Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So
m₁ = 30.0 g = 0.0300 kg
m₂ = 13.0 g = 0.0130 kg
v₁ = + 20.5 cm/s = 0.205 m/s
v₂ = + 15.0 cm/s = 0.150 m/s
and we want to find v₁' and v₂', the final velocities of either object after their collision.
Momentum is conserved throughout the objects' collision, so that
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' and v₂' are the first and second object's velocities after the collision.
Kinetic energy is also conserved, so that
1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²
or
m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²
From the first equation (omitting units), we have
0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'
0.0810 = 0.0300 v₁' + 0.0130 v₂'
81 = 30 v₁' + 13 v₂'
From the second equation,
0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²
0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²
1.55 ≈ 30 (v₁')² + 13 (v₂')²
Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields
v₁' ≈ + 0.172 m/s
and
v₂' ≈ + 0.227 m/s
the push up is dynamic or static
Answer:
Dynamic exercises
Explanation:
what is the specific latent heat of fusion of water?
Answer:
334
Explanation:
Three point charges are placed on the y-axis: a chargeqaty=a,a charge-2qat the origin, and a chargeqaty= -a.Such an arrangement is called an electricquadrupole.(a) Find the magnitude and direction of the electric field at points on thepositive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x >> a.Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.
of the following which is the largest body?
a. the moon
b. Pluto
c. Mercury
d. Ganymede
Answer:
Ganymede is the largest body
Explanation:
it is the satellite of jupiter
 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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What is 1 radian approximately in degrees?
Answer:
180 LUV:)
Explanation:
A wooden barrel full of water has a flat circular top of radius 25.0 cm with a small hole in it. A tube of height 8.00 m and inner radius 0.582 cm is suspended above the barrel with its lower end inserted snugly in the hole. Water is poured into the upper end of the tube until it is full. The density of water is 1.00 × 103 kg/m3.
What is the force with which the water in the barrel pushes up on the top of the barrel?
Answer:
that is all i know
Explanation:
radius= 25.0cm
height= 8m
inner radius= 0.582cm
density= 1.00 × 103kgm³f= m× aThe octopus’s tentacle keeps _ right after it is bitten off ? a. Moving b. Breathing c. Growing
Answer:
The answer is A
Explanation:
The octopus’s tentacle keeps moving right after it is bitten off
What is First Aid.
I mark u brainliest answer
Answer:
First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.
Explanation:
HELP ASAP!!!!! Choose all the answers that apply. Technology A)influences science
B)helps scientists observe fast phenomena
C)is the same as science
D) influences history
E)helps scientists observe slow phenomena
What is Motion ?????
Answer:
the action or process of moving or being moved.
Explanation:
which an object changes its position over time
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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A person is pushing a box with 5 N to the right while another person pushes with 10 N to the left, what is the net force on the box?
Answer:
Fnet = 5 Newton
Explanation:
Fnet = 10 N - 5 N
Fnet = 5 N
A pumpkin is launched in the air and travels at a horizontal velocity of 25 meters per second for 5 seconds. How far does it travel horizontally?
Answer:
30.3 meters, 172 degrees
Explanation:
To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.
Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West
Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North
The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.
Applying the Pythagorean theorem leads to the magnitude of the resultant (R).
R2 = (30.0 m)2 + (4.0 m)2 = 916 m2
R = Sqrt(916 m2)
R = 30.3 meters
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)
tangent(theta) = 0.1333
theta = invtan(0.1333)
theta = 7.59 degrees
This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.
What is the effect of erosion?
A. New land forms at the mouth of a river.
B. New land forms at the top of a mountain.
C. A mountain forms.
D. A fossil is created.
Be able to list the three compounds that are formed as products of highly exothermic
reactions such as detonating nitrogen-based explosives?
Answer:
Ammonium perchlorate NH4ClO4
Ammonium Nitrate
Calcium Cyanamide
When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.
Hopefully this helps :)
I will mark brainlist
A wave is disturbance that transfers energy and matter.
true
false
Answer:
False
Explanation:
A wave is a disturbance that transfers energy from one place to another without transferring matter.
Answer:
I'm pretty sure it true sorry if I'm wrong
Check all true statements about nonmetals. Group of answer choices
All nonmetals are gaseous at room temperature.
