Answer:
The phase constant is 7.25 degree
Explanation:
given data
mass = 265 g
frequency = 3.40 Hz
time t = 0 s
x = 6.20 cm
vx = - 35.0 cm/s
solution
as phase constant is express as
y = A cosФ ..............1
here A is amplitude that is = [tex]\sqrt{(\frac{v_x}{\omega })^2+y^2 }[/tex] = [tex]\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }[/tex] = 6.25 cm
put value in equation 1
6.20 = 6.25 cosФ
cosФ = 0.992
Ф = 7.25 degree
so the phase constant is 7.25 degree
9) Cart 1 has a mass of 4 kg and an initial speed of 4 m/s. It eventually elastically collides with cart 2, whose mass is 6 kg, and which moves at an initial speed of 4 m/s toward cart 1. How fast are the carts moving after their collision? [Enter cart 1's final speed in answer box 1 and cart 2's final speed in answer box 2.]
Answer:
Cart 1 = 4 m/s
Cart 2 = 4 m/s
Explanation:
See attachment
7. Nancy has a mass of 60 kg and sits on the very end of a 3.00 m long plank pivoted in the middle. How much torque must her co-worker provide on the other end of the plank in order to keep Nancy from falling on the ground?
Answer:
Torque = 882Nm
Explanation:
Torque = Mg×distance
But plank's is pivoted ,therefore distance=3/2=1.5m
Mass of Nancy=60jg
Acceleration due to gravity, g=9.8m/s^2
Torque= 60×9.8×1.5
Torque= 882Nm
1. The light from polaris travels through space in the form of energy is called
Radiative energy
Explanation:
The light from polaris travels through space in the form of energy is called
Radiative energy.
This energy is transferred by means of electromagnetic radiation like X-rays, gamma rays, light, heat radiation and so on. It can travel through space in the form of radiation. For instance, we get the heat through the sun, that is located far away from the Earth by means of radiation. Through the electromagnetic waves, sun's heat is transmitted and not through any kind of solid medium, but by means of vacuum.
What is the maximum number of f orbitals that are possible?
Explanation:
The number of subshell present are s, p, d, and f
The number of orbitals in each are as follows -
The s subshell - 1 orbital consists of 2 electrons The p subshell - 3 orbitals consists of 6 electrons The d subshell - 5 orbitals consists of 10 electrons The f subshell - 7 orbitals consists of 14 electronsThe number of orbitals can be calculated by the degeneracy, 2 l + 1 , where l denoted is angular momentum quantum number that determines the shape of an orbital. For s, p, d, and f the angular momentum quantum number is 0, 1, 2 3 respectively.
So, maximum number of f orbitals that are possible can be calculated as -
2 × l + 1 = 2 × 3 + 1 = 7
What is the mass of a bullet moving at 970m/s if the bullet’s KE is 3.9x 10^3J
Answer:
0.08kg
Explanation:
K.E = 1/2 mv^2
v = 970m/s
K.E = 3.9x 10^3J= 3900J
K.E = 1/2 mv^2
3900 = 1/2 m x 970x 970
3900 = 1/2 ×940900m
3900 = 470450m
m = 3900/470450 = 0.00828993516 = 0.008kg
I think the Answer is 0.08kg
You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires. Part A Is the force between the wires attractive or repulsive? View Available Hint(s) Is the force between the wires attractive or repulsive? attractive repulsive Submit Part B What is the force per unit length F/L between the two wires? Express your answer in terms of I, d, and constants such as μ0 and π.
Answer:
a) The force between the two wires is attractive.
b) (F/L) = (μ₀I²)/(2πd)
Explanation:
a) According to Ampere's law, current in the same direction attract, while current in opposite directions repel. So, for this case of two wires carrying curremt in the same direction, the force between the wires is attractive.
b) The force of attraction between two current carrying wires carrying currents of magnitude I₁ and I₂ respectively, at some distance d, apart is given as
F = (μ₀ I₁ I₂ L)/(2πd)
(F/L) = (μ₀ I₁ I₂)/(2πd)
I₁ = I₂ = I
(F/L) = (μ₀I²)/(2πd)
Hope this Helps!!
(a) The force between the two wires is attractive.
