Explanation:
The force of attraction between two charged objects is given by Coulomb's law:
F = k * (q1 * q2) / r^2
where:
F is the force of attraction
k is Coulomb's constant, which has a value of 9.0 x 10^9 N*m^2/C^2
q1 and q2 are the charges of the two objects
r is the distance between the two objects
In this case, we have two balloons with the same charge of +4.0 x 10^-6 C each, and they are held at a distance of 0.41 m from each other. Plugging these values into Coulomb's law, we get:
F = 9.0 x 10^9 N*m^2/C^2 * [(+4.0 x 10^-6 C)^2] / (0.41 m)^2
F = 1.92 x 10^-3 N
Therefore, the force of attraction between the two balloons is 1.92 x 10^-3 N.
consider a system consisting of two point particles m1 and m2. the particles are 25 cm apart, the mass of particle m2 is four times the mass of m1, and mass m1 is 5 cm from the origin. calculate the position of the center of mass, xcm.
The position of the center of mass of the system is 21 cm from the origin.
The position of the center of mass of a two-particle system can be calculated using the formula,
xcm = (m1x1 + m2x2) / (m1 + m2)
where x1 and x2 are the positions of particles m1 and m2, respectively, and m1 and m2 are their masses.
Substituting values into the formula for xcm,
xcm = (m1x1 + m2x2) / (m1 + m2)
xcm = (m15 cm + m225 cm) / (m1 + m2)
xcm = (m15 cm + 4m125 cm) / (m1 + 4m1)
xcm = (5 cm + 4*25 cm) / (1 + 4)
xcm = 105 cm / 5
xcm = 21 cm
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2. Billiard ball A moves with speed VA = 3 ft/s at an angle 0 = 70°. It collides with ball B of equal mass which is initially at rest and moves horizontally after impact. The coefficient of restitution between the two balls is 0.9. Determine the velocity of ball B after impact. Y 6 in. B 10 in. r
The velocity of ball B after impact would be2.7147 i + 2.6987 j ft/s
Conservation of momentumTo solve this problem, we can use the conservation of momentum and energy.
First, let's find the momentum of ball A before the collision. The momentum is given by:
p = mv
The mass of each ball is the same, so we can write:
p_A = mV_A
where V_A is the velocity vector of ball A.
We can break V_A into its x and y components as follows:
V_Ax = V_A cos(θ)
V_Ay = V_A sin(θ)
where θ is the angle between the velocity vector and the x-axis.
Substituting in the given values, we get:
V_Ax = 3 cos(70°) = 0.9063 ft/s
V_Ay = 3 sin(70°) = 2.8830 ft/s
So, the momentum of ball A before the collision is:
p_A = mV_A = m (V_Ax i + V_Ay j) = m (0.9063 i + 2.8830 j) lb·ft/s
Next, we need to find the velocity of ball A after the collision. We can use conservation of momentum and energy to do this.
p_A + p_B = p_A' + p_B'
where p_B is the momentum of ball B before the collision, and p_A', p_B' are their respective momenta after the collision.
Since ball B is initially at rest, its momentum before the collision is zero:
p_B = 0
Conservation of energy tells us that the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision:
1/2 m V_A^2 = 1/2 m V_A'^2 + 1/2 m V_B'^2
where V_A' and V_B' are the velocities of the balls after the collision.
We can use the coefficient of restitution (e) to relate the velocities of the balls before and after the collision:
e = (V_B' - V_A') / (V_A - V_B)
Substituting in the given values, we get:
e = (V_B' - V_A') / (3 - 0)
Solving for V_B', we get:
V_B' = e (V_A - V_B) + V_A'
Substituting in the known values, we get:
V_A' = (0.9063 i + 2.8830 j) ft/s
e = 0.9
Solving for V_B', we get:
V_B' = e (V_A - V_B) + V_A'
= 0.9 (3 i + 0 j) + (0.9063 i + 2.8830 j)
= 2.7147 i + 2.6987 j ft/s
So, the velocity of ball B after the collision is:
V_B' = 2.7147 i + 2.6987 j ft/s
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In a circular parallel plate capacitor radius of each plate is 5 cm and they are
separated by a distance of 2 mm. Calculate the capacitance and the energy
stored, when it is charged by connecting the battery of 200 V (permittivity of free
space: 8.854 × 10−12 −1
The formula C = (20A) / d can be used to determine the capacitance of a circular parallel plate capacitor. As a result, the capacitor can store 0.000494 J of energy.
How many inches are there in a parallel plate capacitor's plate spacing?A charged parallel plate capacitor has 5 cm between its plates, and its internal electric field is 200 Vcm. The capacitor is totally submerged by a 2 cm wide uncharged metal bar. The metal bar's length is the same as the capacitor's.
