Phosphorus bursts into flame when exposed to oxygen. What element might be used when storing pure phosphorus to keep it from reacting?

Part A: The potential on the surface of the charged sphere is [tex]4.5 \times 10^7[/tex] volts.

Part B: The distance is [tex]4.5 \times 10^{10}[/tex] m from the center of the sphere where the potential is 1 mv.

Part C: The energy of the oxygen atom found at the distance of [tex]x = 4.5 \times 10^{10} \;\rm m[/tex] is 3me V.

How do you calculate the potential?

Given that the diameter of the sphere is 2.00 m and the charge of 5.00 mc.

Part A

The potential on the surface of a charged sphere is given below.

[tex]V = k\dfrac {Q}{R}[/tex]

Where V is the potential on the surface, Q is the charge, R is the radius and k is the Coulomb's constant.

[tex]V = 8.99 \times 10^9\times \dfrac {5 \times 10^{-3}}{\dfrac {2}{2}}[/tex]

[tex]V = 4.5 \times 10^7 \;\rm Volts[/tex]

Hence the potential on the surface of the charged sphere is [tex]4.5 \times 10^7[/tex] volts.

Part B

The potential of 1 mv generated at a certain distance x from the centre of the sphere is given below.

[tex]V_x = k \dfrac {Q}{x}[/tex]

[tex]1 \times 10^{-3} = 8.99 \times 10^9 \times \dfrac{5 \times 10^{-3}} {x}[/tex]

[tex]x = 4.5 \times 10^{10} \;\rm m[/tex]

Hence the distance is [tex]4.5 \times 10^{10}[/tex] m from the center of the sphere where the potential is 1 mv.

Part C

Given that an oxygen atom with three missing electrons is released near the van de-Graaff generator. It means that the charge Q = +3e

The electric potential energy of a charged particle located at some point with voltage V is given below.

[tex]U = QV[/tex]

At the distance found in part b, which is [tex]x = 4.5 \times 10^{10} \;\rm m[/tex], the energy of the oxygen atom is,

[tex]U = +3e \times 1 \times 10^{-3}[/tex]

[tex]U = 3me \;\rm V[/tex]

Hence the energy of the oxygen atom found at the distance of [tex]x = 4.5 \times 10^{10} \;\rm m[/tex] is 3me V.

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To find the net force on the donkey, we decompose the angled forces into horizontal and vertical components and add them accordingly. Since the vertical components cancel out, the net force is the sum of Jack's force and the combined horizontal components from Jill and Jane, resulting in a net force of 149 N ahead.

The problem presented requires an analysis of forces and vector addition to calculate the net force exerted on a donkey by three different people. We have Jack applying 97.9 N directly ahead, Jill pulling with 72.7 N at a 45° angle to the left, and Jane pulling with 145 N at a 45° angle to the right. To solve this, we must break down Jill's and Jane's forces into their horizontal and vertical components and then combine these with Jack's force.

Jill's horizontal component: 72.7 N × cos(45°) = 51.4 N to the left

Jill's vertical component: 72.7 N × sin(45°) = 51.4 N up

Jane's horizontal component: 145 N × cos(45°) = 102.5 N to the right

Jane's vertical component: 145 N × sin(45°) = 102.5 N up

Now we can subtract Jill's horizontal component from Jane's because they are in opposite directions: 102.5 N (Jane's right) - 51.4 N (Jill's left) = 51.1 N to the right. The vertical components from Jill and Jane will cancel each other since the donkey is stubborn and not moving vertically, hence we are left with only the horizontal forces to combine with Jack's force.

The total horizontal force exerted by Jack and combined with the net horizontal force from Jill and Jane is 97.9 N (Jack's force)+ 51.1 N (net horizontal force) = 149 N.

Thus, the magnitude of the net force the people exert on the donkey is 149 N directly ahead, assuming no vertical movement.

If the particle is under the Simple Harmonic Motion, the wave is considered to be sinusoidal wave.

Explanation:

A body is considered to be performing the Simple Harmonic Motion if it repeats its motion in a particular path with a specific time period. The two and fro motion of the pendulum about a fixed mean position is termed as the Simple Harmonic Motion.