(b) The force per unit length between the two wires is (μ₀I²)/(2πd)
Force between two parallel current-carrying wires:(a) It is given that two wires carry current in the same direction, so the force between the wires will be attractive.
(b) The magnetic force between the two wires carrying currents I₁ and I₂ separated by a distance d, is given by:
dF = I₁dl × B₂ force on wire with current I₁ due to magnetic field of the wire with current I₂
dF = I₁dl × (μ₀I₂)/(2πd)
F = (μ₀ I₁ I₂ L)/(2πd)
(F/L) = (μ₀ I₁ I₂)/(2πd)
since I₁ = I₂ = I
(F/L) = (μ₀I²)/(2πd)
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While traveling to and from a certain destination, you realized increasing your speed by 30 mph saved 3 hours on your return. If the total distance of the roundtrip was 560 miles, find the speed driven while returning.
Answer 62.78mph
Explanation:
Answer:
The returning speed = 70 miles per hour (mph)
Explanation:
The spacing between two closely spaced oppositely charged parallel plates is decreased. What happens to the electrostatic potential difference between the plates, assuming they form an isolated system: (a) it increases, (b) it decreases, (c) it stays the same, or (d) you can’t tell from the information given?
Explanation:
If the spacing between two closely spaced oppositely charged parallel plates is decreased the electrostatic potential difference between the plates will decrease. An electrostatic potential that is also referred to as the electric field potential or potential drop is the amount of work required to replace a unit of charge from a reference point to a specific point inside the electric field without any change in acceleration. Therefore, if the distance will decrease between oppositely charged plates there will be more affinity to attract which will reduce the amount of work done thus decreasing the electric potential∴ The Correct option is (b)
An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temperature of the wire is 55.0°C. Assume one free electron per atom. Given that at 20 degrees, rhoo = 2.82x10-8 Ωm and α = 3.9x10-3 /C. Determine:
a) the resistivity of the wire.
b) the current density in the wire.
c) the total current in the wire.
d) the potential different that must exist between the ends of a 2m length of wire if the given electric field is to be produced.
Answer with Explanation:
We are given that
Diameter of coil=d=0.115mm
Radius, r=[tex]\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m[/tex]
Using [tex]1mm=10^{-3} m[/tex]
Electric field=E=0.235V/m
T=55 degree C
[tex]T_0=20^{\circ} C[/tex]
[tex]\rho_0=2.82\times 10^{-8}\Omega m[/tex]
[tex]\alpha=3.9\times 10^{-3}/C[/tex]
(a).We know that
[tex]\rho=\rho_0(1+\alpha(T-T_0))[/tex]
Substitute the values
[tex]\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))[/tex]
[tex]\rho=3.2\times 10^{-8}\Omega m[/tex]
(b).Current density,[tex]J=\frac{E}{\rho}[/tex]
Using the formula
[tex]J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2[/tex]
c.Total current,I=JA
Where [tex]A=\pi r^2[/tex]
[tex]\pi=3.14[/tex]
Using the formula
[tex]I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2[/tex]
I=0.076A
d.Length of wire=l=2m
[tex]V=El[/tex]
Substitute the values
[tex]V=0.235\times 2=0.47 V[/tex]
Which of the following physical laws or principles can best be used to analyze the collision between the object and the pendulum bob? Which can best be used to analyze the resulting swing? 1. Newton's first law 2. Newton's second law 3. Newton's third law 4. Conservation of mechanical energy 5. Conservation of momentum
Answer:
4,5
Explanation:
The resulting swing converts potential energy to kinetic energy and kinetic energy to potential energy when the swinging stops. This is line with the law of conservation of mechanical energy which deals with inter conversion of energy forms and energy not being able to get lost.
The conservation of momentum is most suited to the collision. This law states that when a collision occurs the initial momentum before and after the collision is the same(without any external force).
A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with a speed of 10.0 m/s. What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground if the hot-air balloon is rising at 6.0 m/s relative to the ground during this throw?
Answer:
The ball travels at a speed of 11.662 m/s, making a 30.96° angle from the horizontal ground.