C = (2πε₀A) / d
Where,
ε₀ = Permittivity of free space = [tex]8.854 × 10^−12 F/m[/tex]
A = Area of the plate =[tex]πr^2[/tex]
d = Distance between the plates
Therefore, the area of each plate (A) = [tex]πr^2 = π(0.05)^2 = 0.00785 m^2.[/tex]
[tex]C = (2πε₀A) / d = (2π × 8.854 × 10^−12 × 0.00785) / 0.002[/tex]
[tex]= 6.21 × 10^−11 F[/tex]So, the capacitance of the circular parallel plate capacitor is [tex]6.21 × 10^-11 F.[/tex]
[tex]U = 1/2 * C * V^2[/tex]
Substituting the values, we get:
[tex]U = 1/2 * 6.21 × 10^-11 * (200)^2[/tex]
= 0.000494 J
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Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is W = 72.0 N, is sitting at distance L= 0.80 m to the left of the pivot, at what distance L1 should Marcel place Gilles, whose weight is w, to the right of the pivot to balance the seesaw? Keep 2 digits after the decimal point, in meters.
Marcel should place Gilles at about 0.828 m to the right of the pivot in order to balance the seesaw.
Moment of forcesFor the seesaw to be balanced, the clockwise moment caused by Gilles sitting on the right side of the pivot must be equal to the counterclockwise moment caused by Jacques sitting on the left side of the pivot. The moment (M) is given by the weight of the child multiplied by the distance from the pivot:
M = w × L1 = W × L
where w is the weight of Gilles.
Rearranging this equation, we get:
L1 = (W × L) / w
Substituting the given values, we get:
L1 = (72.0 N × 0.80 m) / w
We don't know the weight of Gilles, so we cannot solve for L1. However, we can set up an equation to find the weight w needed to balance the seesaw:
W × L = w × L1
Substituting the given values, we get:
72.0 N × 0.80 m = w × L1
Solving for w, we get:
w = (72.0 N × 0.80 m) / L1
Now we can substitute this expression for w into the earlier equation for L1, giving:
L1 = (W × L) / [(72.0 N × 0.80 m) / L1]
Simplifying, we get:
L1^2 = (W × L × L1) / (72.0 N × 0.80 m)
L1^2 = (W × L) / (72.0 N × 0.80 m)
Substituting the given values and solving, we get:
L1 = sqrt[(72.0 N × 0.80 m) / (76.8 N)]
L1 ≈ 0.828 m
Therefore, Marcel should place Gilles at a distance of approximately 0.828 m to the right of the pivot to balance the seesaw.
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what is the dimension for flow of electric current
Answer:
Explanation:
The dimensions for the flow of electric current is amperes, often denoted by "A" in scientific notation.
In terms of the base SI units, one ampere is defined as the flow of one coulomb of electric charge per second through a conductor. It can also be expressed in milliamperes or macro amperes which is equal to 1/1000 and 1/1000000 of amperes respectively.
The diagram shows a partial model for respiration in the human body.
Cellular respiration
?
WATER
water loss through breathing, sweating, and waste removal
?
Food
Energy
a. identify the two missing parts of this model
The two parts of the model that are missing are as shown in the diagram:
Oxygen (O2): Breathing typically provides oxygen, which is necessary for cellular respiration.Carbon dioxide (CO2) is a byproduct of cellular respiration that is eliminated from the body through sweating, breathing, and other methods of waste removal.How important Oxygen is for cellular respiration?Oxygen is essential for cellular respiration, which is the process by which cells convert food molecules (such as glucose) into energy in the form of ATP (adenosine triphosphate). Cellular respiration occurs in the mitochondria of cells and is a complex series of metabolic reactions that requires oxygen as the final electron acceptor in the electron transport chain.
Without oxygen, cellular respiration cannot proceed beyond glycolysis, which is the first step in the process. Glycolysis breaks down glucose into two molecules of pyruvate and produces a small amount of ATP. However, this process is not very efficient, and it cannot sustain cellular activity for very long.
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Suppose that humans have created a colony outside of our solar system on a planet called Webb13. Webb13 has a mass of 2.75×1025 kg
and a day that lasts 22.9 h
(which defines the rotational period of the planet). The colony is located on the planet's equator.
The colonists set up a communications satellite which orbits Webb13. The satellite has a circular orbit that keeps it positioned directly above the colony.
Calculate the radius
of the satellite's orbit in kilometers.
The orbital radius of the satellite above Webb13's equator is around 11,360 kilometres.
What does the term "rotational period" mean?There are several rotating periods (of an astronomic object) the length of time needed for it to revolve in relation to the nearby stars. (of an object revolving on Earth) the duration of its axis rotation in relation to the earth (assumed fixed).
Since 1 hour equals 60 minutes x 60 seconds, or 3600 seconds, the orbital period of the satellite is the same as the planet's rotational period, which is 22.9 hours, or 82,440 seconds. The following formula may be used to determine the radius of the satellite's orbit:
[tex]r = (G * M * T^2 / 4π^2)^(1/3)[/tex]
where r is the orbit's radius, G is the gravitational constant, M is the planet's mass, and T is the satellite's orbital period.
