The 0.100 kg
sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m
from the center of the 5.00 kg
mass. Assume that the only forces on the 0.100 kg
sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg
and 10.0 kg
spheres are held in place at their initial positions.

What is the speed of the 0.100 kg sphere when it has moved 0.150 m to the left from its initial position?

Answers

Answer 1

As per the given data, the speed of the 0.100 kg sphere when it has moved 0.150 m to the left from its initial position is 0.736 m/s.

Since only the gravitational forces are acting on the 0.100 kg sphere, we can use the conservation of energy principle to find its speed at any position.

We can use the initial position of the sphere as the reference point for potential energy and write the initial total energy as the sum of the potential energy and kinetic energy.

At any other position, the total energy will still be the sum of the potential energy and kinetic energy, but their values will be different.

The initial total energy of the system is:

E_i = m_0gh

Where m_0 is the mass of the 0.100 kg sphere, g is the acceleration due to gravity, and h is the height of the sphere above the reference position. In this case, h = 0.4 m.

The final total energy of the system is:

[tex]E_f = m_0v^2/2 + m_0gh_f[/tex]

Where v is the speed of the sphere, and h_f is the height of the sphere above the reference position at the final position.

Since the system is isolated, the initial and final energies must be equal:

E_i = E_f

[tex]m_0gh = m_0v^2/2 + m_0gh_f[/tex]

Solving for v, we get:

v = sqrt(2gh - 2gh_f)

To find the final height h_f, we can use the fact that the center of mass of the system remains fixed throughout the motion.

The initial center of mass is at a distance of 0.4 m from the center of the 5.00 kg sphere, and the masses of the 5.00 kg and 10.0 kg spheres are 5.00 kg and 10.0 kg, respectively.

Therefore, the initial center of mass is at:

x_cm,i = (0.4*0.1 + 5*0 + 10*0)/(0.1 + 5 + 10) = 0.032 m

where we have taken the x-axis to be horizontal and passing through the centers of the 5.00 kg and 10.0 kg spheres.

At the final position, the center of mass must still be at the same horizontal position:

x_cm,f = (5*0.1*(-0.15) + 10*0)/(0.1 + 5 + 10) = -0.011 m

where we have taken the leftward direction as positive.

The final height of the sphere is then:

h_f = 0.4 - x_cm,f = 0.4 + 0.011 = 0.411 m

Substituting the values of g, h, and h_f in the equation for v, we get:

v = sqrt(2*9.81*0.4 - 2*9.81*0.411) = 0.736 m/s (rounded to three significant figures)

Therefore, the speed of the 0.100 kg sphere when it has moved 0.150 m to the left from its initial position is 0.736 m/s.

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Related Questions

(a) Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) 401.15
(b)How much work (in J) does she do if her center of mass rises 0.260 m?

(c) What is her useful power output (in W) if she does 30 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)

Answers

the force the woman exerts to do a push-up at constant speed is 333 N.

the work the woman does is 152 J.

her useful power output is 76 W.

(a) To calculate the force the woman exerts to do a push-up, we need to use torque methods. The woman is doing a push-up at constant speed, which means that the net torque on her body is zero. The only torque acting on her body is due to her weight W, which acts at the center of mass of her body. The distance between her center of mass and her hands is 0.76 m, and the angle between her body and the horizontal is 45 degrees.

The torque due to her weight about her hands is given by:

τ = r x W = (0.76 m) x (cos 45°)(W)

where r is the distance between her hands and her center of mass and cos 45° is the component of the distance perpendicular to the weight vector. Since the woman is at constant speed, the torque she exerts about her hands must be equal and opposite to the torque due to her weight. Therefore:

τ = (0.76 m)(cos 45°)(W) = (1/2)(W)(0.76 m)

Solving for W, we get:

W = 2(τ/0.76 m) = 2[(0.5)(mg)(0.76 m)/(0.76 m cos 45°)] = 333 N

Therefore, the force the woman exerts to do a push-up at constant speed is 333 N.