Explanation:
If the ball is thrown horizontally with a speed of 10.0 m/s and the hot air balloon is rising at a speed of 6.0 m/s then we can make the following deductions:
1. The ball is ALSO rising at a speed of 6.0 m/s
2. The vertical speed of the ball is 6.0 m/s
Thus the ball travels with 10 m/s horizontal and 6 m/s vertical speed relative to the ground.
To find the total speed and direction:
Total speed = [tex]\sqrt{(V_X)^2+(V_Y)^2}[/tex]
Total speed = [tex]\sqrt{10^2 + 6^2}[/tex]
Total speed = 11.662 m/s
Angle from the horizontal:
Tan( Angle ) = Perpendicular / Base
Tan (Angle) = Vertical Speed / Horizontal Speed
Tan (Angle) = 6/10
Angle = 30.96°
The initial velocity of the ball thrown horizontally from the basket of a hot-air balloon relative to a person on the ground is 10.0 m/s in the horizontal direction.
Explanation:The initial velocity of the ball relative to a person standing on the ground can be found by adding the velocities of the ball and the hot-air balloon together. Since the ball is being thrown horizontally outward, its initial velocity in the horizontal direction will be 10.0 m/s. The vertical velocity of the balloon is 6.0 m/s, but since the ball is not thrown vertically, this velocity does not affect its horizontal velocity.
Therefore, the initial velocity (magnitude and direction) of the ball relative to a person on the ground will be the same as its initial velocity when thrown from the basket, which is 10.0 m/s horizontally.
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Two large, parallel, conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of 15 3.9 10 N − ⋅ acts on an electron if it is placed anywhere between the two plates. (a) Find the electric field magnitude at the position of the electron. (b) What is the potential difference between the plates?
Answer:
a) 2.4×10^4N/C
b) 2.9 ×10^3V
Explanation:
Correct ststement: An electric force of 3.9×10^-15N acts on the electron if it is placed anywhere between the two plates.
a) The electric field magnitude is given by E= F/e
Where F = electric force
e= elementary charge carried by a single proton e= 1.6×10^-19C
E= (3.9×10^-15)/(1.6×10^-19)
E= 2.4×10^4NC
b) Ptential difference is given by:
Change in V= E×change in distance
Potential difference= (2.4×10^4)× (0.12)
Potential difgerence= 2.9×10^3V
A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the distance was 23.74 n. The coefficient of kinetic friction between the tires and the road is μk = 0.29 How fast was the car going when the driver applied the brakes
Final answer:
To find the initial speed of the car, we use the work-energy principle and the equation μk × g × d = ½ × v². By substituting the given coefficient of kinetic friction (0.29) and skid mark distance (23.74 meters), we calculate the initial speed to be approximately 11.68 m/s.
Explanation:
The question involves calculating the initial speed of a car at the moment the driver applied the brakes, which resulted in skid marks on the road. The length of the skid marks and the coefficient of kinetic friction (μk) between the tires and the road are known.
To calculate the initial speed of the car, we use the work-energy principle, which states that the work done by the friction force is equal to the change in kinetic energy of the car:
Work done by friction = Change in kinetic energy
μk × m × g × d = ½ × m × v²
Since mass (m) cancels out and gravitational acceleration (g) is a constant (9.81 m/s²), we can simplify the equation to:
μk × g × d = ½ × v²
Now we can solve for the initial velocity (v):
v = √(2 × μk × g × d)
Plugging in the given values, μk = 0.29 and d = 23.74 meters, we have:
v = √(2 × 0.29 × 9.81 m/s² × 23.74 m)
v ≈ √(2 × 0.29 × 9.81 × 23.74)
v ≈ √(136.4)
v ≈ 11.68 m/s (approximately)
Thus, the initial speed of the car when the brakes were applied was approximately 11.68 meters per second.
In doing a load of clothes, a clothes drier uses 18 A of current at 240 V for 41 min. A personal computer, in contrast, uses 2.2 A of current at 120 V. With the energy used by the clothes drier, how long (in hours) could you use this computer to "surf" the Internet
Answer:
The time taken by the personal computer to 'surf' the internet is 11.13 h.
Explanation:
The expression for the energy in terms of current and time is as follows;
E= Vit
Here, V is the potential, i is the current and t is the time.
According to question, in doing a load of clothes, a clothes drier uses 18 A of current at 240 V for 41 min.
