Answer:
i think the answer is 20.......
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant then changes to k2 = 250 N/m as the spring is stretched to x2 = 65 cm. From x2 = 65 cm to x3 = 89 cm the spring force is constant at F3 = 105 N.
Write an equation for the work done in stretching the spring from x1 to x2.
Calculate the work done, in joules, in stretching the spring from x1 to x2.
Calculate the work, in joules, necessary to stretch the spring from x = 0 to x3.
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
Work done in the springThe work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
where;
k is spring constantΔx is compression of the springW(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
where;
k₁ is first spring constantk₂ is second spring constantF₃ is third force applied to the springW(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
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Need help with this question
Answer:
1. 92, 138, 92
2. 7, 7, 7
3. 17, 18, 17
4. 29, 34, 29
Explanation:
the number on the bottom is always going to be the atomic number, or the number of protons.
the number on the top is going to be the atomic mass number, which is the sum of the number of protons and neutrons.
there is always going to be the same number of electrons as protons, because that was the overall charge of each atom is 0.
for example, for the first one:
there are 92 protons
there are 230 - 92 = 138 neutrons
there are 92 electrons
An object weighs 200N, what is its mass?
Answer:
kg
Explanation:
the highr and jebad to kilogram mizans
Answer:
w=m×g
Explanation:
w is given and take g as 9.8m/s^2
Select. The star cycle that is accurate
One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.
What is the solar system?The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.
The complete question is;
"Choose the correct star life cycle.
A. Supernova, star as well as red giant, the nebula is incorrect.
B. Nebula, white dwarf, planetary nebula That is incorrect.
C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!
D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."
Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.
Hence option C is correct.
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i need help with lab
(a) The equilibrant C for force of vector A and B is 3.43 N.
(b) The equilibrant C for fx of vector A and B is 2.1 N.
(c) The equilibrant C, for fy of vector A and B is 2.12 N.
What is equilibrant force?An equilibrant force is a single force that will bring other bodies into equilibrium.
From configuration 1:Vector A: mass = 0.2 kg, θ = 20⁰
Vector B: mass = 0.15 kg, θ = 80⁰
Fx = mg cosθ
Fy = mg sinθ
where;
m is mass g is acceleration due to gravityVector AForce of A due to its weight
F(A) = mg
F(A) = 0.2 x 9.8 = 1.96 N
Fx = (0.2 x 9.8) cos(20) = 1.84 N
Fy = (0.2 x 9.8) sin(20) = 0.67 N
Resultant forceR = √(0.67² + 1.84²)
R = 1.96 N
Vector BForce of B due to its weight
F(B) = mg
F(B) = 0.15 x 9.8
F(B) = 1.47 N
Fx = (0.15 x 9.8) cos(80) = 0.26 N
Fy = (0.15 x 9.8) sin(80) = 1.45 N
Resultant forceR = √(0.26² + 1.45²)
R= 1.47 N
Equilibrant C of vector A and BEquilibrant force:
Force, C = 1.96 N + 1.47 N
Force, C = 3.43 N
Equilibrant FX:
Fx, C = Fx(A) + Fx(B)
Fx, C = 1.84 N + 0.26 N = 2.1 N
Equilibrant FY:
Fy, C = Fy(A) + Fy(B)
Fy, C =0.67 N + 1.45 N = 2.12 N
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2. Two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M (see figure). M r m L m Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects. Is the rod being compressed, stretched, or neither? If compressed/stretched, calculate the magnitude of this deformation force on the rod. If neither, explain clearly why.
The rod is stretched due to the force of gravitational attraction.
The magnitude of the deformation force on the rod is -GMm×((1/r² + 1/(L+r)²)
What is gravity?The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.
The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.
F = GMm/R²
Given are two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M. Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects.
Let the mass closer to the planet is A and the other is B
FA = Force on planet A = -GMm/r²
FB = Force on planet B = -GMm/(L+r)²
Net Force on the rod is given by the addition of the individual forces.
Fnet= FA + FB = -GMm/r² -GMm/(L+r)²
Fnet = -GMm×((1/r² + 1/(L+r)²)
Thus, the magnitude of the deformation force on the rod is derived above.
A experience more force than B, so A will stretch out from B. Hence the force is stretching the rod.
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The value of acceleration due to gravity (g) on a point 10,000 kilometers above sea level is about 1.49 meters/second2. How much will an object, which weighs 98 newtons on the surface of Earth, weigh on this point? The value of acceleration due to gravity on Earth is 9.8 meters/second2.
