Answer:
Part A
a) F = -16x + 4, b) x = 0.25 m, c) STABLE
Explanation:
Part A
a) Potential energy and force are related
F = [tex]- \frac{dU}{dx}[/tex]- dU / dx
F = - (8 2x -4)
F = -16x + 4
b) The object is in equilibrium when the forces are zero
0 = -16x + 4
x = 4/16
x = 0.25 m
c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.
In this case there is only one equilibrium point
by changing the position a bit
x ’= x + Δx
we substitute
F ’= - 16 x’ + 4
F ’= - 16 (x + Δx) + 4
F ’= (-16x +4) - 16 Δx
at equilibrium position F = 0
F ’= 0 - 16 Δx
we can see that the body returns to the equilibrium position, therefore it is STABLE
PART B
This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved
initial instant. Before the shock
p₀ = m v
final instant. After the crash
p_f = (m + M) v_f
We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic
p₀ = pf
mv = mv ’+ M v_f
in the case of the elastic collision, the kinetic energy is conserved
K₀ = K_f
½ m v² = ½ m v’² + ½ M v_f²
we write the system of equations
mv = mv ’+ M v_f (1)
m (v² -v'²) = M v_f ²
m (v - v ’) = M v_f
m (v-v ’) (v + v’) = M v_f
v + v ’= v_f
we substitute in equation 1 and solve
v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]
v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]
the mechanical energy of the neutron is
initial
Em₀ = K = ½ m v²
final moment
Em_f = K + U = ½ m v_f ² + U
U is the energy lost in the collision
total energy is conserved
Em₀ = Em_f
½ m v² = ½ m v_f ² + U
U = ½ m (v² -v_f ²)
U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]
U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]
U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]
U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]
Let's do the same calculations for the nucleus
initial Em₀ = 0
final Em_f = K + U = ½ M v_f ² + U
Em₀ = Em_f
0 = K + U
U = -K
U = - ½ M v_f ²
U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²
U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]
We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.
b) the fraction of energy lost
f = U / Em₀
f = 4 m M / m + M
c) let's calculate the fraction of energy lost in a collision
m = 1.67 10⁻²⁷ kg
M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg
f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷
f = 5.92 10⁻²⁷ J
the energy of a fast neutron is greater than 1 eV
Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J
Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo
#_collisions = 0.95 Eo / f
#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷
#_collisions = 2.7 10⁷ collisions
Multi-part question If a galaxy moving away from the Earth has a speed of 1000 km/s and emits 656 nm light characteristic of hydrogen (the most common element in the universe). What wavelength would we observe on the Earth
Answer:
658.2 nm
Explanation:
Since the galaxy is moving at relavitistic speed, we use the equation for relativistic Doppler shift of light.
So, the wavelength of light observed on the Earth is λ
λ = λ'([tex]\sqrt{\frac{ 1 + \frac{v}{c} }{1 - \frac{v}{c} } }[/tex])
where λ' = wavelength of light emitted by galaxy = 656 nm, v = speed of galaxy = 1000 km/s (positive since the galaxy is moving away from the Earth) and c = speed of light = 300000 km/s
So, substituting the values of the variables into the equation, we have
λ = λ'(√[{1 + (v/c)}/(1 - (v/c)]
λ = 656 nm(√[{1 + (1000 km/s/300000 km/s)}/(1 - (1000 km/s/300000 km/s)]
λ = 656 nm(√[{1 + 1/300}/(1 - 1/300]
λ = 656 nm(√[{(300 + 1)/300}/{(300 - 1)/300}]
λ = 656 nm(√[{(301)/300}/{(299)/300}]
λ = 656 nm(√[301/299])
λ = 656 nm(√1.0067)
λ = 656 nm × 1.0033
λ = 658.19 nm
λ ≅ 658.2 nm
So, the wavelength observed on Earth is 658.2 nm
Which court would you go to if you wanted to declare bankruptcy?
A.
Civil court
B.
U.S. Court of Appeals
C.
Supreme Court
D.
U.S. District Court
Answer:
A. Cilvil court.........
Civil court would you go to if you wanted to declare bankruptcy. The correct option is A.
Thus, According to the United States Bankruptcy Code, bankruptcy proceedings are typically filed in a federal bankruptcy court.
The federal court system includes these courts, which deal with Chapter 7, Chapter 11, and Chapter 13 bankruptcy cases.
The additional alternatives (U.S. Court of Appeals, Supreme Court, and U.S. District Court) are not the main courts where bankruptcy cases are generally filed.
Thus, Civil court would you go to if you wanted to declare bankruptcy. The correct option is A.
Learn more about Civil court, refer to the link:
https://brainly.com/question/29614273
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An airplane flying at 116 m/s. E, is accelerated uniformly at the rate of 9.2 m/s2, E, for 13 s. What is its final velocity in m/s?