Convert time from minute to second.
[tex]t=\frac{41}{60}s[/tex]
Calculate the energy used by the clothes drier.
Put V= 240 V, [tex]t=\frac{41}{60}s[/tex] and i=18 A in the expression for the energy.
[tex]E= (240)(18)\frac{41}{60}[/tex]
E'= 2937.6 W
Therefore, the energy of a clothes drier is E'= 2937.6 W.
It is given in the problem that a personal computer, in contrast, uses 2.2 A of current at 120 V. With the energy used by the clothes drier, this computer is used to surf the internet.
Calculate the time taken by the computer to 'surf' the internet.
E'=V'i't'
Here, V' is the voltage used by the a personal computer, i' is the current and t' is the time taken by the personal computer.
Put E'= 2937.6 W, V= 120 V and i'= 2.2 A.
2937.6 = (120)(2.2)t'
[tex]t'=\frac{2937.6}{264}[/tex]
t'= 11.13 h
Therefore, the time taken by the personal computer to 'surf' the internet is 11.13 h.
Kalb Enterprises is trying to develop a good brand name for its revolutionary accounting software program. They turn to an external consultant to help them. The consultant gives them a list of characteristics of a "good" brand name. Which characteristic would NOT be on their list
Answer:
The correct answer is Can be pronounced in many ways.
Explanation:
Typically a consultant is hired to solve a major business problem, carry out a major change and increase sales. The consultant, with experience of other companies, with methodology resources listens to the client's need, - the current situation and design an effective plan to implement, thus creating a positive impact on the business in a short time.
What is the fundamental frequency of a 0.003 kg steel piano wire of length 1.3 m and under a tension of 2030 N? Answer in units of Hz. 005 (part 2 of 2) 10.0 points What is the fundamental frequency of an organ pipe 1.38 m in length, closed at the bottom and open at the top? The speed of sound is 340 m/s. Answer in units of Hz.
The fundamental frequency of a piano wire can be calculated using the formula f = sqrt(T/(4L^2*m)). Plugging in the given values, the fundamental frequency of the piano wire is 116.6 Hz.
The fundamental frequency of a piano wire can be calculated using the formula:
f = √(T/(4L2m))
Where f is the fundamental frequency, T is the tension, L is the length, and m is the mass per unit length.
Plug in the given values:
T = 2030 N
L = 1.3 m
m = 0.003 kg/m
f = √(2030/(4*1.32*0.003))
Solving this equation gives the fundamental frequency f = 116.6 Hz.
given the initial ( 0 , − 2 ) and terminal ( 3, 6 ) points of a vector, how do you write a linear combination of the standard unit vector ˆ i and ˆ j ? socratic.org
Answer:
Explanation:
Given that,
The initial point A=(0,-2)
And the final point B=(3,6)
O is a reference point (0,0)
OA=r1=(0,-2)
OB=r2=(3,6)
Then we want to find their combination vector i.e AB.
AB is given as position vector B - position vector A
So, AB=OB-OA
AB=r2-r1
AB=(3,6)-(0,-2). Note -×-=+
Then, AB=(3-0,6--2)
AB=(3,6+2)
AB=(3,8)
In vector form
AB=3•i +8•j
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g = 30~\rm m/s^2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.)a.It will smash his face.b.It will stop well short of his face.c.It will take less time to return to the point from which it was released.d.Its mass will be greater.e.It will take more time to return to the point from which it was released
Answer:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C
Two forces act on a 8.50-kg object. One of the forces is 14.0 N. If the object accelerates at 3.50 m/s2 , what is the greatest possible magnitude of the other force
Answer:
The magnitude of the other force is 43.75 N.
Explanation:
Given that,
Mass of the object, m = 8.5 kg
Force 1, [tex]F_1=14\ N[/tex]
Acceleration of the object, [tex]a=3.5\ m/s^2[/tex]
To find,
The greatest possible magnitude of the other force.
Solution,
Let [tex]F_2[/tex] is the magnitude of other force that is acting on the object. For the greatest force, the two forces must be act in opposite direction such that :
[tex]F_2-F_1=ma[/tex]
[tex]F_2=ma+F_1[/tex]
[tex]F_2=14+8.5\times 3.5[/tex]
[tex]F_2=43.75\ N[/tex]
So, the magnitude of the other force is 43.75 N.