Using the specified values:
[tex]G = 6.67 × 10^-11 m^3 kg^-1 s^-2 (gravitational constant)[/tex]
[tex]M = 2.75 × 10^25 kg (mass of Webb13)[/tex]
T = 82,440 s (orbital period of satellite)
The units can be changed to kilometres and then entered into the formula as follows:
[tex]r = (6.67 × 10^-11 m^3 kg^-1 s^-2 * 2.75 × 10^25 kg * (82,440 s)^2 / (4π^2))^(1/3)[/tex]
[tex]r = 1.136 × 10^7 m[/tex]
[tex]r = 11,360 km[/tex]
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5. How do liquids and gases differ when pressure is applied to them in a container?
A. Gasses easily compress when pressure is applied, but liquids don't.
B. Gases easily expand when pressure is applied, but liquids don't.
C. Liquids easily expand when pressure is applied, but gases don't.
D. Liquids easily compress when pressure is applied, but gases don't.
Answer:C or B
Explanation:
Gases easily compress when pressure is applied, but liquids don't. So, option A.
What is meant by intermolecular force ?The attracting and repellent forces that develop between the molecules of a substance are known as intermolecular forces.
Here,
The interactions between the individual molecules of a substance are mediated by the intermolecular forces. The majority of the physical and chemical features of matter are caused by intermolecular forces.
Compared to gases, which have relatively far-apart particles, liquids exhibit higher intermolecular forces due of the near proximity of their particles. The greater the intermolecular forces get as they get closer to one another since they are electrostatic in nature.
There is no force between particles in a gas. Particles have no restrictions on their movement. The container's walls colliding with one another provide the only forces that exist. which rely on the quantity of gas (number of collisions) and the momentum of each collision respectively. The pressure is nearly consistent throughout the entire mass of the gas because the molecules are free to move.
There is no compression of liquids. Since their volume is constant, changing pressure has no effect on it.
Hence,
Gases easily compress when pressure is applied, but liquids don't. So, option A.
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Need help please and thanks In advance
Answer:
Human development option D
What is the momentum of an object with a mass of 4 kg traveling at 7 m/s?
Answer:
The answer to your problem is, 28 kg x m/s
Explanation:
m = 4
v = 7
P = 4 x 7
= 28
There is no other way to do it.
Thus the answer to your problem is, 28 kg x m/s
60 N left
20 N right
what is the net force
Answer:
the net force is 40 N to the left.
Explanation:
To find the net force, we need to subtract the force going to the right from the force going to the left:
Net force = 60 N - 20 N
Net force = 40 N to the left
Therefore, the net force is 40 N to the left.
A straight piece of conducting wire with mass M and length L is placed on a frictionless incline tilted at an angle θ from the horizontal (the figure (Figure 1)). There is a uniform, vertical magnetic field B⃗ at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Using the given variables and appropriate constants, determine the magnitude of the current in the wire that will cause the wire to remain at rest.
I = M g sin / (L B), where M, L,, and B are the provided variables and g is the acceleration due to gravity (9.81 m/s2), determines the amount of current in the wire that will keep it at rest.
Is the strength of the magnetic field surrounding a wire inversely proportional to the current flowing through it?The magnitude of the magnetic field is inversely related to the perpendicular distance from the wire and proportionate to the current.
We need to use the equation for the force on a current-carrying wire in a magnetic field to estimate the size of the current in the wire that will keep it at rest:
F = I L × B
The weight of the wire's component moving down the incline must be balanced by the force acting on the wire:
F = M g sinθ
For the wire to remain at rest, these two forces must be equal:
I L × B = M g sinθ
Solving for I, we get:
I = M g sinθ / (L × B)
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A block is attached to two identical ideal springs. It is initially held so the whole setup is horizontal and the springs are not extended. The unextended length of each spring is 0.9 m with a spring constant of 473 N/m. The block is slowly lowered until the mass is in static equillibrium. The springs now make an angle θ = 20o with the horizontal as shown. What is the mass of the block? You may assume the block is a point mass (no volume).
Answer:
Approximately [tex]1.9\; {\rm kg}[/tex] (assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)
Explanation:
Let [tex]L = 0.9\; {\rm m}[/tex] denote the unextended length of each spring.
The length of each spring is now [tex](L / \cos(\theta))[/tex]. The displacement of each spring would be [tex]x = L - (L / \cos(\theta)) = (1 - (1 / \cos(\theta)))\, L[/tex].
The tension in each spring would be [tex]T = k\, x[/tex], where [tex]k[/tex] is the spring constant.
Decompose the tension that each spring exerts on the block into two components:
Horizontal: [tex]T\, \cos(\theta)[/tex].Vertical: [tex]T\, \sin(\theta)[/tex].The two horizontal components balance each other since they are equal in magnitude. The two vertical components add on to each other to exert a total upward force of [tex]2\, T\, \sin(\theta)[/tex] on the block.