The motion of the simple pendulum repeats itself after a particular time period and it continuously retraces its path during the motion. Therefore, it is termed as the simple harmonic motion.

The wave depicting the motion of a body performing the simple harmonic motion is shown in the sinusoidal manner. The motion of the body from its mean position to its extreme and then from extreme position to its mean position shows the positive cycle of the sinusoidal waveform.

The motion of the pendulum on the other side of the mean position of the pendulum represents the other half of the sinusoidal waveform.

Thus, If the particle is under the Simple Harmonic Motion, the wave is considered to be sinusoidal wave.

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Answer Details:

Grade: High School

Chapter: Simple Harmonic Motion

Subject: Physics

Keywords:

simple, harmonic, SHM, motion, pendulum, oscillates, mean position, extreme, repeats, time period, sinusoidal.

Answer:

the 3rd choice is the answer

Explanation:

Answer: False

When the diaphragm contracts, the muscles will also contract and pull upward and increase the size of the thoracic cavity thus decreases air pressure inside during inspiration. After the diaphragm contracts, it goes to relaxation, the muscles will also relaxed. It gets looser and return to its original position higher up in the chest. This increase the pressure in the chest, which force the air in the lungs out through the nose.

The resistivity of two copper wires remains the same regardless of their lengths because resistivity is an intrinsic property of the material.

Comparing resistivities of two copper wires of different lengths but the same material, we need to understand that resistivity is an intrinsic property of the material and does not depend on the geometry of the wire. Therefore, regardless of whether one wire is twice as long as the other, the resistivity of both wires remains the same. This is a fundamental concept in the study of electrical resistance and materials.

B.What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"

Solution:

A) A charge q under an electric field of intensity E will experience a force F  equal to:

[tex]F=qE[/tex]

In our problem we have [tex]q=-2.5 nC=-2.5\cdot 10^{-9} C[/tex] and [tex]F=18 nN = 18 \cdot 10^{-9} N[/tex], so we can find the magnitude of the electric field:

[tex]E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m [/tex]

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

[tex]e=1.6\cdot 10^{-19} C[/tex]

Therefore, the magnitude of the force acting on the proton will be

[tex]F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N[/tex]

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

Final answer:

To find the internal resistance of a 12.0-V car battery with specific measurements, use Ohm's Law.

Explanation:

Internal resistance of a battery can be calculated using Ohm's Law, where internal resistance = (emf - terminal voltage) / current. In this case, with an EMF of 12.0 V, terminal voltage of 8.3 V, and current of 89 A, the internal resistance would be calculated as (12.0 V - 8.3 V) / 89 A.

This gives an internal resistance of approximately 0.0427 ohms.

for those on ~ a p e x ~ note that this question asks when the resistors are connected in SERIES

if you have the question for in PARALLEL the answer is:

> The current the battery equals to the sum of the currents in the resistors

wink wonk

If you double the velocity of an object, its momentum doubles. But for the same increase in velocity, the kinetic energy increases four times.

Explanation:

We can prove this statement by looking at the formulas for the momentum and kinetic energy of an object.

Momentum: [tex]p=mv[/tex]

Kinetic energy: [tex]K=\frac{1}{2}mv^2[/tex]

where m is the mass of the object and v its velocity.

From the two formulas, we see that the momentum is directly proportional to the velocity, while the kinetic energy is proportional to the square of the velocity. Therefore, if we double the velocity, the momentum increases by a factor 2, while the kinetic energy increases by a factor [tex]2^2=4[/tex].

Momentum is directly proportional to an object's mass and velocity, while kinetic energy is directly proportional to an object's mass and the square of its velocity. When velocity is doubled, kinetic energy increases four times, while momentum doubles.

Momentum and kinetic energy are both physical quantities that describe the motion of an object. Momentum is directly proportional to an object's mass and velocity, while kinetic energy is directly proportional to an object's mass and the square of its velocity.

