonds between atoms. If you are to use balls and sticks to model a molecule of an element, the model would differ from what is shown because the element is made up of one type of atom. In other words, it is a pure substance that can't bond to other atoms of the same kind. Molecules of elements are usually composed of a single element and are not formed by bonding between two or more different atoms.In contrast to compounds, atoms in an element molecule are joined by a chemical bond called a covalent bond. Furthermore, the same atom is represented by a single-colored ball as there is only one type of atom involved in the element molecule. The sticks represent covalent bonds between the atoms of the same kind in a molecule of the element.For more questions on covalent bonds
https://brainly.com/question/7416042
#SPJ8
A factory is closing in 15 minutes. A box on one of the factory's conveyor belts is 8meters from a storage area and is moving at a constant velocity of 40metersperminute down the conveyor belt. How long will it take the box to get to the storage area?
Answers
The time taken for the box to get to the storage area, given that the box is moving at a constant velocity of 40 meters per minute is 0.2 minutes
How do i determine the time taken?Speed is defined as the distance travelled per time as shown below:
Speed = Distance / time
Cross multiply
Speed × time = distance
Divide both sides by speed
Time = Distance / speed
The following data were obtained from the question:
Distance from storage area = 8 metersConstant speed of box = 40 meters per minuteTime taken =?The time take can be obtained as follow:
Time = Distance / speed
Time = 8 / 40
Time = 0.2 minutes
Thus, the time taken to get the storage area is 0.2 minutes
Learn more about time:
https://brainly.com/question/12777835
#SPJ1
. Acylinder contains 1 mole of oxygen at
a temperature of 27 °C. The cylinder
is provided with a frictionless piston
which maintains a constant pressure
of 1 atm on the gas. The gas is heated
until its temperature rises to 127 °C.
(a) How much work is done by the
piston in the process?
(b) What is the increase in internal
energy of the gas?
(c) How much heat was supplied
to the gas?
(C = 7.03 calmol-¹°C¯¹;
R = 1.99 calmol-¹°C-¹;
1cal = 4.184 J)
Answers
a}The work is done by the piston in the process is 199 cal.
b) The increase in internal energy of the gas is 703 cal
c) The heat was supplied to the gas is 3771 J
(a) To calculate the work done by the piston, we can use the formula:
Work = P * ΔV
Where P is the constant pressure and ΔV is the change in volume. Since the pressure is constant, the work done is given by:
Work = P * (\(V_2 - V_1\))
Since the amount of gas is constant (1 mole), we can use the ideal gas law to calculate the initial and final volumes:
PV = nRT
\(V_1 = (nRT_1) / P_1\)
\(V_2 = (nRT_2) / P_2\)
Here, n is the number of moles (1 mole), R is the gas constant (1.99 cal/mol·°C), T1 is the initial temperature (27 °C + 273 = 300 K), T2 is the final temperature (127 °C + 273 = 400 K), and P1 and P2 are the initial and final pressures, respectively (both 1 atm).
Substituting the values into the equation, we have:
V1 = (1 mol * 1.99 cal/mol·°C * 300 K) / (1 atm) ≈ 597 cal
V2 = (1 mol * 1.99 cal/mol·°C * 400 K) / (1 atm) ≈ 796 cal
Therefore, the work done by the piston is:
Work = 1 atm * (796 cal - 597 cal) = 199 cal
(b) The increase in internal energy of the gas can be calculated using the equation:
ΔU = n * C * ΔT
Where ΔU is the change in internal energy, n is the number of moles (1 mole), C is the molar heat capacity (7.03 cal/mol·°C), and ΔT is the change in temperature (127 °C - 27 °C = 100 °C).
Substituting the values into the equation, we have:
ΔU = 1 mol * 7.03 cal/mol·°C * 100 °C = 703 cal
(c) The heat supplied to the gas can be calculated using the equation:
Q = ΔU + Work
Substituting the values calculated in parts (a) and (b), we have:
Q = 703 cal + 199 cal = 902 cal
Since 1 cal = 4.184 J, the heat supplied to the gas is:
Q = 902 cal * 4.184 J/cal ≈ 3771 J
Know more about work here:
https://brainly.com/question/28356414
#SPJ8
At a certain point in a circuit, 2C of charge have 44J of energy. What is the potential difference at this point?
Answers
The potential difference at the point is 44 V
From the question given above, the following data were obtained:
Energy (E) = 44 J
Charge (Q) = 2 C
Potential difference (V) =?The potential difference can be obtained as follow:
E = ½QV
44 = ½ × 2 × V
44 = 1 × V
V = 44 VTherefore, the potential difference is 44 V
Learn more: https://brainly.com/question/25425391
11. [0/10 Points]
A rectangular block has dimensions 2.9 cm x 2.6 cm x 10.0 cm. The mass of the block is 605.0 g
What is the volume of the block?
4.0
DETAILS
x cm³
What is the density of the block?
4.0
X g/cm³
Submit Answer
PREVIOUS ANSWERS
Answers
Volume of rectangular block is 75.4 cm^3
Density of the rectangular block is 8.02 g/cm^3
Volume is simply defined as the space occupied within the boundaries of an object in three-dimensional space.
It is also known as the capacity of the object.
Volume of rectangular block = length× breadth× height
=2.9 cm × 2.6 cm × 10.0 cm
=75.4 cm^3
Density is defined as the substance's mass per unit of volume.