(b) The work the woman does is equal to the change in her potential energy as her center of mass rises. The woman's mass is not given, so we will assume a value of 60 kg. The gravitational potential energy of the woman is given by:

U = mgh

where m is the mass of the woman, g is the acceleration due to gravity (9.81 m/s^2), and h is the height her center of mass rises (0.26 m). Therefore:

U = (60 kg)(9.81 m/s^2)(0.26 m) = 152 J

Therefore, the work the woman does is 152 J.

(c) The useful power output of the woman is the work she does per unit time, taking into account the work done in lowering her body. Each push-up involves two phases: lifting her body and lowering her body. When she lowers her body, the work done is negative, as the force she exerts is in the opposite direction to the displacement. The work done in lowering her body is equal to the work done in lifting her body, so the total work done in one push-up is zero.

The woman does 30 push-ups in 1 minute, which means she does one push-up every 2 seconds. Therefore, the useful power output of the woman is:

P = (152 J)/(2 s) = 76 W

Therefore, her useful power output is 76 W.

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14. Neglecting air resistance, what maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s?
(a) 98 m
(b) 18 m
(c) 160 m
(d) 63 m​

Answers

Answer:

D

Explanation:

The maximum height reached by a stone thrown straight up with an initial speed of 35 m/s can be found using the kinematic equation:

v^2f = v^2i - 2gh

where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (35 m/s, the magnitude of the velocity with which the stone is thrown upwards), g is the acceleration due to gravity (-9.8 m/s^2), and h is the maximum height reached by the stone.

Rearranging the equation, we get:

h = (vi^2)/(2g)

Substituting the given values, we have:

h = (35 m/s)^2 / (2 * 9.8 m/s^2)

= 62.6 m

Therefore, the maximum height reached by the stone is approximately 63 m.

The answer is (d).

energy transferred from one thing to another when the gulf balls collide?

Answers

This is based on the principle of conservation of energy and momentum.

When the collision of golf balls takes place the energy gets transferred from one ball to the other.

The golf balls experience a force that causes them to change their form and also direction. During the collision, the mechanical energy is converted into heat, sound, and other forms of energy. The rest of energy is used to move in a new direction.

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The cylinder of a heat engine is filled with an air-fuel mixture. Which property of gases is essential to heat engines ability to do work?

Answers

Answer: Pressure of gases.

Explanation:

In a heat engine, the air-fuel mixture is ignited, which causes an increase in pressure of the gases inside the cylinder. This pressure pushes the piston, which is connected to a crankshaft, causing it to rotate and do work. Therefore, pressure is an essential property of gases for the ability of a heat engine to do work.

Diagram A shows a negatively charged conducting rod placed near a light polystyrene ball that is suspended from the ceiling by an insulating thread .Diagram B shows what happens when the ball touches the rod. (a) Explain why the ball is displaced vertically​ in Diagram A (b)Explain what happens after the ball has been allowed to touch the rod (c)Give a reason why the ball has to be coated with a conducting material such as graphite (d) Explain why the polystyrene ball is suspended by an insulated thread and not by a conducting wire​

Answers

When a negatively charged conductive rod is brought close to a lightweight Styrofoam ball suspended from the ceiling by insulating threads (see Figure A), the electrons in the ball are repelled by the rod's negative charge.

What happens when the ball hits the pole?As a result, the electrons in the sphere move away from the rod and spread unevenly over the surface of the sphere. This makes the side of the sphere closest to the stick positively charged and the opposite side negatively charged.When the ball touches the negatively charged wand (see Figure B), the negatively charged electrons in the wand repel the negatively charged electrons in the ball and move away from the contact point. This creates a charge imbalance on the surface of the ball, with excess positive charge on one side and negative charge on the other.As a result of this charge imbalance, the ball experiences an electrostatic force and moves vertically away from the rod. The direction of displacement depends on the relative magnitudes of the electrostatic force and the weight of the sphere. If the electrostatic force is stronger than the weight of the ball, the ball will move up. If the weight of the ball is stronger than the electrostatic force, the ball will move down. In both cases the ball moves vertically.  