In terms of velocity, if you double the velocity of an object, its kinetic energy increases four times (K = (1/2)mv²), but its momentum doubles. This is because momentum is equal to mass times velocity (p = mv), so when velocity doubles, momentum also doubles. However, since kinetic energy is proportional to the square of velocity, when velocity doubles, kinetic energy increases by a factor of four.

Therefore, the statement that best compares momentum and kinetic energy is: If you double the velocity of an object, its kinetic energy increases four times. But for the same increase in velocity, the momentum doubles.

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Answer: The correct answer for the fill in the blank is B.  Mental age involves calculating the chronological age at which a person functions.

Mental age corresponds to the attainment of mental abilities by an individual. It is related to intelligence and based on the chronological age ( calculated from the date on which individual was born) at which any average individual attains the same level of mental abilities.

Therefore, mental age involves calculation of the chronological age at which a person functions and attains mental abilities.

Answer: Opposite direction

Skeletal muscles arise in antagonistic pairs where the muscles contract to produce opposite movements at the same joint. Antagonistic muscles move a body part in one direction by contraction, the other moves that part in opposite direction.

In addition, when a muscle contracts to produced movement, its antagonist relaxes to allow movement to take place such as biceps muscles is a flexor muscle for elbow joint and triceps is the antagonist.

The location in space where Earth's and the Moon's gravitational attractions are equal is closer to the Moon due to its significantly lesser mass compared to Earth. By applying Newton's law of universal gravitation, it can be deduced that this point must be nearer the Moon, as Earth's stronger gravitational force diminishes with increasing distance.

The gravitational attraction between two objects depends on both their masses and the distance between them, according to Newton's law of gravitation. The force of gravity is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers of mass. Based on the data provided, the Earth's gravitational force is much stronger than that of the Moon due to its greater mass. However, as the distance from the Earth increases, its gravitational pull weakens.

Given that the mass of the Moon is about 1/81 of the Earth's mass and the distance from the Earth to the Moon is approximately 3.80×105 km, there exists a point where the gravitational forces exerted by Earth and the Moon on an object are equal, known as the Lagrange point L1. This point is closer to the Moon than to Earth because the Moon's weaker gravitational force requires a shorter distance to match the stronger gravitational force of the Earth. To find this exact point, one would use the formula from Newton's law of gravitation and set the forces equal to each other, solving for the distance from Earth at which this equilibrium occurs.

Considering the information provided about the distances and gravitational forces, we can infer that the space pod mentioned in the question, at the point where Earth's and the Moon's gravity cancel each other out, would indeed be closer to the Moon. This is a consequence of the vast difference in mass between the Earth and the Moon and the inverse-square law of gravity.

Given second-harmonic frequencies and added mass, solve [tex]\(1.5^2 = \frac{m + 1.2}{m}\) to find \( m = 0.96 \)[/tex] kg.

To find the mass [tex]\( m \)[/tex] that results in the given second-harmonic frequencies, we will use the formula for the frequency of standing waves on a wire under tension.

The second-harmonic frequency for a wire is given by:

[tex]\[f = \frac{2}{L} \sqrt{\frac{T}{\mu}}\][/tex]

where:

- [tex]\( f \)[/tex] is the frequency,

- [tex]\( L \)[/tex] is the length of the wire,

- [tex]\( T \)[/tex] is the tension in the wire,

- [tex]\( \mu \)[/tex] is the linear mass density of the wire.

The tension [tex]\( T \)[/tex] in the wire is due to the hanging mass [tex]\( m \)[/tex] and is given by:

[tex]\[T = mg\][/tex]

where [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]).

The second-harmonic frequency is given, so for the initial mass [tex]\( m \)[/tex]:

[tex]\[f_1 = 180 \, \text{Hz}\][/tex]

[tex]\[180 = \frac{2}{L} \sqrt{\frac{mg}{\mu}}\][/tex]

When an additional 1.2 kg is added to the mass, the new mass becomes [tex]\( m + 1.2 \)[/tex] kg and the second-harmonic frequency becomes:

[tex]\[f_2 = 270 \, \text{Hz}\][/tex]

[tex]\[270 = \frac{2}{L} \sqrt{\frac{(m + 1.2)g}{\mu}}\][/tex]