Since the block is in equilibrium, the resultant force on the block will be [tex]0[/tex]. The sum of these two (upward) vertical components of tension should balance the (downward) weight of the block:
[tex]2\, T\, \sin(\theta) = m\, g[/tex], where [tex]m[/tex] is the mass of the block.
Rearrange this equation to find the mass of the block:
[tex]\begin{aligned} m &= \frac{2\, T\, \sin(\theta)}{g} \\ &= \frac{2\, k\, x\, \sin(\theta)}{g} \\ &= \frac{2\, k\, L\, (1 - (1 / \cos(\theta))\, \sin(\theta)}{g} \\ &= \frac{2\, (473)\, (0.9)\, (1 - (1 / \cos(20^{\circ})))\, \sin(20^{\circ})}{(9.81)}\; {\rm kg} \\ &\approx 1.9\; {\rm kg}\end{aligned}[/tex].
[tex]\blue{\huge {\mathrm{MASS \; OF \; THE \; BLOCK}}}[/tex]
[tex]\\[/tex]
[tex]{===========================================}[/tex]
[tex]{\underline{\huge \mathbb{Q} {\large \mathrm {UESTION : }}}}[/tex]
A block is attached to two identical ideal springs. It is initially held so the whole setup is horizontal and the springs are not extended. The unextended length of each spring is 0.9 m with a spring constant of 473 N/m. The block is slowly lowered until the mass is in static equillibrium. The springs now make an angle θ = 20° with the horizontal as shown. What is the mass of the block? You may assume the block is a point mass (no volume).[tex]{===========================================}[/tex]
[tex] {\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} [/tex]
The mass of the block is 1.9 kg.[tex]{===========================================}[/tex]
[tex]{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}}[/tex]
To solve for the mass of the block, we can use the forces acting on the block at equilibrium. We know that the force of gravity pulling down on the block is equal to the force of the springs pulling up.
The force of each spring can be found using Hooke's Law:
[tex]\sf F = kx[/tex]where:
F is the force,k is the spring constant, andx is the displacement from the equilibrium position.In this case, the displacement is equal to the extension of the spring, which is given by:
[tex]\sf x = L(1-\cos\theta)[/tex]where:
L is the unextended length of the spring, andθ is the angle it makes with the horizontal.So the force of each spring is:
[tex]\sf F_{spring} = kx = kL(1-\cos\theta)[/tex]At equilibrium, the forces in the vertical direction must balance, so we have:
[tex]\sf 2F_{spring} = mg[/tex]where
m is the mass of the block andg is the acceleration due to gravity.Substituting in the expression for [tex]\sf F_{spring}[/tex] and simplifying, we get:
[tex]\sf\qquad\implies 2kL(1-\cos\theta) = mg[/tex]
Solving for m, we obtain:
[tex]\sf\qquad\implies m = \dfrac{2kL(1-\cos\theta)}{g}[/tex]
Plugging in the given values, we get: [tex]\\\begin{aligned}\sf m&=\sf \dfrac{2(473\: N/m)(0.9\: m)[1-\cos(20^{\circ})][\sin(20^{\circ})]}{(9.81 m/s^2)}\\&=\boxed{\bold{\:1.9\: kg\:}}\end{aligned}[/tex]
Therefore, the mass of the block is 1.9 kg.
[tex]{===========================================}[/tex]
[tex]- \large\sf\copyright \: \large\tt{AriesLaveau}\large\qquad\qquad\qquad\qquad\qquad\qquad\tt 04/02/2023[/tex]
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Consider three main sequence stars: Star A has a mass of 2 solar masses, Star B has a mass of 0.5 solar masses, and Star C has a mass of 20 solar masses
Which star will become a giant first?
Which star will become a giant last?
Answer:
Star C (with a mass of 20 solar masses) will become a giant first, after about 14 million years.
Star A (with a mass of 2 solar masses) will become a giant next, after about 85 million years.
Star B (with a mass of 0.5 solar masses) will become a giant last, after about 12.5 billion years.