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What happens as a protostar contracts?
A. More hydrogen is produced to become fuel for the star.
B. The temperature rises.
C. The hydrogen fuses into iron.
D. The temperature drops.

Answers

Answer:

As a protostar contract, the temperature rises. This increase in temperature leads to the initiation of nuclear fusion reactions, where hydrogen atoms fuse together to form helium, releasing energy in the process. This energy causes the protostar to heat up and begin to emit light, eventually becoming a stable star.

The answer is B. The temperature rises.

Which one of the following is the longest length?
(a) 100 m
(b) 104 µm
(c) 107 nm
(d) 102 mm​

Answers

Okay, let's convert all the lengths to the same unit to compare:

(a) 100 m = 100 meters

(b) 104 μm = 104 micrometers = 104 × 10^-6 meters = 0.000104 meters

(c) 107 nm = 107 nanometers = 107 × 10^-9 meters = 0.000000000997 meters

(d) 102 mm = 102 millimeters = 102 × 10^-3 meters = 0.0102 meters

The longest length is:

(a) 100 m = 100 meters

The answer is option (a).

Answer: 100 m

Explanation:

1 μm = [tex]10^{-6}[/tex] m = 0,000001 m

1 nm = [tex]10^{-9}[/tex] m = 0,000000001 m

1 mm = [tex]10^{-3}[/tex] m = 0,001 m

∴ 100 m es la mayor longitud

As part of a movie stunt, a full-size remote-controlled car is driven horizontally off a 9.00 m tall cliff at 24.40 m/s. How far (Δx) from the bottom of the cliff does the car land?

Answers

Explanation:

Find the time it takes to hit the  bottom....then multiply this time by the horizontal velocity .......

Time to hit bottom :

       d = 1/2 at^2

        9 m  = 1/2 (9.81 m/s^2) (t^2)    shows  t = 1.35 s

Now the horizontal displacement is

      x = rate * time = 24.40 m/s * 1.35 s = 33.1 m

**NEED ANSWER ASAP**
What are weird properties of quasars that made them difficult for astronomers to understand?

**FAKE ANSWERS WILL BE REPORTED**

Answers

Supermassive black holes that are devouring gas at the center of far-off galaxies are known as quasars.

Since, the quasars were initially identified by astronomers in 1963 as objects that resembled stars but gave off radio waves instead, the term quasar is an abbreviation for quasi-stellar radio source.

Quasars are so bright that they drown out the light from all other stars in the same galaxy. Quasars give off radio waves, X-rays, gamma-rays, ultraviolet rays, and visible light across the entire electromagnetic spectrum. Most of them are larger than our solar system.

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When a massive star, much bigger than our sun, reaches the end of its life cycle, it will expand into a red supergiant and then:
A. lose its corona and enter a second stable phase as a star again.
B. explode into a supernova.
C. collapse into a protostar.
D. collapse into a white dwarf.

Answers

Answer:

When a massive star, much larger than our sun, reaches the end of its life cycle, it will undergo a series of fusion reactions in its core until it forms iron, which cannot undergo further fusion. Without fusion to counteract the force of gravity, the core collapses in on itself, causing the outer layers of the star to rapidly expand and creating a red supergiant. Eventually, the outer layers of the star will be expelled in a supernova explosion, leaving behind either a neutron star or a black hole, depending on the mass of the original star. So, the correct answer is B.

What month is the speed of Earth the fastest?

January
March
June
Decemember

Answers

Answer:

July 29, Earth completed a full spin in about 1.59 milliseconds shorter than its standard timeframe ( 23 hours and 56 minutes).

4. Two identical test tubes are filled with equal volumes of water and
mercury. Which of the following statements is true?
The weight of each liquid is the same.
The bottom area of each test tube is the same.
The pressure at the bottom of each test tube is the same.
All the above.

Answers

The correct answer is: All of the above.

The weight of each liquid is the same because the two test tubes are filled with equal volumes of water and mercury, and the weight of a liquid is directly proportional to its volume.

The bottom area of each test tube is the same because the test tubes are identical, and the bottom area of each test tube is the same.