Mathematically ,density is defined as mass divided by volume.
Density of the block = Mass of block / volume of block
=605.0 g / 75.4 cm^3
=8.02 g/cm^3
To know more about the density here
https://brainly.com/question/6107689
#SPJ1
600 moles of an ideal gas expands at 9 degrees Celsius from 340 L to 970 L isothermally. What is the work done by the gas?
answer with units
Answers
The work done by the gas is \(1.84 × 10^5\) joules (J).
To calculate the work done by the gasWe can use the formula for the work done by an ideal gas during an isothermal process:
W = nRT ln(Vf/Vi)
Where
W is the work done by the gasn is the number of moles of the gasR is the gas constant (8.31 J/mol·K)T is the temperature of the gas in kelvinVi is the initial volume of the gasVf is the final volume of the gasThe temperature must first be converted from Celsius to Kelvin:
T = 9°C + 273.15 = 282.15 K
The values we have can then be plugged in:
W = (600 mol)(8.31 J/mol·K)(282.15 K) ln(970 L/340 L)
W = \(1.84 × 10^5 J\)
Therefore, the work done by the gas is \(1.84 × 10^5\) joules (J).
Learn more about ideal gas here : brainly.com/question/1056445
#SPJ1
Q#4: Ifa particle of mass 3.16 Kg moving at 15.6 m/s to the left collides head on with a particle of mass 2.84 Kg moving at 12.2m/s to the right in laboratory frame of reference a) Find the velocity of the center of mass of the system of two particles after the collision. b) What would be the final velocities of these particles if the motion is observed in the center of mass frame?
Answers
Answer:
1929–1941. The longest and deepest downturn in the history of the United States and the modern industrial economy lasted more than a decade, beginning in 1929 and ending during World War II in 1941.
Explanation:
Hlo mister apke 2 week khatam nhi hue ??
Which type of circuit has
more than one path for
electrons to flow?
A. Parallel
B. Series
C. Neither
Answers
Answer:
A. Parallel connection
Answer:
Parallel circuit to be exact
I'm not sure about how can solve this problem. Please help me!!
Answers
The magnitude of the power dissipated in resistor R4 is approximately 10,028 watts.
How to find magnitude?To find the power dissipated in resistor R4, use the formula:
P = I² × R
where P = power, I = current flowing through the resistor, and R = resistance of the resistor.
The total resistance, Rt, can be calculated using the formula:
1/Rt = 1/R1 + 1/R2 + 1/R3
Substituting the given values:
1/Rt = 1/3 + 1/0.8 + 1/2
Simplifying the equation:
1/Rt ≈ 1.6667
Rt ≈ 0.6 Ω
Next, calculate the total voltage, Vt, by summing the individual voltage sources:
Vt = ε1 + ε2 + ε3
Substituting the given values:
Vt = 9 + 6 + 4
Vt = 19 V
Now calculate the current flowing through resistor R4 using Ohm's Law:
I = Vt / Rt
Substituting the calculated values:
I = 19 / 0.6
I ≈ 31.6667 A
Finally, calculate the power dissipated in resistor R4:
P = I² × R4
Substituting the calculated values:
P = (31.6667)² × 10
P ≈ 10,028 W
Therefore, the magnitude of the power dissipated in resistor R4 is approximately 10,028 watts.
Find out more on magnitude here: https://brainly.com/question/33196976
#SPJ1
The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.
Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.
Answers
The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.
Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.
Answer:
Figure 16.8 The pulse at time
t
=
0
is centered on
x
=
0
with amplitude A. The pulse moves as a pattern with a constant shape, with a constant maximum value A. The velocity is constant and the pulse moves a distance
Δ
x
=
v
Δ
t
in a time
Δ
t
.
The distance traveled is measured with any convenient point on the pulse. In this figure, the crest is used.
If you pull with a constant force of 400n , how much mechanical work does it take to pull pinball launcher back 0.2meters
Answers
If you pull with a constant force of 400 N for 0.2 meters, then the work done will be equal to 80 J.
What is Work?In physics, the word "work" involves the measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied within the direction of the displacement.
The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.
As per the given information in the question,
Force, f = 400 N
Displacement, d = 0.2 meters
\(Work done(W)=Force(f)*Displacement(d)\)
W = 400 × 0.2
W = 80 J.
To know more about Work:
https://brainly.com/question/13662169
#SPJ1
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answers
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
Consider the mass-on-a-spring system as shown in the figure below. The spring has a spring constant of 1.81e+3 N/m, and the block has a mass of 0.988 kg. There is a constant force of kinetic friction between the mass and the floor of 1.79 N. Starting with the spring compressed by 0.172 m from its equilibrium position, how far will the block travel once it leaves the spring? (Assume that block leaves the spring at at the spring's equilibrium position, marked x=0 in the figure.
Answers
K1Answer:
s= 6.5cm
Explanation:
Point 1: just right before the block leaves the spring
Point 2: the block has v2=0 (stops moving)
Apply Kinetic-Work Theorem:
K1 + U1 + Wother = K2 + U2
K1=0
U1= 1/2×k×x² (k= 1.81e+3 N/m and x= -0.172 m)
Wother = F×s= 1.79×s
K2=0
U2=0
=> s= 6.5cm
This is my attempt to solve. Let me know if this isn't right