5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is doubled, how does the electric field change?

Answers

Answer 1

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

Explanation:

From the question we are told that

  The electric field is  [tex]E(r)_1 =  [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]

    Now when the outer diameter is doubled, the radius(b) is also doubled

So

    [tex]E(r)_2 =  [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]

Now

      [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]

=>   [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]

=>  [tex]\frac{E(r)_2}{E(r)_1}  =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]

Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

5. In Investigation 2, If Everything Stays The Same, Except The Diameter Of The Outer Ring Is Doubled,
5. In Investigation 2, If Everything Stays The Same, Except The Diameter Of The Outer Ring Is Doubled,

Related Questions

Hi, Please help.. I have assignments due tonight and I need someone to help me when a question I have generally..

Okay so if Density = Mass divided by Volume and I put that information into a calculator it comes out as
ex. 1.938773646 how do I make it look like something like this 1.4?

Answers

Answer:

did you tried putting it in standard form

Answer:

It may help to round all of the given numbers up to at least 1 or 2 decimal points or you could round up the number you get to 1 or 2 decimal places. For example, for this question round up your answer to 1.9 or 1.94

Explanation:

hope this helps!!

What do mammoths and tigers need energy for

Answers

Muscles
.............

The first ionization potential for calcium (Z = 20, A = 40) is 6.11 eV. Singly-ionized calcium (Ca+) produces two very strong absorption lines in the Sun’s spectrum discovered by Joseph Fraunhofer in 1814, who named them "H" and "K" (he didn’t know they were from calcium, as this was >100 years before the development of quantum mechanics). Both lines always appear together, with lambda subscript H equals 3968 end subscript Å and lambda subscript K equals 3933 Å; hence they are called a "doublet
A. What is the speed of an electron that has just barely enough kinetic energy to collisionally ionize a neutral calcium atom? What is the speed of a calcium ion with this same kinetic energy?
B. What is the temperature T of a gas in which the average particle energy is just barely sufficient to ionize a neutral calcium atom?
C. The lower energy level of both lines is the ground state of Cat. What is the difference in energy in eV) between the two states that correspond to the upper energy levels of the Hand Klines, respectively? How does this compare to the energy of a calcium K photon? Can these two lines can be formed by transitions to upper energy levels with different principal quantum numbers (different n), or do they represent transitions with the same n but some different higher-order quantum number? Explain your reasoning based on your understanding of the general behavior of atomic energy levels (En).

Answers

Answer:

A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B)  T = 4.7 10⁴ K, C) n₂ = 42

Explanation:

A)  For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.

Let's reduce the units to the SI system

        E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J

The kinetic energy of the electron is

        K = ½ m v²

         E₀ = K

         v = √ 2E₀ / m

         v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)

         v = √ (2.14857 10¹²)

         v = 1.47 10⁶ m / s

now the speed of a calcium ion is asked, let's find sum

        m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg

         

        v = √ (2E₀ / M)

         v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)

        v = √ (0.2994457 10⁸)

        v = 0.5426 10⁴ m / s

B) the terminal energy of an ideal gas is

             E = 3/2 kT

              T = ⅔ E / k

              T = ⅔ (9,776 10-19 / 1,381 10-23)

              T = 4.7 10⁴ K

C) To calculate the energy of these lines we use the Planck expression

              E = h f

where wavelength and frequency are related

              c =λ f

              f = c /λ

let's substitute

              E = h c /λ

let's look for the energies

λ = 396.8 nm

  E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹

           E₁ = 5.0126 10⁻¹⁹ J

λ = 393.3 nm

           E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹

           E₂ = 5.0572 10⁻¹⁹ J

The difference in energy between these two states is

          ΔE = E₂ -E₁

          ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J

          ΔE = 0.0446 10⁻¹⁹ J

let's reduce eV

         ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

         ΔE = 2.787 10⁻² eV

Now let's use Bohr's atomic model for atoms with one electron,

               E = -13.606 Z² / n²

where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium

               n = √ (13.606 Z² / E)

λ = 396.8 nm

E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV

               n₁ = √ (13.606 20² / 3.132875)

               n₁ = 41.7

since n must be an integer we take

               n₁ = 42

λ = 393.3 nm

E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV

              n₂ = √ (13.606 20² / 3.16075)

              n₂ = 41.5

Again we take n as an integer

               n₂ = 42

We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin

2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The second hallway is filled with students, and she 4covers its 48.0 m length at a speed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0 m length at a speed of 5.00 m/s. How long does it take Suzette to make to class? Did Suzette beat the bell?

Answers

Answer:

62 secondsno

Explanation:

The total travel time Suzette experiences is the sum of the times in each hallway. Using

  time = distance/speed

we can add the times.

  (35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)

  = 10 s + 40 s + 12 s

  = 62 s

It takes Suzette 62 seconds to get to class. She does not beat the bell.

A crane uses a single cable to lower a steel girder into place. The girder moves with constant speed. The cable tension does work WT and gravity does work WG. Which statement is true

Answers

Explanation:

Work done by a force is given by :

[tex]W=Fd\cos\theta[/tex]

Where

F is force, d is displacement and [tex]\theta[/tex] is the angle between F and d.