Answer:
235.6m/s
Explanation:
you have to use one of the kinematic formulas the best that suits the question given in this case you can use v=u+at.since the time, acceleration and initial velocity have been given in the question
v=116+(9.2)13^2
v=116+119.6
=235.6m/s
I hope this helps
Answer me as much as you can
Answer:
b) Fluorescent powder.3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?
Answer:
Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.
What is an effect of continental drift?
Answer: An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.
i. The lift raises a car to a height of 1.8 m using a force of 5500 N. How much work does the lift
perform? (1 point)
Work = force x distance
Work = 5500 x 1.8
Work = 9900 N
The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.
Work = force x distance
Work = 5500 x 1.8
Work = 9900 N
therefore, work does 9900 N
What is work?Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).
What is work and energy?Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.
To learn more about work, refer
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Thiết bị nào sau đây không phải là nguồn điện
A. Ắc quy
B. Pin
C. Máy phát điện
D. Bóng đèn điện
Which of the globes in Figure 7.9 will light up?
Solve the following numerical problems. a) A load of 400N is lifted up by an effort of 100N. If load distance is 20cm, what will be the effort distance? (Ans: 80cm) b) Two boys, Shrijan having weight 600N and Shrijesh having weight 300N are playing see-saw. If Shrijan is sitting at 2m from fulcrum, where should Shrijesh sit from fulcrum to balance Shrijan?(Ans: 4m) c) A lever of length 1m has been used ttoko lift a load of 600N by applying an effort of 200N. If load is at 20cm from fulcrum, calculate mechanical advantage, velocity ratio and efficiency. (Ans: MA = 3, VR = 4, n=75%) d) Study the figure below and find the value of effort. (Ans: 120N) Muyn) 2.5m 600N 0.5m ? Science and Enyin
Answer:
given,
load = 400 N
effort = 100 N
load distance = 20 cm
we know that ,
E*Ed = L*Ld
=100 N* Ed = 400N * 20 cm
=100N * Ed = 8000N/cm
= Ed =( 8000N/cm ) / 100N
= Ed = 80 cm
b. soln.
given,
load = 600 N
load distance = 2 m
effort = 300 N
effort distance = ?
we know that ,
= E *Ed = L * Ld
= 300N * Ed = 600N * 2 m
= 300N * Ed = 1200N/m
=Ed =( 1200N/m ) / 300 N
= Ed = 4 m
C. soln.
given,
load = 600 N
load distance =20 cm
effort = 200 N
effort distance = ?
M.A = ?
V.R = ?
Efficiency = ?
we know that ,
= E *Ed = L *Ld
= 200N * Ed = 600 N * 20 cm
=200 N *Ed = 12000 N/cm
=Ed = ( 12000 N/cm) / 200 N
= Ed = 60 cm
Also,
M.A = load / effort
=600 N / 200 N
= 3
V.R = Ed/ Ld
= 60 cm / 20 cm
= 4
efficiency = ( M.A / V.R ) 100 %
= ( 3 / 4 ) 100%
= 75 %
d. soln.
given,
load = 600 N
load distance = 0.5 m
effort distance = 2.5 m
effort = ?
we know that ,
= E * Ed = L * Ld
= E * 2.5 m = 600 N * 0.5 m
= E * 2.5 m = 300 N / m
= E = ( 300 N / m ) / 2.5 m
= E = 120 N
Why does the weight of a body differ on different celestial bodies of the universe?
A certain superconducting magnet in the form of a solenoid of length 0.28 m can generate a magnetic field of 7.0 T in its core when its coils carry a current of 80 A. The windings, made of a niobium-titanium alloy, must be cooled to 4.2 K. Find the number of turns in the solenoid.
Answer:
The number of turns in the solenoid is 19506.
Explanation:
Length of solenoid, L = 0.28 m
Magnetic field, B = 7 T
Current, I = 80 A
Let the number of turns is N.
The formula used to find the magnetic field is
[tex]B = \frac{\mu 0N I}{L}\\\\7 = \frac{4 \times 3.14\times 10^{-7}\times N\times 80}{0.28}\\\\N = 19506[/tex]
mention various medium level and higher level human resources related to engineering.
