In Figure (a), both batteries have em - 1.20 V and the external resistance R is a variable resistor. Figure (6) gives the electric potentials V between the terminals of each battery as functions of
Curve 1 corresponds to battery 1, and curve 2 corresponds to battery 2. The horizontal scale is set by R, -0.500 . What is the internal resistance of (a) battery 1 and (b) battery 2?
0.60
V(V)
0
-0.350
(
R(0)
(6)

Answers

Answer 1

Answer:

i know da way ese'

Explanation:


Related Questions

Determine the volume of the ring/tube using Archimedes' Principle and compare your results to the volume of the ring/tube calculated from physical measurements. Do not tie the thread directly to the balance; use the paper clip as a hook. Assume that the density of the water is 1.0 g/cm. Neatly show all work and provide all necessary data. If the TA cannot duplicate your results from the data that you provide, your score will be drastically reduced.
Volume of ring/tube via Archimedes' Principle (A): ___________
Volume of ring/tube via Physical Measurement (B): __________
Percent Difference-=(A-B)/ (A+B)/2 x100%-=_________
Percent Difference <-3% <-5% <= 10% <--15% <-20% | >20%
Points 50 45 40 25 10 0

Answers

Answer:fuafnshf dj en jz

Explanation:

Can. Nd I do j dj cdj an man Jaz jxn nah an b

Since water is much denser than air, deep-sea divers experience a much higher ambient pressure underwater. Each 10 meters of depth underwater adds another 1 atm to the ambient pressure experienced by the diver. (Note: this is in addition to the 1 atm ambient pressure at the surface of the water!) What pressure, in psi, is experienced by a diver 50.0 meters below the surface of the water

Answers

Answer:

If you are at sea level, each square inch of your surface is subjected to a force of 14.6 pounds. The pressure increases about one atmosphere for every 10 meters of water depth. At a depth of 5,000 meters the pressure will be approximately 500 atmospheres or 500 times greater than the pressure at sea level.

Explanation:

At sea level, a force of 14.6 pounds is applied to every square inch of your surface. For every 10 meters of sea depth, the pressure rises by approximately one atmosphere. The pressure will be about 500 atmospheres, or 500 times more than the pressure at sea level, at a depth of 5,000 meters.

What is pressure?

Pressure is defined as the force applied perpendicularly to an object's surface divided by the surface area over which it is applied.

Pressure is the physical amount of force exerted on a particular area.

Pressure can be expressed as

Pressure = Force / area

There are three types  of pressure.

Absolute pressureGauge pressureDifferential pressure

Thus, at sea level, a force of 14.6 pounds is applied to every square inch of your surface. For every 10 meters of sea depth, the pressure rises by approximately one atmosphere. The pressure will be about 500 atmospheres, or 500 times more than the pressure at sea level, at a depth of 5,000 meters.

To learn more about pressure, refer to the link below:

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What has to happen for a star to join the main sequence ?

1) Nuclear Fusion

2) Shell Heating

3) Use up most of its available fuel

4) Hydrostatic Equilibrium

Answers

Answer:

Nuclear fusion

Explanation:

This is because main sequence of star is powered by fusion of hydrogen to helium atoms together and this process releases energy. The energy released when the gas collapse into a protostar make the center of the protostar to be extremely hot. When the core becomes very hot, nuclear fusion can start.

Your friend says an appliance uses energy. How would you correct his statement?

Answers

Answer:

Not all appliances run on energy. Some of them run on gas. Some both. It just depends on the age of the appliance, the make of the appliance, and the company who made it.

An astronaut on the Moon has a weight of 128 N and a mass of 80 Kg. What
is the gravitational field strength of the moon?

Answers

Answer:

1.6 g

Explanation:

Use the formula: W =  m× g

       ∴ g =  W / m

Here, W = 128N

         m = 80 Kg

     ∴ g = 128 ÷ 80

           = 1.6 N (Ans)

Lee and Leigh are twins. At their first birthday party, Lee is placed on a spaceship that travels away from the earth and back at a steady 0.714 c . The spaceship eventually returns, landing in the swimming pool at Leigh's eleventh birthday party. When Lee emerges from the ship, how old is he?

a. He is still only 1 year old
b. He is 8 years old
c. He is also 11 years old
d. He is 18 years old

Answers

Answer:

b. He is 8 years old

Explanation:

We will use Einstein's formula for time dilation, to calculate the age of Lee. Because Lee was traveling comparable to the speed of light, his age must be lesser than Leigh.