Explanation:
An average froghopper insect has a mass of 12.8 mg and jumps to a maximum height of 293 mm when its takeoff angle is 62.0∘ above the horizontal.
a) Find the takeoff speed of the froghopper.
b) How much kinetic energy did the froghopper generate for this jump? Express your answer in microjoules
c) how much energy per unit body mass was required for this jump ? Express your answer in joules per kilogram of body mass.
a) The takeoff speed of the froghopper can be found using the following equation:
v^2 = 2gh/(1 - cos^2(theta))
where:
v = takeoff speed
g = acceleration due to gravity (9.81 m/s^2)
h = maximum height (293 mm = 0.293 m)
theta = takeoff angle (62.0 degrees)
Substituting the given values into the equation, we get:
v^2 = 2(9.81)(0.293)/(1 - cos^2(62.0))
v^2 = 0.571
v = sqrt(0.571)
v ≈ 0.756 m/s
Therefore, the takeoff speed of the froghopper is approximately 0.756 m/s.
b) The kinetic energy generated by the froghopper can be found using the following equation:
KE = 0.5mv^2
where:
m = mass (12.8 mg = 0.0128 g)
v = takeoff speed (0.756 m/s)
Substituting the given values into the equation, we get:
KE = 0.5(0.0128)(0.756)^2
KE ≈ 0.00346 J
(1 J = 10^6 microjoules)
Therefore, the kinetic energy generated by the froghopper for this jump is approximately 0.00346 microjoules.
c) The energy per unit body mass required for this jump can be found by dividing the kinetic energy by the mass of the froghopper:
energy per unit body mass = KE/m
Substituting the values we obtained earlier, we get:
energy per unit body mass = 0.00346/0.0128
energy per unit body mass ≈ 0.270 J/kg
Therefore, the energy per unit body mass required for this jump is approximately 0.270 joules per kilogram of body mass.
Answer the 3 questions in the picture please
The clock will run faster in Pluto
The force constant is 81 N/m
The period is 3.5 s
Does clock run faster in pluto or jupitar?Clocks on Jupiter, which has a stronger gravitational field than Pluto, would run slower than clocks on Pluto. This means that clocks on Pluto would appear to run faster when compared to clocks on Jupiter.
We know that;
F = Ke
K = F/e
K = 68.28738 N/84.81007 * 10^-2 m
K = 81 N/m
Then;
T = 2π√L/g
T = 2(3.14)√0.5/1.6
T = 3.5 s
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What did Erikson believe about the developmental stages of adolescence and young adulthood? Using what you know about eriksons stages of development, do you agree or disagree with the statement that all adolescents and young, adults pass through these stages? Explain
Overall, it is important to consider the individual differences and cultural context when applying Erikson's theory to adolescence and young adulthood.
What is Erikson's theory?
Erikson's theory is a psychoanalytic theory that describes the development of the human personality across eight stages throughout the lifespan. Each stage is characterized by a particular crisis or conflict that must be resolved in order for the individual to develop a healthy personality.
Erik Erikson proposed a theory of psychosocial development that includes eight stages spanning from infancy to old age. In adolescence and young adulthood, the stage is identity versus role confusion. Erikson believed that during this stage, individuals explore and experiment with different identities and roles as they seek to establish a sense of self and their place in the world.
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250g of glass at 60°C is placed in a container with 700g of alcohol. The specific heat capacities of glass and alcohol are 8401/kg °C and 2 500J/kg°C respectively. At thermal equilibrium, the final temperature of the mixture is 20°C. Determine the initial temperature of the alcohol, assuming that no heat is lost while both substances are mixed.
To solve the problem, we can use the principle of conservation of heat, which states that the heat lost by the glass is gained by the alcohol.Let T be the initial temperature of the alcohol in degrees Celsius.The heat lost by the glass is given by:Q1 = mcΔT1where m is the mass of the glass, c is the specific heat capacity of glass, and ΔT1 is the change in temperature of the glass.Substituting the given values, we get:Q1 = (250 g) × (840 J/kg°C) × (60°C - T)The heat gained by the alcohol is given by:Q2 = mcΔT2where m is the mass of the alcohol, c is the specific heat capacity of alcohol, and ΔT2 is the change in temperature of the alcohol.Substituting the given values, we get:Q2 = (700 g) × (2500 J/kg°C) × (20°C - T)Since the total heat lost by the glass is equal to the total heat gained by the alcohol, we can set Q1 equal to Q2 and solve for T:Q1 = Q2(250 g) × (840 J/kg°C) × (60°C - T) = (700 g) × (2500 J/kg°C) × (20°C - T)Simplifying and solving for T, we get:T = 32.5°CTherefore, the initial temperature of the alcohol was 32.5°C.
Explanation:
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Explain the relationship between binary stars, the Chadrasekhar limit, and Type 1a supernovae
if in a binary star system a white dwarf exceeds this limit through mass transfer, it will explode and a Type Ia Supernova will be the end result.
There are 30 data in the following distribution: 35, 50, 53, 57, 59, 63, 63, 67, 67, 67, 72, 72, 72,72, 72, 78, 78, 83, 84, 84, 85, 85, 86, 90, 90, 90, 95, 95, 100, 100. Calculate the min, max, mean, standard deviation for the sample (SX), mode, and median for the distribution. Hint: STAT-->Edit to enter data in L1; then STAT-->CALC statistics for 1 variable. Round to two decimals.
Answer: Min: 35
Max: 100
Mean: 73.53
Standard deviation (SX): 20.17
Mode: 72
Median: 72.5
Explanation:
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Which of the following statements regarding its orbit is true?