The pressure at the bottom of each test tube is the same because the pressure at a given depth in a fluid depends only on the height of the fluid column and the density of the fluid, and not on the shape or size of the container. Therefore, since the test tubes are identical and filled with the same fluids to the same height, the pressure at the bottom of each test tube is the same.

Sanjay and Ting, each with a mass of 25 kg, are riding opposite each other on the edge of a 150 kg, 3.0-m-diameter playground merry-go-round that's rotating at 15 rpm. Each walks straight inward and stops 35 cm from the center.

What is the new angular velocity, in rpm?
Express your answer in revolutions per minute.

Answers

The merry-go-round's new angular velocity is 0.321 rpm.

Calculation-

the system's overall angular momentum is:

L = Iω

The moment of inertia of a solid disk rotating about its centre is given by:

[tex]I = (1/2)mr^2I = (1/2)(150 kg)(1.5 m)^2 = 168.75 kg·m^2[/tex]

initial angular momentum of the system

L = Iω = [tex](168.75 kg·m^2)(15 rpm)(2π/60 s) = 52.36 kg·m^2/s[/tex]

The new angular velocity is determined by:

L = I'ω'

where L represents the system's initial angular momentum.

The system's new moment of inertia is:

[tex]I' = I - 2mr^2[/tex]

we get:

[tex]I' = 25 kg - 2(168.75 kg/m2)(0.35 m)^2 = 162.88 kg·m^2[/tex]

We get the following by substituting into the conservation of angular momentum equation:

L = I'ω'

[tex](162.88 kg/m2) / 52.36 kg/m2[/tex]

ω' = 0.321 rpm

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A car is traveling at a speed of 30m/s and leaves the ramp at a 37 degree angle. What is the total hang time of the car?

Answers

Explanation:

INITIAL  Vertical velocity is given by

30 m/s * sin 37

 then gravity begins to slow it down

30 sin37  -  (9.81) t   = vertical velocity at  t

    when v = 0 , the car is at its apex and will fall back down in the same amount of time

  0 = 30 sin37 - 9.81 t      shows t = 1.84 seconds to peak

    then another 1.84 seconds to fall to the ground   total = 3.7 seconds

A ball (0.410 kg) is kicked at an angle of 44.0° above the horizontal axis (above the +x-axis).  The initial speed of the ball is 24.2 m/s.  Ignoring air resistance, determine the momentum of the ball just before it hits the ground.

Answers

Answer:

9.922

Explanation:

When ignoring the air resistance and the ball is kicked and fell into the same plane. the initial speed is the speed when it is stricken on the ground. the initial momentum magnitude equals the final momentum magnitude only the direction is changed.

∴ Momentum=  mass*velocity  

                      = o.41kg*24.2

                      = 9.922[tex]kgms^{-1[/tex]

what type of of electrical energy is produced by batteries

Answers

The type of electrical energy produced by batteries is chemical potential energy. One or more cells that transform chemical energy into electrical energy make up a battery.

What is electrical energy?

A type of energy known as electrical energy is connected to the movement of electric charge. It is the energy that is transported across a conductor by moving electrons in an electric circuit.

Direct current (DC) batteries generate electrical energy. In contrast to alternating current (AC), which occasionally flips direction, direct current is a type of electrical energy that only flows in one direction through a circuit. A chemical reaction that takes place inside the battery itself produces the electrical energy that a battery generates.

An electrical potential difference between the battery's positive and negative terminals is produced by a chemical reaction that occurs inside each cell. When a circuit is attached to a battery, electrons move from the battery's negative terminal through the circuit and out to the positive terminal, causing an electrical current to flow.

Little electronic devices like calculators and lamps can be powered by batteries, as well as bigger applications like electric vehicles and backup power systems. Batteries have a finite capacity, though, and over time as their chemical reactions wear out, they will eventually get depleted.