Nonmetals have low melting and boiling points. ✅
Nonmetals are brittle and have a relatively low density. ✅
Nonmetals are good at thermal and electrical conduction.
Using the colored periodic table below, match the group or area to its color.
alkali metals red alkaline ✅
earth metals orange ✅
transition metals white ✅
halogens purple ✅
Nobel gasses teal✅
How many of each subatomic particle are in a neutral atom of Potassium-39? What is its mass number? Potassium is in group 1 and has the symbol "K".
protons 19 ✅
neutrons 20 ✅
electrons 19 ✅
mass number 39✅
Match the ion with its charge.
11 protons, 12 neutrons, 10 electrons = +1 ✅
Groups 17, period 2 = -1 ✅
31 protons, 39 neutrons, 28 electrons = +3✅
Group 2, period 5 = +2
Which best describes a metal such as Silver (Ag)?
Answer: Lustrous, malleable, and forms cations.
Determine if the "quoted" word(s) makes each statement True or False.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". False ✅
If matter has a higher temperature, that means there will be "more" molecular motion. True✅
If you put a balloon in a freezer, its volume would "increase". False ✅
If you put the balloon into a chamber where there is half the pressure, its volume will "double". True ✅
If you cool down a propane tank, there will be "less" pressure in it. True ✅
The average atomic mass of Aluminum (Al) is "16.00 amu".
False✅
Oxygen's atomic number is "8" therefore it has "8" protons in its nucleus. True✅
An "electron" has the same mass as a neutron. False✅
The nucleus contains "protons and neutrons", virtually all the mass of an atom.✅
Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.
Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.
Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.
Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.
The following is an accurate matching of the ions;
11 protons, 12 neutrons, 10 electrons = +1 - Na^+
31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+
Group 2, period 5 = +2 - Sr^2+
Silver is a lustrous metal.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". FalseIf matter has a higher temperature, that means there will be "more" molecular motion. TrueIf you cool down a propane tank, there will be "less" pressure in it. TrueIf you put a balloon in a freezer, its volume would "increase". FalseIf you put the balloon into a chamber where there is half the pressure, its volume will "double". TrueThe average atomic mass of Aluminum (Al) is "16.00 amu". FalseOxygen's atomic number is "8" therefore it has "8" protons in its nucleus. TrueAn "electron" has the same mass as a neutron. FalseThe nucleus contains "protons and neutrons", virtually all the mass of an atom - True
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Answer:
If your answers are the green check marks all of them are correct
Explanation: I took the test
PLZ HELPPPP!!
this question is about popping microwave popcorn:
If you turn the microwave on for two minutes, is the rate of popping always the same, or does it change? Explain.
No,the poping of microvave is not same there time is different
In an oscillating LC circuit, when 81.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor
Answer:
21
Explanation:
9+10=21
A 500 kg cart is rolling to the right at 1.3 m/s. a 60 kg man is standing on the right end of the cart. what is the speed of the cart if tha man suddenly starts running to the left with a speed 10.0 m/s relative to the cart
Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton
5.08 m
Explanation:
The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:
[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]
where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get
[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]
Taking the square root of r^2, we then get the distance as
[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]
The values are given as follows:
[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]g = 9.8\:\text{m/s}^2[/tex]
[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]
[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]
Putting in all of these values in our equation for r,
[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]
what were your preparetion before going the different physical fitness test?
Answer:
Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
We have seen in earlier readings how to determine the speed of a wave on a string. What will happen to the wavelength of a sinusoidal wave on a string if the tension in the string is increased (assuming we keep the frequency of the wave the same)
This question involves the concepts of tension in a string and the wavelength of the wave in a string.
The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.
The speed of a wave on a string is given by the following formula:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where,
v = speed of wave = fλ
f = frequency of the wave
λ = wavelength of the wave
T = tension force
μ = linear mass density of the string
Therefore,
[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]
It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,
[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]
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i need help with these please
Answer:
Explanation:
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
There are 270 students and teachers going on a field trip to a science center. If each school bus holds 54 people, how many buses are needed?
Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.
Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.
Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:
[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]
Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:
[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]
Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:
[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]
Finalmente, reemplazamos los valores para obtener:
[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]
Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.
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