In this problem, a crane is moving in downward direction, the force gravity is in downward direction and the tension is in upward direction.

We know that if force and displacement is in same direction, work is positive while if force and displacement is in oposite direction, work is negative.

I would mean that, [tex]W_g[/tex] is positive, because gravity is parallel to the displacement and [tex]W_t[/tex] is negative, because the tension is opposite to the displacement.

what phase changes take place when you are adding energy to the substance

Answers

Answer:

During a phase change, a substance undergoes transition to a higher energy state when heat is added, or to a lower energy state when heat is removed. Heat is added to a substance during melting and vaporization. Latent heat is released by a substance during condensation and freezing. Explanation:

Help me Please!!!!!!!

Answers

The speed is equal to the area under the line up to the point where t .=15 s.
Do find the area of the triangle and that if the rectangle and add them together.
The area of the triangle is 25 and the rectangle is also 25 so the speed is 50 m/s

The precision value of measuring tape is

1)0.1cm

2)0.1mm

3)1cm

4)0.01cm

Answers

C.1cm

Explanation:

precision is how close two or more measurements are to each other

4?

Explanation:

sorry im not sure but

You can always take a metre ruler as a starting point. Your metre ruler has the same precision as your 15.0cm or 30.0cm ruler, so bring a ruler during exams as they'll come in handy ;)

The order goes like this:

rulers: 0.1cm or 1mm

measuring tape: 0.01cm or 0.1mm

vernier calipers: 0.001cm or 0.01mm

micrometer screw gauge: 0.0001cm or 0.001mm

((if i'm not wrong))

Please help me with this!

Answers

hope this helps good luck

A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.

Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?

How far does block two travel, d2 in meters, before coming to rest after the collision?

Answers

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:

Friction

Given that,

[tex]m_{1}[/tex] = 18.5 kg

d = 2.3m

To find,

Acceleration of block 1:

∑[tex]F = ma[/tex]

⇒ -m₁gμ = m₁a

⇒ a = -gμ

⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]

∵ a [tex]= -2.16 m/s^2[/tex]

Now,

To determine the velocity of block one prior to the collision:

We know,

The initial velocity of block 1 = 8.25 m/s

⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx

⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]

∵ [tex]v = 7.63 m/s[/tex]

We also know,

[tex]m_{2}[/tex] = 7.25 kg

Now,

The velocity of block 2 post collision:

⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision

Through this,

⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]

∵[tex]v = 19.5 m/s[/tex]

The distance can be found through:

⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]

⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]

∵ Δ[tex]x = 87.8 m[/tex]

Thus, 19.5 m/s and 87.8 m are the correct answers.

Learn more about "Friction" here:

brainly.com/question/13357196

Pressure and temperature ______ with depth below Earth’s surface.

Answers

Answer:

Pressure increases as you move deeper below earth's surface.

Tempurature  increases as you move deeper below earth's surface.

Hope this helps!

Explanation:

If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?

Answers

Answer: f= M×A

1.75kg×24= 42N

Explanation:

Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!

The plate is a model for how sunlight hits Earth’s surface. Which parts of Earth are most similar to the plate with an axis angle of 0°?

Answers

Answer: The parts closer to the equator.

Explanation:

The parts of the earth closer to the equator are similar to the plate, because they receive more sunlights than other parts. Which makes them hotter than any other regions in the earth. Example of such countries are Gabon, Uganda, Kenya, Maldives, the democratic republic of Congo, sao tome and principle  e.t.c

Is the answer clockwise (CW) or counter clockwise (CCW) ?

Answers

I believe it’s counter clockwise hun

A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?

Answers

Answer:68.15m/s

Explanation:

Given:

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

Formula:

v₁²=v₁²+2a (x)

Set up:

=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]

Solution:68.15m/s

An object at rest starts accelerating.
If it travels 20 meters to end up going 10 m/s.
what was its acceleration?
Variables:
Equation and Solve:

Answers

Answer:

We are given:

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

Solving for 'a'

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration? (68.15 m/s)

Answers

v² - u² = 2 ax

where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.

So

v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)

v² = 4645 m²/s²

v ≈ 68.15 m/s

Why is it better to use the metric system, rather than the English system, in scientific measurement?

A. The English system uses one unit for each category of measurement.
B. The metric system uses one unit for each category of measurement.
C. The English system uses consistent fractions that are multiples of 10.
D. The metric system utilizes a variety of number conversions.

Answers

A. The English system uses one unit for each category of measurement.

Answer:

A

Explanation:

Question 4
Which of the following is unique for any given element?
O the mass of a neutron
o the number of neutrons
o the charge on the electons
O the number of protons

Answers

the number of protons

it's unique for any element because it's determined by the atomic number and no two elements have the same atomic number

Which would increase the speed of a sound wave?
O A wave passes from a solid to a liquid while remaining the same temperature.
The medium increases in temperature while remaining in the same phase.
The medium decreases in temperature while remaining in the same phase.
O A wave passes from a liquid to a gas while remaining the same temperature.