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Example 9:
3.
Figure 5.24 shows a barrel of weight 1500 N
and radius 0.5 m that rests against a step of
height 0.2 m.
0.2 m
0.5 m
▲ Figure 5.24
(
What is the smallest horizontal force F th
the centre O needed to push the barrel over
the step

Answers

To drive the barrel over the step, the least horizontal force F necessary is 2366.16 N.

How to find horizontal force?

To push the barrel over the step, the minimum force required should overcome the force of gravity acting on the barrel and the force of static friction between the barrel and the surface.

The perpendicular component of the weight is given as N = mgcosθ, where m = mass of the barrel,

g = acceleration due to gravity, and

θ = angle between the weight and normal to the surface.

In this case, θ as the inverse tangent of the ratio of the height of step to the distance from the edge of the step to the center of the barrel:

θ = tan⁻¹(0.2/0.5) = 0.39 radians

Therefore, the normal force is N = (1500)(9.81)cos(0.39) = 1443.6 N.

The force of static friction can be found as f = μsN,

where μs = coefficient of static friction.

Assume the coefficient of static friction between the barrel and the surface is 0.6.

f = (0.6)(1443.6) = 866.16 N.

The minimum force required to push the barrel over the step should overcome both these forces. Then, the smallest horizontal force that can push the barrel over the step is:

F = force of gravity + force of static friction

F = 1500 + 866.16

F = 2366.16 N.

Therefore, the smallest horizontal force F required to push the barrel over the step is 2366.16 N.

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Penguins in Gold Harbour love to communicate with other members of their penguin family. Here in Sanford, the speed of sound in air is about 344.0 m/s. Calculate the speed of sound (in m/s) in Gold Harbour, on a day when the air temperature is -2.7 °C.

Round to the nearest hundredth.
Please show all work!!!!

Answers

The speed of sound in Gold Harbour, on a day when the air temperature is -2.7 °C, is 331.5 + 0.6 * (-2.7) = 328.65 m/s.

What is Gold Harbour?

Gold Harbour is a small settlement located on Antarctic's King George Island. It is home to a Chilean research base, which is operated by the Chilean Antarctic Institute. It also acts as a summer base for the Chilean Navy, and provides support for the scientific research conducted by other countries, including the United States, United Kingdom, and Russia. The area is known for its stunning natural beauty, with mountains, glaciers, and icebergs all in close proximity. It is also an important habitat for several species of wildlife, including penguins, seals, and sea birds.

The speed of sound in air is affected by temperature, and the formula for calculating the speed of sound in air is v = 331.5 + 0.6 * (air temperature in °C).
Therefore, the speed of sound in Gold Harbour, on a day when the air temperature is -2.7 °C, is 331.5 + 0.6 * (-2.7)
= 328.65 m/s.

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If I want to convert 6,785 mg into grams. I would need to move the decimal places to the

Answers

You divide by 1000 so it would be 6.785g

String 1 in the figure has linear density 2.60 g/m and string 2 has linear density 3.30 g/m. A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. She wants both pulses to reach the ends of the strings simultaneously.

What should the string length L1 be?

What should the string length L2 be?

Answers

Answer:

Here is your answer please change up some words to remain plagraism free.

Explanation:

To determine the required lengths of strings 1 and 2 so that pulses sent in both directions reach the ends of the strings simultaneously, we need to apply the principle that the time it takes for a wave pulse to travel a distance on a string is equal to the distance divided by the wave speed.

The wave speed, in turn, is determined by the tension in the string and the linear density of the string according to the formula:

v = sqrt(T/μ),

where v is the wave speed, T is the tension, and μ is the linear density.

Let L1 be the length of string 1 and L2 be the length of string 2. Since the wave speed is the same for both strings, we can set up the following equations:

L1/v = L2/v

sqrt(T1/μ1)*L1 = sqrt(T2/μ2)*L2

where T1 and T2 are the tensions in strings 1 and 2, respectively.

