1 identify two ways (factors) you can change in order to the charge the objects experience

Answers

Answer 1

Answer:

Follows are the solution to this question:

Explanation:

Please find the correct question in the attached file:

In point (a), The one factor is the charge on each of the objects. In point (b), at this point, the second factor is used as the distance between the charges.
1 Identify Two Ways (factors) You Can Change In Order To The Charge The Objects Experience
Answer 2

The two ways to charge an object are Friction and Induction.

Charging an object:

To charge and object generally we can think of three ways, Friction, Conduction, and Induction. Here I am going to explain charging by friction and induction.

(i) Charging by Friction:

In this case, two objects are rubbed against each other. The friction between the two objects creates enough energy for the transfer of charges  ( electrons) from one object's surface to the other. So the imbalance in the electrons produces a charge on the objects.

(ii) Charging by induction:

In this case, an already charged object is brought near a neutral object but there is no contact between the objects. The charged object attracts or repels the electrons of the neutral object, depending upon the type of charge it has. Thus creating a charge imbalance in the neutral object.

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Related Questions

m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.

Answers

Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor:  T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

T₂ = 49.3°C

Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answer the following questions. You hold a small metal ball of mass a height above the floor. You let go, and the ball falls to the floor. Choose the origin of the coordinate system to be on the floor where the ball hits, with up as usual. Express all results in terms of , , and . Just after release, what are and

Answers

Answer:

y(i) = h

v(y.i) = 0

Explanation:

See attachment for elaboration

An electromotive force E(t) = 200, 0 ≤ t ≤ 50 0, t > 50 is applied to an LR-series circuit in which the inductance is 50 henries and the resistance is 5 ohms. Find the current i(t) if i(0) = 0.

Answers

Answer:

The answer is below

Explanation:

An LR series circuit has a differential equation in the form of:

[tex]L\frac{di}{dt}+iR=E(t)\\ \\Given\ that\ 50H,R=5\ ohms,E(t)=200,for\ 0 \leq t \leq\ 50. Hence:\\\\50\frac{di}{dt}+5i =200\\\\\frac{di}{dt}+0.1i =4\\\\Solving\ the\ differential\ equation:\\\\The\ integrating\ factor(I)=e^{\int\limits {0.1} \, dt }=e^{0.1t}. The\ DE\ becomes:\\\\e^{0.1t}\frac{di}{dt}+e^{0.1t}(0.1i) =4e^{0.1t}\\\\e^{0.1t}i=\int\limits {4e^{0.1t}} \, dt \\\\e^{0.1t}i=40e^{0.1t}+A\\\\i(t)=40+Ae^{-0.1t}\\\\but\ i(0)=0\\\\0=40+Ae^{-0.1(0)}\\\\A=-40\\\\i(t)=40-40e^{-0.1t}\\\\[/tex]

At 50 seconds:

[tex]i(50)=40-40e^{-0.1*50}\\\\i(50)=40-40e^{-5}[/tex]

[tex]L\frac{di}{dt}+iR=E(t)\\ \\Given\ that\ 50H,R=5\ ohms,E(t)=0,for\ t> 50. Hence:\\\\50\frac{di}{dt}+5i =0\\\\\frac{di}{dt}+0.1i =0\\\\\frac{di}{dt}=-0.1i\\\\\frac{di}{i}=-0.1dt\\\\\int\limits {\frac{di}{i}} =\int\limits {-0.1} \, dt\\ \\ln(i)=-0.1t+A\\\\taking\ exponential:\\\\i=e^{-0.1t+A}\\\\i=e^{-0.1t}e^A\\\\i(t)=Ce^{-0.1t}\\\\i(50)=40-40e^{-5}=Ce^{-5}\\\\C=40(e^5-1)\\\\i(t)=40(e^5-1)e^{-0.1t}\\\\[/tex]

[tex]i(t)=\left \{ {{40-40e^{-0.1t}\ \ \ \ 0 \leq t \leq 50 } \atop {40(e^5-1)e^{-0.1t}\ \ \ \ t>50}} \right.[/tex]

A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?

Answers

Answer:

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

Explanation:

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.


Objects want to keep doing the same thing is a way of stating ....

Answers

Answer:

Objects want to keep doing the same thing is a way of stating Newtons First Law.

The correct option is (a) inertia. "Objects want to keep doing the same thing" it implies a connection to the principle of inertia in physics. Inertia is the property of matter that resists changes in its state of motion or rest.

According to Newton's first law of motion, an object will remain at rest or continue moving in a straight line at a constant velocity unless acted upon by an external force. This property is commonly known as "the law of inertia." In other words, objects tend to maintain their current state of motion (whether at rest or in motion) unless influenced by an external force.