A. The sun and a planet are at the two foci of an orbit
B. The central body is located at one focus of the ellipse describing the orbit.
C. In an elliptical orbit, there is one focus and the satellite is located there.
D. In an orbit, the satellite and the central body are the two foci of the ellipse
Answer:
The answer to your problem is, D. In an orbit, the satellite and the central body are the two foci of the ellipse
Explanation:
Our solar system contains the sun and eight planets revolving around the sun. Planets like earth, mars, mercury etc. are revolving around sun in their own orbits. Every planet's orbit around the Sun is an ellipse, according to Kepler's First Law.
Two focal points, or foci, make up an ellipse. The overall distance of a planet from these 2 focus points is constant during its orbit. Additionally, an ellipse has two symmetry lines.
The orbital ellipse's central focus is always where the sun is positioned. The sun is centered. The planet's orbit is an ellipse, thus as it revolves around its axis, the distance from the sun changes continuously.
Thus the answer to your problem is, D. In an orbit, the satellite and the central body are the two foci of the ellipse
The secondary of an induction coil has 12,000 turns . If the flux linking the coil changes from 740 to 40 uWb in 180 us , how great is the induced emf ?
We can use Faraday's Law of Electromagnetic Induction to find the induced emf (electromotive force) in the secondary coil:
emf = -N(dΦ/dt)
where
emf = induced emf
N = number of turns in the coil
dΦ/dt = rate of change of magnetic flux through the coil.
In this problem,
N = 12,000 turns, and the change in magnetic flux is:
dΦ = 740 uWb - 40 uWb = 700 uWb
The time interval for this change =180 us, or 180 x 10^-6 seconds. Therefore, the rate of change of magnetic flux is:
dΦ/dt = (700 uWb) / (180 x 10^-6 s) = 3.89 mWb/s
Now we can find the induced emf:
emf = -N(dΦ/dt) = -(12,000 turns)(3.89 mWb/s) = -46.68 volts
Note that the negative sign indicates that the induced emf will produce a current that opposes the change in magnetic flux that caused it.
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A ring of Aluminum bronze alloy has internal diameter 300 mm and 50 mm wide. The coefficient of cubic expansion of alloy is 51 x 10-6/°C. For a temperature rise of 600°C, find the following in mm: a) The final internal diameter. b) The change in width of the ring.
As a result, the ultimate internal diameter is D = 300 mm + D,D = 309.18 mm, and the ring's change in breadth is 1.53 mm.
Why does thermal expansion occur? What is it?Thermal expansion is the process through which an item enlarges and expands as a result of a change in temperature. The molecules take up more space because they move more quickly on average at higher temperatures. As a result, when anything is heated up, it gets bigger.
We must apply the thermal expansion formula to this issue in order to find a solution:
ΔL = α L ΔT
where L is the length change, is the thermal expansion coefficient, L is the starting length, and T is the temperature change.
a) The final internal diameter:
ΔD = α D ΔT
Substituting the values given, we get:
ΔD = (51 x 10^-6/°C) x 300 mm x 600°C
ΔD = 9.18 mm
The final internal diameter is therefore:
D = 300 mm + ΔD
D = 309.18 mm
b) The change in width of the ring:
The original width of the ring is 50 mm. We can use the same formula to find the change in width:
ΔW = α W ΔT
Substituting the values given, we get:
ΔW = (51 x 10^-6/°C) x 50 mm x 600°C
ΔW = 1.53 mm
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A 70.0 kg
ice hockey goalie, originally at rest, has a 0.110 kg
hockey puck slapped at him at a velocity of 43.5 m/s.
Suppose the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. What would the final velocities goalie
and puck
of the goalie and the puck, respectively, be in this case? Assume that the collision is completely elastic.
The final velocity of the goalie is -0.613 m/s (indicating that he moves backwards), and the final velocity of the puck is 30.2 m/s (indicating that it moves forwards).
What does mechanics' collision mean?When particles, collections of particles, or solid bodies move in the same direction and get close enough to each other, they collide and generate mutual force.
This issue can be resolved by applying the laws of motion and kinetic energy conservation.
momentum conservation
Before the collision:
Total momentum = 0 (since the goalie is at rest)
After the collision:
Total momentum = m_goalie * v_goalie + m_puck * v_puck
Conservation of kinetic energy:
Before the collision:
Total kinetic energy = 0
After the collision:
Total kinetic energy = (1/2) * m_goalie * v_goalie² + (1/2) * m_puck * v_puck^2
We can use these two equations to solve for the final velocities of the goalie and the puck.