An object, which weighs 98 N on the surface of Earth, weigh on this point is 2.23 x 10²⁴ kg.
What is gravity?The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.
Given, the Earth has the acceleration due to gravity, g = 9.81 m/s².
Force of gravity W = mass x acceleration due to gravity
98N = m x 9.8m/s²
m = 10 kg
The value of acceleration due to gravity (g) on a point 10,000 km above sea level is about 1.49 m/s².
The acceleration due to gravity and mass is related as
g = GM/R²
where G = gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg² and R is the distance between two masses.
Substituting the values, we get
1.49 m/s² = 6.67 x 10⁻¹¹ N.m²/kg² x M/ (10000 x 10³ m)²
M = 2.23 x 10²⁴ kg
Therefore, an object will weigh on this point is 2.23 x 10²⁴ kg
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Two identical 3.0-kg cubes are placed on a horizontal surface in
contact with one another. The cubes are lined up from left to
right and a force F₁ is applied to the left side of the left cube
causing both cubes to move at a constant speed v. If the
coefficient of kinetic friction between the cubes and the
surface is 0.3, what is the magnitude of the force exerted by the
left cube on the right cube?
(3
The magnitude of the force exerted by the left cube on the right cube is 17.64N.
What is frictional force?When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.
Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.
From the equilibrium of forces in vertical direction
Normal force N= 2m x g
friction force f = μN =μ(2m)g
From the equilibrium of forces in horizontal direction
F₁ =ma =0
using Newton's third law of motion, we get
F₁ - f =0
F₁ =f = μ(2m)g
Put the values, we get
F₁ = 17.64N
Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.
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Question 2 of 25
What is the term for a pair of forces described by Newton's third law?
A.
Normal-friction pair
B.
Action-reaction pair
C.
Kinetic-static pair
D.
Inertia-movement pair
The term for a pair of forces described by Newton's third law is action-reaction pair. Details about Newton's law can be found below.
What does Newton's third law state?Newton's third law of motion states that for every action, there is an equal and opposite reaction.
This law proposed that when an object 1 acts on object 2 with a force of a particular magnitude, object 2 also acts on object 1 with an opposite force of same magnitude.
Therefore, it can be said that the term for a pair of forces described by Newton's third law is action-reaction pair.
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. Two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as u(r) = − Gm R1 −r − Gm R2 −r . This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets. (a) Draw a graph of u as a function of position (r) along the line between the two planets. (b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2. What is the minimum speed v which will permit the projectile to reach station Beta?
(a) A graph of u as a function of position (r) along the line between the two planets is attached below.
(b) The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
What is gravitational potential energy?If an object is lifted, work is done against gravitational force. The object gains energy.
Given are two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as
u(r) = − (Gm /R1 −r) − (Gm / R2 −r) .
This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets
(a) A graph of u (r) versus position (r) along the line between the two planets is attached in answer.
(b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.
The range of the projectile is given by R = v²sin2θ / g
g = gravitational acceleration of Earth
If g = g(p) for planet , range R = v²sin2θ / g(p)..................(1)
The gravitational force of attraction = weight force
Gm² /d² = m g(p)
g(p) = Gm/d².........................(2)
For R = d/3, from equation (1), we have
d/3 = v²sin2θ / g(p)
Plug the expression for g(p) , we get
v = √ [Gm/3dsin2θ ]
For velocity to be minimum, sin2θ =1
So, the minimum velocity will be
v = √ [Gm/3d]
Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
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A 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal. How much work is done in pulling the sled? What is the acceleration in the x direction, assuming that friction is negligible? Assuming the sled started at rest, how long does it take to pull the sled 185 m?
The work done in pulling the sled is 445,930.2 Joules.
The acceleration in the x direction, assuming that friction is negligible, is 20.85 m/s².
Time taken to pull the sled 185 m is 4.21s.
What is work done?Work done is equal to product of force applied and distance moved.
Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.
Work = Force x Distance x cos(angle)
W= 1225 x 385 x cos 19°
W = 445,930.2 Joules
Thus, the work done in pulling the sled is 445,930.2 Joules
From the Newton's second law of motion, we have'
F = ma
acceleration, a = 1225cos19° / ( 545 /9.81)
a = 20.85 m/s²
Thus, the acceleration in the x direction is 20.85 m/s²
Using the second equation of motion, we get
s = ut+ 1/2 at²
Substitute the values, we have
185 m = 0 +1/2 x 20.85 x t²
t = 4.21 s
Thus, the time taken to pull the sled 185 is 4.21s.