When we say "Objects want to keep doing the same thing," we're drawing an analogy between this scientific principle and human behavior. It suggests that objects, like humans, have a natural inclination to resist change and continue their current course of action. This analogy helps convey the idea that objects exhibit a tendency to persist in their existing state.

"Objects want to keep doing the same thing" to inertia emphasizes the notion that objects, like humans, tend to maintain their current state of motion or rest unless influenced by an external force.

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The correct question is:

Objects want to keep doing the same thing is a way of stating ......

a) inertia

b) force

c) power

d) moment

PLEASE HELP 25 POINTS WHAT IS THE ANSWER

Answers

The answer is letter C vro

thanks me.later

1.
Which of these is a relative adverb?

where

lovely

who

Answers

Who is the right answer

Answer:

Where

Explanation:

Similar to Hippocrates, modern scientists who study etiology believe that
sickness is caused by spiritual events such as demons entering the body.
diseases come from natural causes such as bites from infected insects.
all diseases can be cured by the right combinations of different treatments.
the best treatment for a germ is a calming practice such as yoga or massage.

Answers

In  Medicine, "etiology" means the origin or cause of a disease.

Similar to Hippocrates, modern scientists who study etiology believe that   diseases come from natural causes such as bites from infected insects.

Answer:diseases come from natural causes such as bites from infected insects

Explanation: just took test

until a train is a safe distance from the station, it must travel at 5 m/s. once the train is on open track, it can speed up to 45m/s. if it takes 8 seconds to reach 4 m/s, what is the acceleration of the train?

Answers

Answer:

5 meters per second squared

Explanation:

We calculate the acceleration using the formula:

a = (vf - vi) / t

where "vf" is the final velocity, "vi" the initial velocity, and "t" the time it took to change from the initial velocity to the final one.

In our case:

a = (45 - 5) / 8 = 40 / 8 = 5 m/s^2

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the projectile will be ("E4" means "*10^4") *
1 point
4E4 J
1E4 J
1E3 J
None of the above

Answers

Answer:

[tex]K=4\times 10^4\ J[/tex]

Explanation:

Given that,

Mass of cannon is,M =  2000 kg

Mass of projectile, m = 2 kg

Velocity of the projectile, v = 200 m/s

We need to find the kinetic energy of the projectile. We know that the formula for the kinetic energy of an object is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\\text{Putting all the values, we get}\\\\K=\dfrac{1}{2}\times 2\times (200)^2\\\\K=40000\ J[/tex]

or

[tex]K=4\times 10^4\ J[/tex]

So, the kinetic energy of the projectile will be [tex]4\times 10^4\ J[/tex].

The volume of a cube is found by multiplying length times

width times height. If an object has a volume of 1.44 m?,

what is the volume in cubic centimeters? Remember to mul-

tiply each side by the conversion factor.

Answers

Answer:

The volume in cubic centimeters is 1,440,000

Explanation:

You know that an object has a volume of 1.44 m³  and you want to calculate the volume in cubic centimeters cm³.

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other), the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So, [tex]x=\frac{c*b}{a}[/tex]

Since 1 m³ is equal to 1,000,000 cm³ ( this is the conversion factor) , then you can apply the following rule of three: if 1 m³ is equal to 1,000,000 cm³, 1.44 m³ is equal to how many cm³?

[tex]cm^{3} =\frac{1.44m^{3}*1,000,000 cm^{3} }{1m^{3} }[/tex]

cm³=1,440,000

The volume in cubic centimeters is 1,440,000

Consider the four scenarios involving visible light. In scenario A, visible light has a wavelength of 715.3 nm. Determine its frequency, energy per photon, and color.

Answers

Answer:

1. frequency (f) = 4.194 × 10^14Hz

2. Energy of photon (E) = 2.779 × 10^-19 J

3. Color is RED

Explanation:

1. Based on the information in this question, a visible light has a wavelength of 715.3 nm.

λ = v/f

Where; λ = wavelength (nm)

v = speed of light (3 × 10^8m/s)

f = frequency (Hz)

f = v/λ

f = 3 × 10^8/715.3 × 10^-9

f = 0.004194 × 10^(8+9)

f = 0.004194 × 10^17

f = 4.194 × 10^14Hz

2. Energy per photon is calculated thus;

E = hf

Where; E = energy of photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency (Hz)

E = 6.626 × 10^-34 × 4.194 × 10^14

E = 27.789 × 10^(-34+14)

E = 27.789 × 10^-20

E = 2.779 × 10^-19 J

3. Based on the wavelength range of the visible spectrum, wavelength of 715.3nm falls between 625-740nm, which is RED color of light. Hence, the color of the visible light is RED.