Answer:
Higher-level human resources related to engineering are those who:
Assist other engineers with solving complex problems. Work with management to keep the project on track. Mentor younger engineersMedium Level engineers will normally comprise the following:
They are responsible for developing engineering plans and performing calculations with respect to the same. Project workflow management Engineering Support facilitation Project Reporting Coordination of Team MembersCheers
A train travels 600 kilometers in 1 hour. What is the train's velocity in meters/second?
here's the answer to your question
A 250g object hangs from a spring that has a spring constant of 48.0 N/m and oscillates with an amplitude of 5.42cm
1)The magnitude of the objects acceleration when the displacement is 4.27 cm (down) is __ m/s^2
2)Given that the object has an amplitude of 5.42 cm the maximum speed that the object is __m/s
Explanation:
Given that,
Mass of an object, m = 250 g = 0.25 kg
Spring constant, k = 48 N/m
The amplitude of the oscillation, A = 5.42 cm = 0.0542 m
1. At equilibrium,
ma = kx
Where
a is the acceleration of the object
So,
[tex]a=\dfrac{kx}{m}\\\\a=\dfrac{48\times 0.0542}{0.25}\\\\a=10.4\ m/s^2[/tex]
2. The maximum speed of the object is :
[tex]v=A\omega\\\\v=A\sqrt{\dfrac{k}{m}}\\\\v=0.0542\times \sqrt{\dfrac{48}{0.25}}\\\\v=0.75\ m/s[/tex]
Hence, this is the required solution.
A tray containing 0.20kg of water at 20degree celsius is placed in a freezer. The temperature of the water drops to 0degree celsius in 10 minutes. Calculate
a) The energy lost by the water asit cools to 0 degree celsius.
b) The average rate at which the water is losing energy in J/s.
c) Estimate the time taken for the water at 0 degree celsius to turn completely into ice.
d) state any assumptios you make
Answer:
a. Energy lost, Q = 16,800 Joules.
b. Power = 28 J/s
c. Time, t = 2357.14 seconds
d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
Explanation:
Given the following data;
Mass = 0.20 kgInitial temperature, T1 = 20°CFinal temperature = 0°CTime = 10 minutesa. To find the energy lost by the water as it cools to 0 degree celsius;
Mathematically, heat capacity is given by the formula;
[tex] Q = mcdt [/tex]
Where;
Q represents the heat capacity or quantity of heat.M represents the mass of an object.C represents the specific heat capacity of water.dt represents the change in temperature.dt = T2 - T1
dt = 20 - 0
dt = 20°C
We know that the specific heat capacity of water is equal to 4200 J/kg°C
Substituting the values into the formula, we have;
[tex] Q = 0.20 * 4200 * 20 [/tex]
Energy lost, Q = 16,800 Joules.
b. To find the average rate at which the water is losing energy in J/s by using the following formula;
[tex] Power = \frac {energy}{time} [/tex]
First of all, we would have to convert the value of time in minutes to seconds.
Conversion:
1 minute = 60 seconds
10 minutes = X seconds
Cross-multiplying, we have;
X = 60 * 10
X = 600 seconds
Substituting the values into the formula, we have;
[tex] Power = \frac {16800}{600} [/tex]
Power = 28 J/s
c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;
We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.
Mathematically, the latent heat of fusion is calculated by using the formula;
Energy, Q = ml = pt
Substituting the values into the formula, we have;
0.20 * 3.3 * 10⁵ = 28 * t
0.20 * 330000 = 28t
66000 = 28t
[tex] t = \frac {66000}{28} [/tex]
Time, t = 2357.14 seconds.
d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
What do is mean by environment friendly behaviour?
Calculate the efficiency of the following appliances:
1. A radiator that converts 1000) of electrical energy into 900J
of heat energy and 100J of light energy
2. A torch that converts 100J of chemical energy into 35) of
light energy and 65J of heat energy
3. A car that converts 10,000J of chemical energy into 6000) of
kinetic energy and 4000J of heat energy.
4. An energy saver light converts 1,000J of electrical energy
into 7003 of light energy and 300J of heat energy.
5. A speaker converts 100J of electrical energy into 50J of
sound energy and 50J of heat and kinetic energy.
we
Answer:
1. The efficiency of the radiator is 90 %
2. The efficiency of the torch is 65 %
3. The efficiency of the car is 40 %
4. The efficiency of the energy saver is 70 %
5. The efficiency of the speaker is 50 %
Explanation:
Efficiency = (Useful energy out ÷ Total energy in) × 100 J
1. Useful energy = 900 J
The total energy in = 1000 J
The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %
2. Useful energy = 65 J
The total energy in = 100 J
The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %
3. Useful energy = 4,000 J
The total energy in = 10,000 J
The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %
4. Useful energy = 700 J
The total energy in = 1,000 J
The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %
5. Useful energy = 50 J
The total energy in = 100 J
The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %
Pls help me with this fast. I will mark brainiest
Answer:
a) 70, 95
b) 95-70= 25cc
c) density= mass/volume
102/25
=4.08g/cc
Why is velocity proportial to pressure?