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A body moves in an x-y plane with a velocity vm/s. The component of v are Vx=5m/s, Vy=7m/s along x and y axis respectively. What is the magnitude and direction of the velocity
the answer is in the picture ok
good luck don't worry it's a correct
#carry on learning
What is the acceleration of Karla's I Phone being thrown from Mr. Higley's classroom at 0m/s if it hits the wall 1.2 seconds later going 35 m/s?
The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
What is acceleration?The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.
The given data in the problem is given by ;
u is the initial speed = 0 m/sec
v is the final speed= 35 m/sec
t is the time interval= 1.2 second
a is the acceleration=? m/sec²
The formula for acceleration is;
[tex]\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2[/tex]
Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
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What is the force of gravity in water
A 0.15-kilogram baseball moving at 43 meters per second was stopped by the catcher’s glove in 0.050 seconds. What was the average force exerted on the ball by the glove? Round your answer to two significant figures.
The average force exerted on the ball by the glove will be 129 N.
What is force?Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
Given data;
Mass,(m)=0.15-kilogram
Initial speed,u=43 meters per second
Time period,t= 0.050 seconds
Average force exerted,F=?
The average force exerted on the ball by the glove is;
[tex]\rm F = m\frac{v-u}{t} \\\\ F=0.15( \frac{43-0}{0.050} )\\\\ F=129 N[/tex]
Hence, the average force exerted on the ball by the glove will be 129 N.
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The glucometer applies a potential difference of 0.90 volts across a blood sample. The glucose concentration of the blood sample is 0.98 grams/litre. Determine the current in the blood sample.
The amount of current in the blood sample is determined as 0.12 A.
Current in the blood
The current in the blood is calculated by applying Ohm's law as shown below;
V = IR
where;
V is voltageI is currentR is resistanceFrom resistivity chart corresponding to 0.98 g/L, resistance, R = 7.4
I = V/R
I = 0.9 /7.4
I = 0.12 A
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Two masses, m and 2m, approach each along a path at right angles to each other. After collision, they stick together and move of at 2m/s at angle 37⁰ to the original direction of the mass m. What where the initial speeds of the two particles?
The initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.
Initial speed of the two massesThe initial speed of the two masses is calculated by applying the principle of conservation of linear momentum as follows;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
let m be in horizontal directionlet 2m be in vertical directionangle between the two masses = 90 degreesin y - direction[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1sin(0) + 2mu_2sin(90) = (m + 2m)(2)sin(37)\\\\2mu_2 = 3.61m\\\\u_2 = \frac{3.61 m}{2m} \\\\u_2 = 1.81 \ m/s[/tex]
in x - direction[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1cos(0) + 2mu_2cos(90) = (m + 2m)(2)cos(37)\\\\mu_1 = 4.79 m\\\\u_1 = 4.79 \ m/s[/tex]
Thus, the initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.
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Why is it important to remember "association, not
causation" when discussing correlations?
It is important to remember "association", instead of "causation" when discussing correlations because associations can be used to make predictions.
What is the value of associations in science?The associations among variables have a fundamental value in science because they can determine how such variables behave in hypothetical situations.
Moreover, correlation only indicates a type of association between variables and therefore it is less informative.
In conclusion, it is important to remember "association", instead of "causation" when discussing correlations because associations can be used to make predictions.
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What is the frequency of light with a wavelength of 2.10 × 10-7 m?
Answer:
V = ?
C = 3 × 10^8
wavelength = 2.10 × 10^-7
wavelength = C/V
V = C/wavelength
V = 3 × 10^8/2.10 × 10^-7
V = 1.43 × 10^ 15
A person who falls between introversion and extroversion is
Answer:
the answer is ambivert have a good fourth of July
Answer:
i believe the answer is ambivert
Water flows with a volume flow rate of 1.44 m³/s in a pipe. Find the water speed where the pipe radius is 0.430 m.
Answer:
0.6192
Explanation:
1.44 x 0.430=0.6192
A resistor has a resistance of 8.2 kΩ. If a voltage of 15.9 V were placed across it, what would be the current, in mA? Give the answer to two decimal places; don't worry if the computer adds zeroes.
The value of the electric current for the given conditions willl be 1939.02 A.