If you weighed 100 lb on Earth, what would you weigh at the upper atmosphere of Jupiter? For reference, Jupiter has a mass that is about 300 times Earth’s mass and a radius that is 10 times Earth’s radius.
Answer:
The answer to the question is
A 100 lb person would weigh 300.33 lbf at the upper atmosphere of Jupiter
Explanation:
To solve the question we note that
Mass of object = 100 lb = 45.35924 kg
Mass of Jupiter = 300×Mass of Earth = 300×5.972 × 10²⁴ kg =1.7916×10²⁷ kg
Radius of Jupiter = 10× Radius of Earth = 10×6,371 km = 63710 km
Gravitational constant, G = 6.67408 × 10⁻¹¹ m³ kg-1 s-2
Gravitational force is given by [tex]F_G= \frac{Gm_1m_2}{r^2}[/tex]
Plugging in the values we get
[tex]F_G[/tex] = [tex]\frac{6.67408*10^{-11}*45.35924*1.7916*10^{27}}{63710^2}[/tex] = 1335.93 N
Converting into lbf gives 1335.93 N *0.2248 lbf/N = 300.33 lbf
On Jupiter, you would weigh 2.5 times more than on Earth due to its stronger gravity. Therefore, if you weighed 100 lb on Earth, you would weigh approximately 250 lb in the upper atmosphere of Jupiter though your weight may slightly vary with altitude. It is important to note that while weight changes with location, mass remains constant.
Explanation:If you weighed 100 lb on Earth, it is interesting to calculate what you would weigh in the upper atmosphere of Jupiter. Since gravitational force is what we perceive as weight, and gravity varies with the mass of the planet and the distance from its center, we can determine the comparative weight. Jupiter has a mass approximately 318 times that of Earth and a radius about 11 times greater. However, for simplicity, the student question references Jupiter's mass as 300 times that of Earth and a radius 10 times that of Earth. This would normally impact the surface gravity calculation significantly.
Now, since it's stated that on Jupiter you would weigh 2.5 times more than on Earth, if you weigh 100 lb on Earth, you would weigh 250 lb on Jupiter. But note that in the upper atmosphere, your weight might be slightly less than this because you would not be at Jupiter's surface, and gravity decreases with altitude.
Lastly, remember the important conceptual distinction: While your weight changes depending on the gravity of the celestial body you are on, your mass remains constant. Your mass represents the amount of matter within you and does not change with location.
Ling lives 2 miles from school. It took him 15 minutes to bike from school to home. The first half of the distance he biked at a speed of 12 mph. What was his speed for the remaining distance
Explanation:
As we know that
time = distance/speed
The time used for firs half of the trip was
(1 mi)(12 mi/hour) = 1/12 hours = 5 minutes
The last half of the trip will took 10 minutes, 1/6 hour.
Speed = distance/ time
(1 mil) = (1/6h) = 6 mil/h
so the speed for last half of the trip was = 6mph
the average speed was
(2mil)(1/4 hour) = 8 mil/hour
So the ling's average speed was 8mph.
Imagine that you know the mass of a nearby star and you know that there is a planet orbiting around the star with a mass much smaller than the mass of the star. Explain with a sentence or two how, using the Doppler effect technique, you can measure the semi-major axes of a planet orbiting the star.]
Answer:
By measuring the time taken for the stars line of sight velocity to cycle from peak to Peak, and by calculation using newtons version of Kepler's third law
Explanation:
The motion of orbiting planets using planet-hunting techniques can rely on doppler effect. The light from the stars they orbit, as seen from Earth. As the star moves back and forth, the Doppler shift causes a slight change in its apparent colour which can be detected using spectroscopy. The blue shift and the red shift.
The Doppler technique of blue shift and the red shift can be used to estimate the semi major axis of the planets orbit by
Measuring the time it takes for the stars line of sight velocity to cycle from peak to Peak, and using newtons version of Kepler's third law
Which of the following classes of biomaterials would be most appropriate for use to fabricate an artificial tendon, a tissue that must sustain substantial deformation at low forces and return rapidly to its original dimensions upon release of the stress why?1. metals2. ceramics3. polymers
Final answer:
The most appropriate biomaterials for artificial tendons are polymers, due to their high toughness, elasticity, and ability to mimic natural extracellular matrices.