To find [tex]\( m \)[/tex], we will set up the ratio of the two frequencies and solve for [tex]\( m \)[/tex]:

[tex]\[\frac{f_2}{f_1} = \frac{270}{180} = 1.5\][/tex]

Using the ratio of the frequencies:

[tex]\[\frac{270}{180} = \frac{\sqrt{\frac{(m + 1.2)g}{\mu}}}{\sqrt{\frac{mg}{\mu}}}\][/tex]

Squaring both sides to eliminate the square roots:

[tex]\[\left(\frac{270}{180}\right)^2 = \frac{(m + 1.2)g}{mg}\][/tex]

[tex]\[\left(1.5\right)^2 = \frac{(m + 1.2)}{m}\][/tex]

[tex]\[2.25 = \frac{m + 1.2}{m}\][/tex]

Multiplying both sides by [tex]\( m \)[/tex]:

[tex]\[2.25m = m + 1.2\][/tex]

Solving for [tex]\( m \)[/tex]:

[tex]\[2.25m - m = 1.2\][/tex]

[tex]\[1.25m = 1.2\][/tex]

[tex]\[m = \frac{1.2}{1.25}\][/tex]

[tex]\[m = 0.96 \, \text{kg}\][/tex]

Therefore, the mass [tex]\( m \)[/tex] is [tex]\( 0.96 \)[/tex] kg.

for the whole quiz of intro to psychology ;)

1. Stimulus--> Physiological changes--> emotion

2. Darwin

3.You experience physiological changes and a feeling of fear simultaneously

4. Cannon-Bard theory

5. James-Lange theory

6. David G. Myers

7. Happiness

8. Acting happy

9. negative and dysfunctional aspects of emotion and behavior.

10. an aspect of consciousness characterized by a certain physical arousal involving facial and bodily changes, brain activation, and tendencies toward action, all shaped by cultural rules.

I do this cause i wished someone did this for me. So stress no more from falling behind and Ace that quiz. your welcome comrades

-10/11/18

According to the Cannon-Bard theory, physiological changes and the feeling of fear occur simultaneously.

The Cannon–Bard theory is also called the thalamic theory of emotion which is related to the thalamus. It is a part of the brain that deals with sensory and motor functions. The main concepts of this theory are that emotional expression arises from the function of hypothalamic structures while emotional feeling arises from stimulation of the dorsal thalamus.

This theory states that the lower part of the brain controls the experience of emotion while the upper part of the brain, called the cortex, controls the expression of emotion. These two parts of the brain react together

In this theory, stimulating events trigger emotions and physiological responses that occur at the same time. For example, seeing a bear in the woods can cause both a feeling of fear (an emotional response) and a faster heartbeat (a physical response).

Thus, according to the Cannon-Bard theory, physiological changes and the feeling of fear occur simultaneously.

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The horizontal component of the velocity of a stone thrown upward at an angle remains constant during its flight, as there is no horizontal force to affect it. This is due to the principles of projectile motion in physics. Gravity only affects the vertical velocity, not the horizontal.

In physics, the motion of a projectile thrown at an angle can be separated into its vertical and horizontal components. Both of these components operate independently of each other. When a stone is thrown upward at an angle, the horizontal component of its velocity remains constant throughout the flight, under ideal conditions, because there's no horizontal force to cause acceleration or deceleration.

So, as the stone rises and falls, the horizontal velocity remains unchanged because gravity only affects the vertical component of the velocity. In this process known as projectile motion, gravity pulls the object downward, causing the vertical component of velocity to decrease as the stone rises, and increase as it falls, but the horizontal component is unaffected by gravity. The effect of air resistance is typically ignored in basic physics problems, but in reality, it would gradually decrease the stone's velocity in both directions.

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Here we’re solving a problem where a ball is projected horizontally from a height of h=1.8 m with a horizontal velocity of Vx. At the impact with the ground, the ball has travelled 0.5 m horizontally.

Solution:

We will need kinematic equations

V1^2-V0^2=2aS ………………….(1)

S=V0*t + (1/2)at^2………………..(2)

Where

S=displacement (distance), m

V0=initial velocity, m/s

V1=final velocity, m/s

a=acceleration, m/s^2

t=time, seconds

At the point of impact, there a vertical velocity (downwards) of Vy.