[tex]T = \frac{T_o}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]

where,

T₀ = Time on Earth = ?

T = Relative Time = 10 years

v = relativistic speed of Lee = 0.714 c

c = speed of light = 3 x 10⁸ m/s

Therefore,

[tex]10\ years = \frac{T_o}{\sqrt{1-\frac{(0.714\ c)^2}{c^2} } } \\\\[/tex]

T₀ = 7 years

Hence, the age of Lee will be:

[tex]Lee's\ Age = 1\ year + 7\ years = 8\ years[/tex]

b. He is 8 years old

Help please due tomorrow

Answers

Hope this helps!!!!

A negatively charged rod briefly touches a neutral metal ball. The metal ball will now be ____________

Answers

Answer:

let say it will be positive

So the the greater the height, the
farther something can fall, the greater
the potential energy.
True
False

Answers

The heavier the object and the higher it is above the ground, the more gravitational potential energy it holds.

People travel from all over the world to see more than 30 glaciers at Glacier National Park in Alaska.
Which factor causes glacial movement downhill?

A. gravity
B. oceans
C. snow
D. wind

Answers

The answer I believe is c
Explanation
During the day when temperatures are higher, the snow melts and water enters the cracks in the rock. When the temperature drops below 0°C the water in the crack freezes and expands by about 9 per cent. This makes the crack larger.
The answer is c

Glaciers move by sliding over bedrock or underlying gravel and rock debris. With the increased pressure in the glacier because of the weight, the individual ice grains slide past one another and the ice moves slowly downhill.

What is the force holding you down?

Answers

Answer:

The force holding you down is gravity.

Explanation:

Gravity is a force between two objects with mass. It pulls things together. You have mass, and the Earth has mass, so gravity tries to pull you and the Earth together. The gravitational force is much bigger for more massive objects.

Answer:

Gravitational Force

Explanation:

Gravitational Force holding everything down

-TheUnknownScientist

HELP!!!!

A student did an experiment to determine the
specific heat capacity of an unknown metal.
She heated 1.00 x 10- kg of the metal to 225°C
and quickly placed it in an insulated container
(negligible specific heat capacity) that contained
0.0900 kg of water at a temperature of 18.0°C.
What is the final temperature of the water if the
specific heat capacity of the metal is
2.11 x 102 J/kg.°C?

Answers

Answer:

T₂ = 16.83°C

Explanation:

Applying the law of conservation of energy principle here in this situation we get the following equation:

[tex]Energy\ Lost\ by\ Metal = Energy\ Gaine\ by\ Water\\m_{metal}C_{metal}(T_2-T_{1metal}) = m_{w}C_{w}(T_2-T_{1w})[/tex]

where,

T₂ = Final Temperature of Water = Final Temperature of Metal = ?

C_metal = Specififc Heat Capacity of the metal = 2.11 x 10² J/lg.°C

T_1metal = Initial Temperature of Metal = 225°C

m_metal = mass of metal = 1 x 10⁻²[tex](0.01\ kg)(211\ J/kg.^oC)(T_2-225^oC) = (0.09\ kg)(4184\ J/kg.^oC)(T_2-18^oC)\\2.11 T_2 - 474.75 = 376.56T_2 - 6778.08\\374.45T_2 = 6303.33\\[/tex] kg (exponent assumed due to missing info in question)

C_w = Specififc Heat Capacity of the water = 4184 J/lg.°C

T_1w = Initial Temperature of water = 18°C

m_w = mass of water = 0.09 kg

Therefore,

[tex](0.01\ kg)(211\ J/kg.^oC)(T_2-225^oC)=(0.09\ kg)(4184\ J/kg.^oC)(T_2-18^oC)\\\\2.11 - 474.75T_2 = 376.56 - 6778.08T_2\\[/tex]

T₂ = 16.83°C

When the electrons reach the collector, they flow towards the positivly charged grid. The resulting current is measured. Note that as the electrons accelerate from the cathode toward the grid, they collide with the mercury atoms. Assume that these collisions are completely elastic. How does the collected current vary if the ΔVgridΔVgrid is slowly increased? View Available Hint(s)

Answers

Answer:

We can conclude by saying that in the beginning current will increase but after sometime, it becomes saturated.