Explanation:
For fabricating an artificial tendon, which requires the capacity to sustain substantial deformation at low forces and quickly return to its original dimensions after stress release, polymers would be the most appropriate class of biomaterials. Biopolymers like collagen are noted for high toughness and elasticity, vital properties for mimicking the natural structure and function of tendons. Modern bioengineering strategies also involve creating synthetic extracellular matrices (ECM) which utilize peptides and polymer conjugates to mimic natural ECM, providing both structural support and facilitating tissue integration and regeneration.
Metals and ceramics, while being strong and rigid, do not offer the same flexibility and elastomeric properties required by tissues such as tendons that undergo repeated and significant strain. Semi-flexible polymers and polymer gels on the other hand, have been engineered to possess a balance of strength and flexibility, similar to that of natural tendon tissue.
If an object oscillates in simple harmonic motion, with the position described by the equation: x(t) = 42.5*cos(21t) What is the angular frequency of oscillation w ?
Answer:
The angular frequency of oscillation w = 21
Explanation:
To solve the question, we note that x(t) is the point in the motion of the object. Therefore to find the angular frequency of oscillation, we find the relationship between the angular velocity and time
The angular frequency, ω is a scalar quantity used to depict the rate of rotation per unit time
When there is a function in simple harmonic motion (SHM) with the following equation then ω is the angular frequency
x(t) = A·cos (2πft) = A·cos(ωt) which is similar to 42.5*cos(21t)
then 21 = angular frequency
What is the sum of the kinetic energies of the alpha particle and the new nucleus?
Answer: The total energy created by the Alpha decay.
Explanation: The sum total of the kinetic energy of the alpha particle and the new nucleus is the total energy created by the alpha decay.
Consider the decay of a radioactive nuclide by the spontaneous emission from its nuclei of alpha particles. An alpha particle which is composed of two protons and two neutrons and has a charge of +2. With an appreciable mass and its ejection from the nuclide creates a certain amount of recoil energy in the nucleus. The total energy (Ex) created by alpha decay is therefore the sum of the kinetic energy of the particle, the recoil energy given to the new nucleus, and the total energy of any emitted gamma rays.
The sum of the kinetic energies of an alpha particle and the new nucleus after alpha decay equals the initial energy (Q-value) of the reaction, most of which is carried by the alpha particle due to mass differences.
Explanation:The sum of the kinetic energies of the alpha particle and the new nucleus after an alpha decay can be determined by considering the conservation of energy and momentum. The energy released during the decay, known as the Q-value, is primarily carried away by the alpha particle due to its relatively smaller mass compared to the daughter nucleus.
The energy of the alpha particle can be measured experimentally, which then allows us to determine the energy of the new nucleus. According to conservation of energy, the sum of the kinetic energies of the alpha particle and the new nucleus is equal to the Q-value of the reaction. If the Q-value (Qa) is known to be 4.3 MeV, and assuming the potential energy is zero in the final state, this energy will be distributed between the alpha particle and the new nucleus.
Using the example values provided, if the kinetic energy of the new nucleus (KEnucleus) is calculated to be 23.3 eV, then the remainder of the Q-value minus the kinetic energy of the new nucleus will represent the kinetic energy of the alpha particle. Since the alpha particle carries away most of the kinetic energy, the sum of the kinetic energies will be very close to the initial Q-value.
A hollow aluminum cylinder 19.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0°C. (The average linear expansion coefficient for aluminum is 2.4 x10^-5/°C, and the average volume expansion coefficient for turpentine is 9.0 x10^-4/°C.)
(a) How much turpentine overflows?
in cm^3?
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)
in cm^3?
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede?
in cm?
Answer:
(a) 0.1134 L
(b) 2.009 L
(c) 0.07 cm from the top
Explanation:
The volume of the aluminum container and the turpentine is 2.000 L at 23 °C.
(a) When the entire system is heated to from 23 °C to 91 °C. The volume of the aluminum container is:
Vf = Vi*(1 + 3α*ΔT)
where:
Vi is the initial volume, Vf the final volume, α is the linear expansion coefficient and ΔT is the temperature change.