The frequency is "[tex]419.5\times 10^{12} \ s^{-1}[/tex]", energy per photon is "[tex]2.780\times 10^{-19} \ J[/tex]" and the color is "red".

Given:

Wavelength,

[tex]\lambda = 715.1 \ nm[/tex]

or,

           [tex]= 715.1 \ nm\times (10^-9 \ m/nm)[/tex]

As we know the formula,

→ [tex]Frequency = \frac{Speed \ of \ light}{wavelength}[/tex]

By substituting the values, we get

                    [tex]= \frac{(3\times 10^8) }{715.1\times 10^{-9} }[/tex]

                    [tex]= 419.5\times 10^{12} \ s^{-1}[/tex]  

Now,

The energy per photon will be:

→ [tex]E = Planck's \ constant\times Frequency[/tex]

      [tex]= 6.626\times 10^{-34}\times 419.5\times 10^{12}[/tex]

      [tex]= 2.780\times 10^{-19} \ J[/tex]

Thus the above answer is correct.

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un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración necesaria para detenerlo?

Answers

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

Vf: Velocidad final Vo: Velocidad inicial a: Aceleración d: Distancia recorrida

En este  caso:

Vf: 0 m/s, porque el avión se detieneVo: 50 m/sa: ?d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

[tex]a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}[/tex]

a= - 10.42 m/s²

La aceleración necesaria para detener el avión es - 10.42 m/s².

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.9 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 16.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?

Answers

Answer:

The velocity is [tex]v_b = 20.17 \ m/s[/tex]

Explanation:

From the question we are told that

   The mass of the ball is  [tex]m = 0.245 \ kg[/tex]

   The radius is  [tex]r =  59.8 \  cm  =  0.598 \ m[/tex]

   The force is  [tex]F =  30.9 \ N[/tex]

   The speed of the ball is  [tex]v = 16.0 \ m/s.[/tex]

Generally the kinetic energy at the top of the circle is mathematically represented as

    [tex]K_t  =  \frac{1}{2} *  m  *  v^2[/tex]

=> [tex]K_t  =  \frac{1}{2} *  0.245   *  16.0 ^2[/tex]  

=> [tex]K_t  =  31.36 \ J[/tex]  

Generally the work done by the force applied on the ball from the top to the bottom  is mathematically represented as

       [tex]W =  F *  d[/tex]

Here  d is the length of  a semi - circular arc which is mathematically represented as

       [tex]d =  \pi * r[/tex]

So

      [tex]W =  30.9 *  0.598  [/tex]

      [tex]W = 18.48 \ J [/tex]

Generally the kinetic energy at the bottom is mathematically represented as

      [tex]K_b  =  \frac{1}{2} *  m *  v_b^2[/tex]

=>    [tex]K_b  =  \frac{1}{2} *  0.245  *  v_b^2[/tex]

=>   [tex]K_b  =  0.1225  *  v_b^2[/tex]

From the law of energy conservation

     [tex]K_t +  W  =K_b[/tex]

=>    [tex]31.36+  18.48 = 0.1225  *  v_b^2[/tex]

=>    [tex]v_b = 20.17 \ m/s[/tex]

A force of 40 N is applied tangentially to the rim of a solid disk of radius 0.10 m. The disk rotates about an axis through its center and perpendicular to its face with a constant angular acceleration of 145 rad/s2. Determine the mass of the disk.

Answers

Answer:

 m = 2,759 kg

Explanation:

For this exercise we use the torque relationship

           τ = I α

         

the moment is

          τ= F r sin θ

since the force is tangential to the ring, the angle is 90º sin 90 = 1

          τ = F r

the moment of inertia of a ring is given by

         I = m r²

let's substitute

           F r = m r²α

            m = F / r α

let's calculate

          m = 40 / (0.10 145)

          m = 2,759 kg

A 66-N ⋅ m torque acts on a wheel with a moment of inertia 175 kg ⋅ m2. If the wheel starts from rest, how long will it take the wheel to make one revolution?

Answers

Answer:

t = 5.77 s

Explanation:

This exercise will use Newton's second law for rotational motion

            τ = I α

             α = τ / I

             α = 66/175

             α = 0.3771   rad/s²

now we can use the rotational kinematics relations, remember that all angles must be in radians

 

            θ = 1 rev = 2π radians

               

            θ = w₀ t + ½ α t²

as the wheel starts from rest w₀ = 0

             t = √ (2θ/α)

let's calculate

             t = √ (2 2π / 0.3771)

             t = 5.77 s

what is a proper way and a safe way to dispose batteries

Answers

Answer:

throw them away............