We can solve for L1 and L2 by combining these two equations and solving for each variable. Substituting the given linear densities of strings 1 and 2, we get:

sqrt(T1/2.60)*L1 = sqrt(T2/3.30)*L2

Squaring both sides and simplifying, we get:

(T1/T2) = (3.30/2.60) * (L1/L2)^2

Substituting the condition that the pulses reach the ends of the strings simultaneously, we know that the total time for a pulse to travel down string 1 and back up to the knot is equal to the time for a pulse to travel down string 2 and back up to the knot. This condition implies that the total length of string 1 (2L1) is equal to the total length of string 2 (2L2):

2L1 = 2L2

Solving this equation for L2 and substituting it into the expression for T1/T2 derived above, we get:

T1/T2 = (3.30/2.60) * (L1/2L1)^2 = 1.25

Solving for L1, we obtain:

L1 = sqrt(T1/μ1) * (2L2/v) = sqrt((1.25)*(2.60/3.30)) * (2L2)

Simplifying this expression, we get:

L1 = (2/3) * sqrt(2.60/3.30) * L2

Therefore, the required length of string 1 is (2/3) * sqrt(2.60/3.30) times the length of string 2. We can substitute the given length of string 2, say L2 = 1 meter, into this expression to obtain the required length of string 1:

L1 = (2/3) * sqrt(2.60/3.30) * 1 meter ≈ 0.693 meter.

Therefore, the required length of string 1 is approximately 0.693 meter and the required length of string 2 is 1 meter.

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Problem 2.3. (5 pts) A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a frictionless, horizontal air track. (a) Calculate the maximum speed of the cart if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the cart when the position is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm​

Answers

The maximum speed of the cart if the amplitude of the motion is 3.00 cm is  0.036 m/s

The velocity of the cart when the position is 2.00 cm is 0.1414 m/s.

The kinetic and potential energies of the system when the position of the cart is 2.00 cm​ is 5×10⁻³ J and 4×10⁻³ J resp.

a) To find maximum speed potential energy of the spring gets converted into kinetic energy of the cart in the oscillator motion, Hence,

1/2mv² =1/2kA²

1/2×0.5×v² = 1/2 ×20× 0.03²

v² = 40×9×10⁻⁴

v = 0.036 m/s

b) For the velocity at a given point is calculated by the formula,

v = ±ω√(A² - x²)

v = ±ω√(0.03² - 0.02²)

v = ±ω × 0.0223

v = ±√k/m × 0.0223

v = ±√20/0.5× 0.0223

v = 0.1414 m/s

c)

kinetic energy of the system,

K = 1/2 mv²

K = 1/2 ×0.5×0.1414²

K = 5×10⁻³ J

Potential energy of the system

P = 1/2 kx²

P = 1/2 × 20× 0.02²

P = 4×10⁻³ J

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15 A car of mass 750 kg is accelerating up a slope of a certain angle to the horizontal where sin theta = 1/70 at 1.5 m/ s². Ignoring any road resistance, find the tractive force of the engine.​

Answers

The tractive force of the engine is 1020 N.

Mass of the car, m = 750 kg

Acceleration of the car, a = 1.5 m/s²

Traction, also known as tractive force, is the force used to produce or create motion by using dry friction between a body and an inclined surface.

Weight of the car, W = mg

W = 750 x 9.8

W = 7350

The coefficient of friction is the ratio or percentage of the opposing frictional force to the normal force pressing the two surfaces into contact and motion.

The tractive force of the engine,

F' = ma - mg sinθ = m(a - g sinθ)

F' = 750[1.5 - (9.8/70)]

F' = 750 x 1.36

F' = 1020 N

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2. There are many factors that play a role in body image. How might a person's culture
and family background shape their body image? You may choose to examine your own
ethnicity or discuss a variety of different ethnicities

Answers

A person's culture and family background can significantly shape their body image. Different cultures have unique standards of body ideals, which can influence how people perceive their own bodies and others.

What is culture?

In some cultures, a larger body size is considered desirable and associated with prosperity, fertility, and beauty, while in other cultures, a thinner body size is considered more attractive and associated with success, health, and discipline.

Family background can also play a role in shaping body image. For instance, if family members constantly make negative comments about weight or appearance, this can negatively impact a person's body image and self-esteem. Additionally, if a person grows up in a family where unhealthy eating habits are normalized or encouraged, this can contribute to the development of unhealthy body image and eating behaviors.