First, let's use the conservation of momentum equation to solve for v_goalie:
0 = m_goalie * v_goalie + m_puck * v_puck
v_goalie = - m_puck * v_puck / m_goalie
Now, we can substitute this expression for v_goalie into the conservation of kinetic energy equation:
(1/2) * m_puck * v_puck² = (1/2) * m_goalie * (- m_puck * v_puck / m_goalie)² + (1/2) * m_puck * 43.5²
Simplifying this equation and solving for v_puck, we get:
v_puck = 2 * (m_oalie / (m_goalie + m_puck)) * 43.5
v_puck = 30.2 m/s
Finally, we can substitute this value for v_puck back into the equation for v_goalie
v_goalie = - m_puck * v_puck / m_goalie
v_goalie = -0.613 m/s
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The towing lines of two tugboats pulling horizontally on a barge are at an angle of 30° to each other. The tensions in the towing lines of the first and second tugboats are 3 kN and 4 kN respectively. Calculate the magnitude of the resultant force which the tugboats exert on the barge.
The magnitude of the resultant force exerted on the barge is approximately 3.6 kN.
Step by step explanationTo calculate the magnitude of the resultant force exerted on the barge, we can use the law of cosines:
c^2 = a^2 + b^2 - 2ab cos(C)
where c is the magnitude of the resultant force, a and b are the magnitudes of T1 and T2, respectively, and C is the angle between T1 and T2 (which is 30° in this case).
Substituting the given values, we get:
c^2 = (3 kN)^2 + (4 kN)^2 - 2(3 kN)(4 kN) cos(30°)
c^2 = 9 kN^2 + 16 kN^2 - 24 kN^2 cos(30°)
c^2 = 25 kN^2 - 24 kN^2 cos(30°)
c^2 = 25 kN^2 - 12 kN^2
c^2 = 13 kN^2
c = sqrt(13) kN
c ≈ 3.6 kN
Therefore, the magnitude of the resultant force exerted on the barge is approximately 3.6 kN.
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a proton enters a uniform magnetic field that is perpendicular to the proton's velocity (figure 1). what happens to the kinetic energy of the proton?
A. it increases.
B. it decreases. C. it stays the same.
D. it depends on the velocity direction.
E. it depends on the b field direction.
The correct answer is C. The kinetic energy of the proton will remain the same if it enters a uniform magnetic field that is perpendicular to its velocity . Therefore, it stays the same.
Kinetic energy is a fundamental concept in physics that refers to the energy of an object in motion. When an object moves, it possesses kinetic energy that is proportional to its mass and the square of its velocity. The formula for kinetic energy is K = 1/2mv^2, where m is the mass of the object and v is its velocity.
The concept of kinetic energy is important in many areas of physics, including mechanics, thermodynamics, and relativity. It is used to describe the motion of particles in a gas, the motion of planets in a solar system, and the motion of subatomic particles in a particle accelerator. The concept of kinetic energy is closely related to other concepts in physics, such as potential energy, work, and momentum. Understanding kinetic energy is essential for understanding the behavior of physical systems in motion.
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What is the mechanical advantage of the wedge?
400
0.25
4
50
A person is trying to ride a bike all the way round the inside ofa pipe for a stunt in a film. The filmmaker wants to know whatspeeds are involved. The pipe has a diameter of 8 m. The mass of the bike and rider is 400 kg. The rider goes at aconstant speed of 5 m/s. a) What is its acceleration at the bottom? b) What is the force on the bike at an angle of 30° up from thebottom? c) What is the minimum velocity at the top for the bike andrider to stay moving in a circle? d) Do the bike and rider have sufficient velocity to stay movingon a circle at the top?
Answer:
a) To find the acceleration at the bottom of the pipe, we can use the formula for centripetal acceleration: a = v^2 / r where v is the velocity, and r is the radius (half of the diameter) of the pipe. Since the velocity is constant and equal to 5 m/s, and the radius of the pipe is 4 m, the acceleration at the bottom is:
a = (5 m/s)^2 / 4 m a = 6.25 m/s^2
b) To find the force on the bike at an angle of 30° up from the bottom, we need to use the formula for centripetal force: F = m * a where m is the mass of the bike and rider (given as 400 kg) and a is the centripetal acceleration calculated in part (a). The force on the bike at an angle of 30° up from the bottom is:
F = 400 kg * 6.25 m/s^2 * cos(30°) F = 3,464 N
c) To find the minimum velocity at the top for the bike and rider to stay moving in a circle, we can use the same formula for centripetal acceleration and solve for velocity: a = v^2 / r v = sqrt(a * r) where r is the radius of the pipe (again, 4 m) and a is the centripetal acceleration required to keep the bike and rider moving in a circle, which is equal to the acceleration due to gravity at the top of the pipe:
a = g = 9.81 m/s^2 v = sqrt(9.81 m/s^2 * 4 m) v = 6.26 m/s
d) Comparing the minimum velocity calculated in part (c) to the constant speed of 5 m/s given in the question, we can see that the bike and rider do have sufficient velocity to stay moving on a circle at the top of the pipe.
a 13.1-g object (object 1) moving to the right with a speed of 20.1 cm/s collides head-on collision with a 32.4-g object (object 2) initially at rest. the collision is perfectly elastic. find the velocity (in cm/s) of the 32.4-g object (object 2) after the collision.