The horizontal velocity Vx remains constant since projection till impact.

Vertical velocity Vy:

Using equation (1),

V0=0 (projected horizontally, so vertical velocity=0)

S=1.8 m (downwards)

a=9.81 m/s^2 (acceleration due to gravity, downwards)

=>

Vy=V1=sqrt(V0^2+2*a*S)=sqrt90+2*9.81*1.8)=5.9427

Horizontal velocity, Vx:

ball travelled 0.5m in time t it took ball to hit ground.

Using equation (2),

S=1.8m

V0=0

a=9.81

=>

1.8=0*t+(1/2)(9.81)t^2

Solve for t

t=sqrt(2*1.8/9.81)=0.60578 s

Horizontal velocity, Vx = 0.5/0.60578 = 0.82538 s

Speed of ball on impact is the vectorial sum of Vx and Vy:

Speed = sqrt(Vx^2+Vy^2)=sqrt(5.9427^2+0.82538^2)=5.99977 m/s, say 6.0 m/s.

Answer: 0.82

On ED2020

The amount of flow of charge per unit time is known as the current. The largest current that this resistor can take safely without burning out will be 0.6 amperes.

The process of losing power in the form of heat as a result of the main activity is known as power dissipation. Dissipation of power is a natural occurrence.

All of the circuit's resistors that have a voltage drop across them will dissipate power. Due to the conversion of electrical energy to thermal energy, all resistors will have a power rating.

The power dissipated on a resistor is given as

P = I²R

I is current  flowing through it  

R is the resistance

[tex]\rm {P = I^2R}\\\\I=\sqrt{\frac{P}{R} }\\\\I=\sqrt{\frac{10}{27.36\\}[/tex]

I = 0.6 A

Hence the largest current that this resistor can take safely without burning out will be 0.6 amperes.

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Answer: Option D

Explanation: The fertilizers consists of the phosphates and nitrates. These aids in the growth of the plant. Fertilizers used by the plants are sometimes carried away due to the rain.

The run off from the agricultural field contains phosphates that is transferred from the field to the nearby water bodies.

The answer i got was 50 miles per hour

The weight of an astronaut standing on the moon can be calculated using the equation weight = mass × acceleration due to gravity. By calculating the weight using the ratio of acceleration due to gravity on the moon compared to that on Earth, we find that an astronaut with a weight of 210 lb on Earth would weigh approximately 158.18 lb on the moon.

The weight of an astronaut standing on the moon can be calculated using Newton's second law of motion. The formula to calculate weight is weight = mass × acceleration due to gravity. Since the acceleration due to gravity on the moon is approximately 1/6 that at the surface of the earth, we can calculate the weight using this ratio.

First, we need to convert the weight on earth from pounds to kilograms. Since 1 kilogram is approximately 2.2 pounds, we divide the weight on earth (210 lb) by 2.2 to get the weight in kilograms.

Next, we multiply the weight in kilograms by the acceleration due to gravity on the moon (1/6 of 9.8 m/s^2) to find the weight of the astronaut on the moon.

This results in a weight of approximately 158.18 lb on the moon for an astronaut whose weight on earth is 210 lb.

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Answer:

C) TURBINE

Explanation:

As we know that wind powered system is used to generate electricity using wind energy.

Here wind powered system has different components

1) Blades: these are long plates connected to the turbine. when these plates encountered high speed wind they start rotating on its axle.

This will produce kinetic energy in the axle which is transferred to the turbine.

2) Turbine: It is used to convert mechanical energy of the wind into useful electrical energy.

This is continuously driven by the kinetic energy of wind

So here correct answer is

C) Turbine

a dense substance that is hard and incompressible

Answer:

B.A circuit that has little or no resistance

Explanation: a circuit is said to be short circuit if it has little or no resistance.

Shirt circuiting is very dangerous and can be done intentionally or unintended

D could also be correct if an amount of wire is ought to be used and you use a little wire. Knowing that wires have internal resistance too.

It could lead to short circuit