Overall, it is important to recognize the influence that culture and family background can have on body image, and to promote a more diverse and inclusive standard of beauty that celebrates different body types and sizes.

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OK, once again we have a pendulum, this time of length 1.06 m, which you release from rest at an angle of 41.2 degrees to the vertical. What will be the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical?

Answers

The speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical is 3.02 m/s.

A pendulum is a weight suspended from a fixed point that swings back and forth due to the force of gravity.

Based on the given information, we have a pendulum of length 1.06 m and it is released from rest at an angle of 41.2 degrees to the vertical. We need to find the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical.

To solve this problem, we can use the conservation of mechanical energy. At the highest point of the pendulum's swing, all of its energy is in the form of potential energy, and at the lowest point of its swing, all of its energy is in the form of kinetic energy. Therefore, we can write:

PE_max = KE_min

where PE_max is the potential energy at the maximum height and KE_min is the kinetic energy at the lowest point.

The potential energy of a pendulum is given by:

PE = mgh

where m is the mass of the pendulum, g is the acceleration due to gravity, and h is the height above some reference point.

The kinetic energy of a pendulum is given by:

KE = (1/2)mv^2

where v is the speed of the pendulum.

First, we need to find the vertical height difference between the pendulum's highest and lowest points. To do this, we can use trigonometry:

h = L(1 - cosθ)

where L is the length of the pendulum and θ is the initial angle to the vertical. Substituting the given values, we get:

h = 1.06(1 - cos(41.2°)) = 0.654 m

Next, we can use the conservation of mechanical energy to find the speed of the pendulum at the lowest point of its swing. At this point, all of the potential energy has been converted into kinetic energy, so we can write:

PE_max = KE_min

mgh = (1/2)mv^2

Canceling out the mass, we get:

gh = (1/2)v^2

Solving for v, we get:

v = sqrt(2gh)

where g is the acceleration due to gravity. Substituting the given values, we get:

v = sqrt(2(9.81 m/s^2)(0.654 m)) = 3.78 m/s

Finally, we need to find the speed of the pendulum when it reaches an angle of 20.6 degrees above the vertical. At this point, the pendulum has a potential energy of:

PE = mgh' = mgh cos(20.6°)

where h' is the height of the pendulum at this point. To find h', we can use trigonometry:

h' = L(1 - cosθ')

where θ' is the angle above the vertical. Substituting the given values, we get:

h' = 1.06(1 - cos(20.6°)) = 0.242 m

Substituting the values for h' and solving for the kinetic energy, we get:

KE = PE_max - mgh' = mgh - mgh'

Substituting the known values, we get:

KE = (1 kg)(9.81 m/s^2)(0.654 m) - (1 kg)(9.81 m/s^2)(0.242 m) = 5.11 J

Now, we can solve for the speed at this point:

KE = (1/2)mv^2

5.11 J = (1/2)(1 kg)v^2

v = sqrt((2)(5.11 J)/(1 kg)) = 3.02 m/s

Therefore, The pendulum is moving at a speed of 3.02 m/s when it reaches an angle of 20.6 degrees above vertical.

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Which of the following sets of two charges is experiencing the strongest
attraction?
Charges of +2 C and -2 C, separated by 1 m.
Charges of +1 C and -3 C, separated by 1 m.
Charges of +2 C and +2 C, separated by 1 m.
Charges of +1 C and +3 C, separated by 1 m.

Answers

The force of attraction between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

What is a square ?

A square is a two-dimensional geometric shape that has four sides of equal length and four right angles (90-degree angles) between them. The sides of a square are parallel to each other and perpendicular to its adjacent sides. All four corners of a square are also known as vertices, and the diagonals of a square bisect each other at right angles.

The area of a square is calculated by multiplying the length of one of its sides by itself. The perimeter of a square is calculated by adding up the lengths of all four sides. The properties of a square make it useful in various applications, such as in geometry, architecture, and construction.