You need to throw them away

Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years

Answers

Answer:

T = 1.4696 10⁴ years

Explanation:

For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law

               F = ma

               G m M / r² = m a_c = m v² / r

                G M / r = v²

the speed of the circular orbit is

               v = 2π r / T

           

we substitute

               G M / r = 4π² r² / T²

               T² = (4π² / G M)  r³

Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse

               T² = (4π² /GM)  a³

the constant is worth

                (4π² / GM) = 2.97 10⁻¹⁹    s² / m³

let's reduce the distance to SI units

AU is the distance from the Earth to the Sun

               a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)

               a = 8.976 10¹³ m

               T² = 2.97 10⁻¹⁹ (8.976 10¹³)³

               T² = 21.4786 10²²

               T = 4.63 10¹¹ s

Let's reduce to years

               T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)

               T = 1.4696 10⁴ years

What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

Answers

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

A tennis ball is dropped from the top of a tall building. a second tennis ball is thrown down from the same building make a statement about the acceleration of each tennis ball.

A) The first ball falls with a greater acceleration

B) The second ball falls with a greater acceleration.

C) they both are with the same acceleration because they started from the same height.

D) they both are the same acceleration because they are in free fall.

E) more information is required

Answers

The two balls have the same acceleration because they are in free fall.

The two balls projected downwards from the top of the build are influenced by gravity. Both ball experience free fall due to gravity.

The equation of motion of the two ball is given as;

v² = u² + 2gh

where;

v is the final velocity of each ballu is the initial velocity of each ballh is the distance traveled by each ballg is acceleration due to gravity = 9.8 m/s²

Thus, we can conclude that the two balls have the same acceleration because they are in free fall.

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What determines the paths of electrons around a nucleus?

a. the fundamental quantum of energy

b. interaction with other atoms

c magnetism

d. the number of neutrons in the nucleus

Answers

Answer:

* positive repulsion, which makes the electrons move as far away as possible

* Attraction with the core atomizes

he analysis is the foundations of quantum theorizing

Explanation:

Electrons are deposited in an atom following two principles.

* positive repulsion, which makes the electrons move as far away as possible

* Attraction with the core atomizes

The mathematical model used for the analysis is the foundations of quantum theorizing

If a projectile hits a stationary target, and the projectile continues to travel in the same direction, the mass of the projectile is less than the mass of the target. the mass of the projectile is equal to the mass of the target. the mass of the projectile is greater than the mass of the target. nothing can be said about the masses of the projectile and target without further information. this is an unphysical situation and will not actually happen.

Answers

The correct arrangement of the question is;

If a projectile hits a stationary target, and the projectile continues to travel in the same direction,

A) the mass of the projectile is less than the mass of the target.

B) the mass of the projectile is equal to the mass of the target.

C) the mass of the projectile is greater than the mass of the target.

D) nothing can be said about the masses of the projectile and target without further information.

E) this is an unphysical situation and will not actually happen.

Answer:

Option C: The mass of the projectile is greater than the mass of the target.

Explanation:

We want to find what will happen when a projectile continues in motion after it hits a target.

Now, for the projectile to keep moving in that direction after it hits the target, it means it had a force bigger than the force of the target to overpower it and force it to move with it.

Now, from law of inertia, Force = ma.

But in this case acceleration is 0 because the speed of the projectile is constant.

Thus, the force depends on the mass. So for a higher force, the mass of the projectile has to be more than that of the stationary object.

Thus, option C is correct

What is the final velocity (in m/s) of a hoop that rolls without slipping down a 3.50-m-high hill, starting from rest

Answers

Answer:

8.29m/s

Explanation:

Given

Height of hill H = 3.50m

Initial velocity u = 0m/s

Required

Final velocity v

Using the equation of motion

v² = u²+2gH

v² = 0²+2(9.81)(3.5)

v² = 0+7(9.81)

v² = 68.67

v = √68.67

v = 8.29m/s

Hence the final velocity of the hoop is 8.29m/s

Basalt is a rock that cooled quickly after lava erupted through a volcano.

What is the best description of its texture?

Answers

when the lava erupted from the volcano basalt was the rock that cooled quickly or basalt was the only one rock that cooled quickly after the lava erupted from the volcano

Explanation:

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A 1200-kg car pushes a 2100-kg truck that has a dead battery to the right. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N. Which, if any, of the following gives a correct acceleration constraint for this problem? The variable ac is the acceleration of the car, and the variable t is the acceleration of the truck.

a. ac=at
b. ac=−at
c. ac=21001200at
d. ac=12002100at
e. There are no acceleration constraints for this problem.