Answer:
Pressure and velocity are inversely proportional to each other because if pressure increase, the velocity decrease to keep the algebraic sum of potential energy, kinetic energy and pressure constant.
What's an area of the earth that suggest that " the Earth is like one giant jigsaw" idea could be true.
an athlete had lifts a load with a mass of 150kg.
1) calculate the gravitational potential energy gained
2) if the mass of the load is increased to 200kg, calculate the gravitational potential energy gained by the load.
3) based on answers in 1&2, state the relationship between the mass of the load and the gravitational potential energy.
[tex]\\ \sf\longmapsto P.E=mgh[/tex]
[tex]\\ \sf\longmapsto P.E=150(2)(10)[/tex]
[tex]\\ \sf\longmapsto P.E=3000J[/tex]
In 2nd case
Mass =m=200kgHeight=h=2mg=10m/s^2[tex]\\ \sf\longmapsto P.E=200(2)(10)[/tex]
[tex]\\ \sf\longmapsto P.E=4000J[/tex]
We can observe that
If mass of body increases gravitational potential energy will increase.A tennis player hits a ball hard and 0.80 seconds later hears the echo from a wall. The speed of sound in air is 340m/s how far away is the player from the wall
Answer:
136 m.
Explanation:
From the question given above, the following data were obtained:
Time (t) = 0.80 s
Speed of sound (v) = 340 m/s
Distance (x) =?
The distance of the player from the wall can be obtained as follow:
v = 2x / t
340 = 2x / 0.80
Cross multiply
2x = 340 × 0.80
2x = 272
Divide both side by 2
x = 272 / 2
x = 136 m
Thus, the distance of the player from the wall is 136 m
what you filling your heart with
oxygen and blood
Answer:
Explanation:
The right side of your heart receives oxygen-poor blood from your veins and pumps it to your lungs, where it picks up oxygen and gets rid of carbon dioxide. The left side of your heart receives oxygen-rich blood from your lungs and pumps it through your arteries to the rest of your body.
#I AM ILLITERATE
4 A student says that he has made a magnetic field with some iron filings. What would
you say to him?
Answer:
Yes but no
Explanation:
Here the kid cant say that he has created s magnetic field as iron fillings do not have magnetic properties
define one standard kg
Answer:
The mass of platinum iridium rod whose diameter and height are equal an kept at international bureau of weight and measurement in Paris of France is called as one standard kg.
Discuss the role of globalization in the development of sI unit
Answer:
Sharing of informationExplanation:
The development of SI unit has helped in the sharing of scientific as well as techical information internationally.Answer:
It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.
pls someone help me pls…. and pls explain to me how
Answer:
1.12 × 10⁴ m/s
Explanation:
The escape velocity of the object v = √(2GM/R) where G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of the Earth = 6 × 10²⁴ kg and R = radius of the Earth = 6.4 × 10⁶ m
Since v = √(2GM/R)
Substituting the values of the variables into the equation, we have
v = √(2GM/R)
v = √(2 × 6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)
v = √(13.34 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)
v = √(80.04 × 10⁻¹¹ × 10²⁴Nm²/kg/6.4 × 10⁶ m)
v = √(80.04 × 10¹³Nm²/kg ÷ 6.4 × 10⁶ m)
v = √(80.04 ÷ 6.4 × 10¹³ ÷ 10⁶Nm/kg)
v = √(12.50625 × 10⁷ Nm/kg)
v = √(125.0625 × 10⁶ Nm/kg)
v = 11.18 × 10³ m/s
v = 1.118 × 10 × 10³ m/s
v = 1.118 × 10⁴ m/s
v ≅ 1.12 × 10⁴ m/s
Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=4.00×103 kg and the second a mass of m2=7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?
Answer:
Their final relative velocity is 0.190 m/s
Explanation:
The relative velocity of the satellites, v = 0.190 m/s
The mass of the first satellite, m₁ = 4.00 × 10³ kg
The mass of the second satellite, m₂ = 7.50 × 10³ kg
Given that the satellites have elastic collision, we have;
[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]
[tex]v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]
Given that the initial velocities are equal in magnitude, we have;
u₁ = u₂ = v/2
u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s
v₁ and v₂ = The final velocities of the satellites
We get;
[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]
[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]
The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂
∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190
The final relative velocity of the satellite, [tex]v_f[/tex] = 0.190 m/s
Which of the following processes occurs in a battery?
Batteries convert mechanical energy into chemical energy.
Batteries convert chemical energy into kinetic energy.
Batteries convert mechanical energy into electric energy.
Batteries convert chemical energy into electric energy.
Answer:
Batteries convert chemical energy into electric energy.Explanation:
As seen, batteries have chemical energy in them stored which when used in any electronics produces electricity.
Thus, we can conclude that a battery converts chemical energy into electric energy.