Answers

Answer:

The medium increases in temperature while remaining in the same phase

Explanation:

The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.

What is meant by the temperature coefficient of sound wave?

The temperature coefficient of a sound wave is defined as the measure of increase in the velocity of the sound wave per unit rise in its the temperature.

Here,

The speed of a sound wave is affected by various factors in the medium through which it propagates. Among them, speed of the wave is mostly influenced by the temperature of the medium.

The speed of a sound wave in a medium is directly proportional to the square root of its absolute temperature. So,

       v [tex]\alpha[/tex] √T

where v is thee speed of the sound wave and T is the absolute temperature.

Therefore,

As the temperature of the medium increases, the kinetic energy of the wave particles increases. Thus the speed of the sound wave is increased. As a result, the sound waves will move faster.

Hence,

The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.

To learn more about temperature coefficient of sound, click:

https://brainly.com/question/29006600

#SPJ7

N₂ + H₂
NH3
how do i balance this equation?

Answers

Answer:

N2 + 3H2 ----->  2NH3

Explanation:

Reactants side:

2 Nitrogen

5 Hydrogen

Products Side:

2 Nitrogen

5 Hydrogen

The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be

Answers

This question is incomplete

Complete Question

m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.

a) if the system is released from rest what will be its acceleration

Answer:

0.7 m/s²

Explanation:

The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.

(a) if the system is released from rest what will be its acceleration

g = acceleration due to gravity = 9.81 m/s²

Coefficient of Kinetic Friction = μk = 0.30

m1 = 10kg

m2 = 4.0kg

The formula to solve question a is given as:

a = acceleration at rest

m2g- μk m1g = (m1+ m2) a

Making a the subject of the formula:

a = (m2g- μk×m1g )/(m1+ m2)

a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)

a = 0.7 m/s²

Which of the following is the closest to the scientific fact
A. A hypothesis
B. A theory
C. An opinion
D. A prediction

Answers

B - A theory seems to be the closest

A theory is the closest to a scientific fact (Option B).

A scientific theory is a well-sustained scientific idea that has been verified using the scientific method.

A scientific theory can be refuted by the emergence of new lines of evidence against some aspect of this scientific statement.

A hypothesis is a given explanation about a question that emerged by observing the natural world.

In conclusion, a theory is the closest to a scientific fact (Option B).

Learn more in:

https://brainly.com/question/2375277

What is the answer to this ?

Answers

Symbolic interactionism

The smokestack of a stationary toy tra in consists

of a vertical spring gun chat shoots a steel ball a meter

or so straight into the air-----so straight that the ball

always falls back into the smokescack. Suppose the

train moves at constant speed along the straight track. Do you think the ball will still return to the smoke-

stack if shot from the moving train? What if the train

gains speed along the straight track? What if it moves at a constant speed on a circular track? Discuss why your answers differ,

Answers

Answer:

i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack

ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack

iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track

Explanation:

The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack

The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack

The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track

An electron moving in the direction of the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the

Answers

Answer:

-z

Explanation:

The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it  due to a magnetic field pointing in the +z direction.

As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.

Engineers are using computer models to study train collisions to design safer
train cars. They start by modeling an elastic collision between two train cars
traveling toward each other. Car 1 is traveling north at 20 m/s and has a mass
of 12,745 kg. Car 2 is traveling south at 15 m/s and has a mass of 4,125 kg.
After the collision, car 1 has a final velocity of 3 m/s north. What is the final
velocity of car 2?
A. 56 m/s north
B. 56 m/s south
C. 38 m/s south
D. 38 m/s north

Answers

Answer:

did you get the answer???

Answer: 38 m/s north

Explanation:

The light bulbs are identical. Initially both bulbs are glowing. What happens when the switch is closed

Answers

Answer:

They turn off

Explanation:

At an amusement park, a swimmer uses a water slide to enter the main pool. You may want to review (Pages 234 - 241) . Part A If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.81 m , what is her speed at the bottom of the slide

Answers

Answer:

Her speed at the bottom of the slide is 7.42 m/s

Explanation:

From the question,

The swimmer starts at rest, that is, her initial speed, u is 0 m/s.

Since she slides without friction and descends through a vertical height, then it is a free fall motion (due to gravity).

Also, from the question,

She descends through a vertical height of 2.81 m.

To determine her speed at the bottom of the slide, that is her final speed,

From one of the equations of motion for freely falling bodies

v² = u² + 2gh

Where v is the final speed

u is the initial speed

g is acceleration due to gravity (g = 9.8 m/s²)

and h is height

From the question,

u = 0 m/s

h = 2.81 m

Putting the values into the equation

v² = u² + 2gh

v² = 0² + 2×9.8×2.81

v² = 55.076

v =√55.076

v = 7.42 m/s

Hence, her speed at the bottom of the slide is 7.42 m/s.

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground

Answers

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

Other Questions
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