Answers

Answer:

Option A: a_c = a_t

Explanation:

Let's denote the force acting on the car due to the truck as F_ct

Similarly, the force acting on the truck due to the car is denoted as F_tc.

Now, F_ct = m_c × a_c

Where;

m_c is mass of car

a_c is acceleration of car

Also, F_tc = m_t × a_t

Where;

m_c is mass of truck

a_c is acceleration of truck

Now, since the car is the one pushing the truck, from Newton's laws, we can say that they stick together and move in the same direction at the same acceleration.

Thus, the truck will move with an acceleration equal to that of the car pushing it.

Thus; a_c = a_t

If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by the ordinary matter (Mord = 1.90 ✕ 1041 kg), find the speed of the star. Assume a circular orbit and assume all the Galaxy's matter is concentrated at the center.

Answers

Answer:

The value is  [tex]v = 1.309*10^{5}\ m/s[/tex]

Explanation:

The radius is [tex]r = 7.40 *10^{20} \  m[/tex]

The mass of the ordinary matter is [tex]M_{rod} =  1.90 *10^{41}\  kg[/tex]

Generally the speed of the star is mathematically represented as

         [tex]v = \sqrt{\frac{G * M}{r} }[/tex]

Here G is the gravitational constant with a value

        [tex]G = 6.67384 * 10^{-11}[/tex]

So

      [tex]v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }[/tex]

=>    [tex]v = 1.309*10^{5}\ m/s[/tex]

If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p

Answers

Answer:

8.93*10^13 N.

Explanation:

Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:

       [tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]

where, m1 = mass of  the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:

       [tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]

Fg = 8.93*10^13 N.

A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .

Answers

Answer:

The magnitude is [tex]B = \frac{450}{7.9* 10^5}[/tex]

The direction is  the positive x axis

Explanation:

From the question we are told that

   The  electric field is  E = 450 V/m in the negative y ‑direction

   The speed of the proton is  [tex]v= 7.9* 10^5\  m/s[/tex] in the positive z direction

Generally the overall force acting on  the proton is mathematical represented as

              [tex]F_E =  q(\vec E + \vec v  * \vec B)[/tex]

Now for the beam of protons to continue un-deflected along its straight-line trajectory then  [tex]F_E =0[/tex]

So

          [tex] 0  =  q( E (-y) + v(z)  * \vec B)[/tex]

=>      [tex]E\^y = v \^ z  * \vec B[/tex]

Generally from unit vector cross product vector multiplication

         [tex]\^ z  \ *  \  \^ x  =  \^  y[/tex]

So the direction of  B (magnetic field must be in the positive x -axis )

So

       [tex]E\^y = v \^ z  *  B\^ x [/tex]

=>     [tex]E\^y = vB ( \^ z  *  \^ x) [/tex]

=>     [tex]E\^y = vB ( \^y) [/tex]        

=>      [tex]E = vB [/tex]  

=>      [tex]B = \frac{450}{7.9* 10^5}[/tex]

=>      [tex]B =  0.0005696 \ T [/tex]  

   

   

1pt During a phase change, as energy is added, the temperature of a substance
O A. increases
O B. is constant
O C. decreases
O D. first increases and then decreases to its original value

Answers

Answer:

D

Explanation:

Because first phase thr it will follow the law of thermodynamics as it will reach the equilibrium point then decrrases afterwards

The sun produces energy via nuclear fusion at the rate of 4×1026 J/s . Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.)

Answers

Answer:

2 x 10^10 years.

Explanation:

Given that the sun produces energy via nuclear fusion at the rate of 4×1026 J/s . Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.)

Let us first calculate energy from hydrogen gas.

4(1.007825) + 2(0.00549) - 4.002603 = 0.029795 amu

Since 4.0313 amu H+ > 0.029795 amu =  ratio of 1 to 0.00739

dE = ((2*10^30 kg) x 0.8 x 0.25)(3.00x10^8 m/s)^2 = 3.6x10^46 kg x m^2/s^2  

3.6x10^46 kg x m^2/s^2 x 0.00739 = 2.6604 x 60^44 kg x m^2/s^2

4x10^26 J/s x s = 2.6604x10^44 kg x m^2/s^2  solve for s.

s = 6.651x10^17 seconds

6.651x10^17 seconds x 1 min/60 s x 1 hr/ 60 min x 1 day/ 24 hr x 1 yr / 365 days = 2.1x10^10 years

 Based on the proposed overall fusion equation, it will therefore take the sun shine 2 x 10^10 years before it exhausts its hydrogen fuel.

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