Vf = 2*[1 + 3*2.4 *10^-5*(91 - 23)] = 2.009 L
The volume of the turpentine is
Vf = Vi*(1 + β*ΔT)
where:
β is the volume expansion coefficient
Vf = 2*[1 + 9.0*10^-4*(91 - 23)] = 2.1224 L
The overflow is:
2.1224 - 2.009 = 0.1134 L
(b) the volume of turpentine remaining in the cylinder at 91 °C is the same volume of the hollow aluminum cylinder, that is, 2.009 L
(c) Cooling the system back to 23 °C means the aluminum container is back to 2.000 L. The volume of turpentine is:
Vf = 2.1224*[1 + 9.0*10^-4*(23-91)] = 1.9925 L
The following proportion must be satisfied:
Volume of turpentine / Volume of container = deep turpentine / deep of container
1.9925 / 2.000 = deep turpentine/19
deep turpentine = (1.9925 / 2.000)*19 = 18.93 cm
This is 19 - 18.93 = 0.07 cm from the top
A sample of radioactive waste has a half-life of 10 years and an activity level of 2 curies. After how many years will the activity level of this sample be 0.25 curie?
To find out after how many years the activity level of a sample with a half-life of 10 years decreases from 2 curies to 0.25 curie, we calculate that it takes 30 years.
The question asks, "A sample of radioactive waste has a half-life of 10 years and an activity level of 2 curies. After how many years will the activity level of this sample be 0.25 curie?" To solve this problem, we follow the basic principle of half-lives, which states that the amount of a radioactive substance decreases to half its initial amount after one half-life period.
In this example, the activity decreases from 2 curies to 0.25 curies. To go from 2 curies to 1 curie, it takes one half-life (10 years). To go from 1 curie to 0.5 curies takes another half-life (adding up to 20 years in total). Finally, to decrease from 0.5 curies to 0.25 curies requires a third half-life period, adding another 10 years, for a total of 30 years.
Therefore, it will take 30 years for the activity level of the sample to reduce to 0.25 curie.
Nine-year-old Dakoda realizes that the quantity of water in a glass remains the same, even when the water is poured into a different shaped glass and appears to look like more or less. In Piaget’s terms, Dakoda has developeda) Conservationb) Assimilationc) Accomodationd) Object Permanencee) Centration
Answer:
Conservation
Explanation:
She has observation conservation because If the temperature of the liquids stays constant and the container is insulated and not heat or cool the liquid much would not change the density of the liquid very much so that it's original volume could remain constant.
The interesting thing is not that the child assumes the taller glass holds more liquid but that they fail to understand conservation: the fact that the water from one glass is going to be the same amount after being poured into any other container. It's as if they did not realize the water came from the same glass.
Projectile Motion: A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will
Answer:
Explanation:
The package had the same velocity as the plane when it was dropped. Newton's 1st Law says that "an object in motion tends to stay in motion, at the same velocity, in a straight line unless acted on by an outside force".
There only outside force acting on the package was its weight -- that force is straight down. The horizontal velocity that the plane gave the package continued (as Newton said it would), so as it fell, horizontally it kept pace with the plane.
If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the coefficient of friction must be so the car doesn’t slide of the road?
Answer:
The minimum friction coefficient required is 0.3
(friction coefficients have no units)
Explanation:
To find a force we need to know something about a mass, and we haven't been told the mass of the car. Let's just call it 'm' and leave it at that for the moment, because it will cancel out in the end.
The centripetal force is given by F = ma = mv2/r
We have values for the velocity and the radius, so:
Fcent=m×6 × 6/13.5 = 2. 667m N
The frictional force must be equal to or greater than this force in order for the car to successfully make it around the curve without sliding out.
The frictional force will be given by:
Ffrict = μFnorm
Where Fnorm is the normal force, equal to mg.
We can equate these two forces, the frictional force and the centripetal force:
Fcent = Ffrict
2.667m=μmg
We can cancel out a factor of m in both sides and rearrange to make μ the subject:
μ = 2.667g
Substituting in the value g=9.8 ms−2,
μ = 2.667/9.8 = 0.27
Approximately = 0.3
Explanation:
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