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In which of the following substances will sound travel the fastest?
air
O iron
water

Answers

Iron. Is the correct answer
Iron. Is the correct answer

Answer:

iron

Explanation:

iron is a solid where the particles are closer together, so sound will vibrate the particles faster than in a liquid or gas where the particles are further spaced out.

Answer:

iron

Explanation:

iron is a solid where the particles are closer together, so sound will vibrate the particles faster than in a liquid or gas where the particles are further spaced out.

A ball of mass 2 kg is moving with a velocity of 12m/s collides with a stationary ball of mass 6 kg and comes to rest. Calculate velocity of ball of mass 6kg after collision.

Answers

Answer: 4 m/s

Explanation:

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision.

Let's denote the initial velocity of the 2 kg ball as "v1i", the initial velocity of the 6 kg ball as "v2i", the final velocity of the 2 kg ball as "v1f", and the final velocity of the 6 kg ball as "v2f".

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:

m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f

where m1 is the mass of the 2 kg ball, m2 is the mass of the 6 kg ball, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities.

Given:

m1 = 2 kg

m2 = 6 kg

v1i = 12 m/s (initial velocity of the 2 kg ball)

v2i = 0 m/s (initial velocity of the 6 kg ball, as it is stationary)

v1f = 0 m/s (final velocity of the 2 kg ball, as it comes to rest)

Plugging in the given values into the conservation of momentum equation:

2 * 12 + 6 * 0 = 2 * 0 + 6 * v2f

24 = 6 * v2f

Dividing both sides by 6:

v2f = 24 / 6 = 4 m/s

So, the velocity of the 6 kg ball after the collision is 4 m/s.

An overtone is louder than the fundamental tone.
O True
False

Answers

The given statement, "An overtone is louder than the fundamental tone" is False.

An overtone is a higher-frequency vibration that occurs simultaneously with the fundamental frequency of a sound wave. These vibrations are multiples of the fundamental frequency and contribute to the overall timbre or tone quality of a sound.

In some cases, overtones can be louder than the fundamental tone, depending on the specific harmonic series present in the sound wave. This phenomenon is known as overtone prominence, and it can be heard in certain musical instruments like bells or cymbals, where the higher harmonics of the sound are emphasized.

However, in most cases, the fundamental tone is perceived as the strongest and most dominant sound in a given sound wave.

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which system best illustrates attractive forces

Answers

Answer:

A

Opposite poles will attract

A the opposite are eventually going to attract to each other

lucy finished 1/4 of her homework at an average speed of 15 questions per hour then she finished the remaining 45 questions at another speed. if the total time spent on the homework was 2.5 hours, what was the amount of time she spent on the remaining 45 questions

Answers

Therefore, Lucy spent 1.875 hours on the remaining 45 questions.

How is  the amount of time she spent on the remaining 45 questions?

Let's start by finding the total number of questions in Lucy's homework. If she finished 1/4 of her homework at an average speed of 15 questions per hour, then the total number of questions must be:

1/4 x Total number of questions = Number of questions finished at 15 questions per hour

1/4 x Total number of questions = 15 questions per hour

Solving for the total number of questions, we get:

Total number of questions = 60 questions

Now we know that Lucy finished 60 - 45 = 15 questions at an average speed of x questions per hour. Let's use the formula:

time = distance / speed

to find the amount of time she spent on the remaining 45 questions.

For the first part of the homework, Lucy spent:

time = distance / speed

time = 15 questions / hour

time = 1/4 x 2.5 hours

time = 0.625 hours

So, she spent 0.625 hours on the first 15 questions.

For the remaining 45 questions, we have:

time = distance / speed

time = 45 questions / x questions per hour

We know that the total time spent on the homework was 2.5 hours, so:

0.625 hours + 45 questions / x questions per hour = 2.5 hours

Solving for x, we get:

x = 18 questions per hour

Now we can use this speed to find the time spent on the remaining 45 questions:

time = distance / speed

time = 45 questions / 18 questions per hour

time = 2.5 hours - 0.625 hours

time = 1.875 hours

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