What is ohm’s law?Ohm's law claims that the voltage across a conductor is directly proportional to the current flowing through it.
Ohm's law claims that the voltage across a conductor is direct to the current flowing through it. This current-voltage connection may be expressed mathematically as,
V=IR
15.9 V = I × 8.2 × 10³ Ω
I = 1939.02 A
Hence, the value of the electric current for the given conditions willl be 1939.02 A.
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Lucia wants to change the motion map shown so that it’s shows uniform circular motion. What change should Lucia make ?
Answer:
The last one - each vector pointing towards the center of the circle must be the same length for uniform circular motion
what type of energy does a pv cell create ?
Answer: Solar Photovoltaic (PV) cells generate electricity by absorbing sunlight and using that light energy to create an electrical current. There are many PV cells within a single solar panel, and the current created by all of the cells together adds up to enough electricity to help power your school, home and businesses.
--Cited from Solar Schools
The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel from 20 °C to 66 °C. Now imagine those same joules were used instead to accelerate the same mass of nickel from rest. What would be the final speed, in m/s?
Hints:
The formula for kinetic energy is: Ek=1/2 mv^2
For joules needed for heating, use the mass in grams, but for kinetic energy, convert the mass to kilograms.
Round your answer to the nearest whole number.
The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.
Final speed of the nickelApply the principle of conservation of energy.
Q = mcΔθ
Q = (18)(0.444)(66 - 20)
Q = 367.63 J
Q = K.E = ¹/₂mv²
2K.E = mv²
v = √(2K.E/m)
where;
v is the final speedv = √(2 x 367.63)/(0.018))
v = 202.1 m/s
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A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If
the bond length is 2x10 -10 m and the magnitude of the charges involved is
approximately 1.60x10 -20C, what is the force between the molecules involved in the bond?
The force between the molecules involved in the bond is 6. 426 *10^-11 Newton
How to determine the force
Using the formula:
F = K[q1 x q2]/D^2
where K is coulombs constant =9 *10 ^9 Nm^2/C^2.
q1 and q2 = charges = 1.60x10 -20C
d = distance between the charges = 2x10 -10 m
Substitute the values into the formula
F = [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]
F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]
F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]
F = [tex]6. 426 * 10^-11[/tex] Newton
Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton
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Need help with this question!
Answer:(note that values will be on top in small)
cobalt :- 60Co
potassium :- 40K
neon :- 24Ne
lead :- 208Pb
A pendulum swings to and fro every 2.0 s. The period of the swing is.
O 2 Hz
2 s
O 0.5 s
O 0.5 Hz
Answer:
Explanation:
Solution
In physics, a period is a measurement of time. The answer must be either B or C. It is a direct measurement of time, not its reciprocal. So the answer is B. A and D is a measurement of the actual cycle that it takes before the motion begins to repeat itself.
Answer
B
Two force of 20N and 40N act at a
Point and the angle between them is 50 degrees.
Find the resultant force
Find the direction using sine rule
R=√20²+40²+2.20.40 cos 50
R=55 N
R/sin β = F/sin α
55/sin 50 = 40/sin α
α = 33°
The following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s. Find
A, the average velocity of the particle in the time interval t₁=2sec and t₂=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
will give brilliance
A. The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 4 i m/s.
B. The velocity and acceleration at any time t is 4t i +1 j and 4i.
C. The average acceleration in the time interval given in part (a) is 4
What is acceleration?Acceleration can be defined as the rate change of velocity with time.
acceleration a = (Δv) / (Δt)
Given the following equation shows the position of a particle in time t, x=at² i + bt j
where t is in second and x is in meter. a=2m/s², b=1m/s.
A. Velocity is time rate of change of displacement.
V = dx/dt
V = d/dt (2t² i + 1t j)
V = 4t i +1 j
Velocity in time interval,
At t₁=2sec , V₂= 4x2 i + 1 j = 8 i + 1 j
At t₂=3sec, V₃ = 4x3 i + 1 j = 12 i + 1 j
Average velocity in time interval t₁=2sec and t₂=3sec is
, V₃ - V₂ = (12 i + 1 j) - ( 8 i + 1 j )
Average velocity = 4 i
B. Velocity at time t, is
V(t) = V = 4t i +1 j
Acceleration a = dV/dt
a(t) = d/dt (4t i +1 j)
a(t) = 4 i +0 j
C. Average acceleration in time interval t₁=2sec and t₂=3sec is
a = 4 m/s²
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