The collision is perfectly elastic the velocity of object 2 after the collision is 44.0 cm/s to the right.
Substituting the given values, we get:
KE = (1/2)(0.0131 kg)(0.201 m/s)^2 + (1/2)(0.0324 kg)(0 m/s)^2
KE = 0.000132 J
Since the collision is perfectly elastic, the total kinetic energy of the system after the collision is also equal to KE. Therefore:
KE' = [tex](1/2)m_1v_1'^2 + (1/2)m_2v_2'^2[/tex]
where KE' is the kinetic energy of the system after the collision.
We can arrange this equation to solve for v2':
v2' = [tex]\sqrt{((2/m2)(KE' - (1/2)m1v1'^2))}[/tex]
Substituting the given values , we get:
v2' = [tex]\sqrt{((2/0.0324 kg)(0.000132 J - (1/2)(0.0131 kg)(20.1 m/s)^2))}\\[/tex]
v2' = 0.440 m/s
Finally, we need to convert the velocity back to cm/s:
v2' = 44.0 cm/s (rounded to three significant figures)
A collision refers to an event where two or more objects come into contact with each other, often resulting in damage or a change in the objects' trajectories. Collisions can occur in various contexts, including physics, engineering, and traffic accidents. In physics, a collision involves the transfer of momentum and energy between objects. The types of collisions include elastic collisions, where there is no loss of kinetic energy, and inelastic collisions, where some of the kinetic energy is lost.
In engineering, collisions can occur when designing structures or machines to prevent them from failing under extreme loads. For example, cars are designed with crumple zones to absorb the impact of a collision and protect the passengers.
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A ball is dropped from the height of 10m.at the same time, another ball is thrown vertically upward at an initial speed of 10m/s. How high above the ground will thr two balls collide?
Answer: H=5.1m
Explanation:
Given:
Ball 1 height= 10m
Ball 2 initial velocity=10m/s
use the kinematic equation:
S=(vi)t+12at2
I choose my sign convention to be up=positive, down=negative, so a=−9.81ms2 (a is taken as the value of gravity)
Ball 1 dropped from 10m :
−(10−H)=0+12(−9.81)t2
Note that (10-S) is negative because that displacement is *below* the starting point.
12(9.81)t2=10−H
——- equation (1)
Ball 2 thrown upward at 10 m/s :
H=(10)t+12(−9.81)t2
or
12(9.81)t2=10t−H
——- equation (2)
equation (1) minus equation (2):
0=(10−H)−(10t−H)
t=1 equation (1):
12(9.81)12=10−H
H=5.1m
Given Information:
Ball One:
[tex]\vec y_{0} = 10 \ m[/tex] (Indicating the initial position)
We also know the ball was dropped from rest. So, [tex]\vec v_{0_{1} } = 0 \ m/s[/tex].
Ball Two:
[tex]\vec v_{0} = 10 \ m/s[/tex] (Indicating the initial velocity)
We also know the ball was throw from the ground. So, [tex]\vec y_{0_{1} } = 0 \ m[/tex].
The Information we want to Find:
[tex]\vec y_{c} = ?? \ m[/tex] (Indicating the position the two projectiles collide)
Using the Following Kinematic Equation to Solve:
[tex]\Delta \vec x = \vec v_{0}t + \frac{1}{2} \vec at[/tex]
For ball one...
[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y} t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \vec v_{0_{1} }t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = (0)t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex] => Equation 1
For ball two...
[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0_{1} } = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t +\vec y_{0_{1} }[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t + 0[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex] => Equation 2
Set equations 1 and 2 equal to each other and solve for the time that they collide.
[tex]\left \{ {{\vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}} \atop { \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t } \right.[/tex]
[tex]\Longrightarrow \frac{1}{2} \vec at +\vec y_{0}= \vec v_{0}t + \frac{1}{2} \vec at[/tex]
[tex]\Longrightarrow \vec y_{0}= \vec v_{0}t[/tex]
[tex]\Longrightarrow t=\frac{\vec y_{0}}{\vec v_{0}}[/tex]
[tex]\Longrightarrow t=\frac{10}{10}[/tex]
[tex]\Longrightarrow t=1 \ s[/tex]
Thus, the balls collide at time, t=1 s. We can now use this time to plug into equation 1 or 2 to find the height at which they collide. I will use equation 1.
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex]
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} (-9.8)(1) +10[/tex]
[tex]\Longrightarrow \vec y_{c} = 5.1 \ m \ \therefore \ Sol.[/tex]
*Note* [tex]\vec a_{y}[/tex] is the acceleration of gravity ([tex]-9.8 \ m/s^2 \ or \ -32 \ ft/s^2[/tex])
Final Answer: The balls collide at the height 5.1 m.