Explanation:

Note: No information on change in number of electron generated.

Since there is a collision, the electrons emitted will not reach the collector at same time. As the voltage is increased, the the speed with which the electrons will reach the collector starts to increase. Due to this, electric current will first increases till all the emitted electrons reach the collector. Since we are not provided with the information that number of electrons generated are changing, after increasing voltage current will increase for some time and then reaches a saturated state.

We can conclude by saying that in the beginning current will increase but after sometime it becomes saturated.

PLEASE HELP ME WITH THIS ONE QUESTION
A diverging lens has a focal length of –10 cm. Suppose an object is placed 15 cm in front of the lens. a) What is the image distance? b) Is the image real or virtual?

Answers

Answer:

b

Explanation:

For given diverging lens(Concave) ; a) The distance of the image is -6 Cm, b) The image is virtual.

What is diverging lens?

The lens that diverge incident parallel beam of light after refraction from the lens, is called diverging or concave lens.

The images formed by it are always smaller, upright, towards object and between focus and optical center of the lens.

Given data:

a)

u = -15 Cm, f = -10 Cm, v= ?by using lens formula,

        [tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\[/tex]

On putting the values,

        [tex]\frac{1}{-10} = \frac{1}{v} - \frac{1}{-15}\\\frac{1}{-10} = \frac{1}{v} + \frac{1}{15}[/tex]

        [tex]\frac{1}{v} = \frac{1}{-10} -\frac{1}{15} \\\frac{1}{v} = \frac{-3-2}{30} \\[/tex]

         [tex]v = -6 Cm[/tex]

b) Magnification ,

m= ([tex]\frac{v}{u}[/tex])

on putting values,

[tex]m =(\frac{-6}{-15} )=\frac{2}{5} =0.4[/tex]

since m is positive it means the image is virtual

Hence a) distance of image is -6 Cm and b) the image is virtual.

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A baseball was thrown off of a 35 meter high building. It lands 50 meters from the base of the building.

Answers

Wonderful.

Look out below !

Was the FAA notified ?

Learning Task 1: Analyze the figure and answer the questions that follow.

1. Which one is the charged object?

2. What made the hair of the girl to rise?​

Answers

Answer:

1) Van der rcf generator

2) the charge is distributed among all the hairs, as they all have the same potential,

the charges are of the same sign repel each other

Explanation:

1.) The object is a Van der rcf generator, which is loaded by friction,

The girl has no load

2) when the girl touches the sphere of the generator part of the electrons of this is transferred to the girl, when this charge reaches the hair, the charge is distributed among all the hairs, as they all have the same potential,

the charges are of the same sign repel each other

Why one won't save a document as a text file

Answers

Answer:

i dont know

Explanation:

do you mean to save the answer or to save your question

Calculate the equivalent resistance for each of the following circuits.

Answers

Answer:

5. 60 Ω

6. 60 Ω

7. 10 Ω

8. 0.625 KΩ

Explanation:

5. Determination of the equivalent resistance.

Resistor 1 (R₁) = 10 Ω

Resistor 2 (R₂) = 20 Ω

Resistor 3 (R₃) = 30 Ω

Equivalent Resistance (R) =?

Since the resistors are arranged in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 10 + 20 + 30

R = 60 Ω

Thus, the equivalent resistance is 60 Ω

6. Determination of the equivalent resistance.

Resistor 1 (R₁) = 10 Ω

Resistor 2 (R₂) = 35 Ω

Resistor 3 (R₃) = 15 Ω

Equivalent Resistance (R) =?

Since the resistors are arranged in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 10 + 35 + 15

R = 60 Ω

Thus, the equivalent resistance is 60 Ω

7. Determination of the equivalent resistance.

Resistor 1 (R₁) = 6 Ω

Resistor 2 (R₂) = 4 Ω

Equivalent Resistance (R) =?

Since the resistors are arranged in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂

R = 6 + 4

R = 10 Ω

Thus, the equivalent resistance is 10 Ω

8. Determination of the equivalent resistance.

Resistor 1 (R₁) = 10 KΩ

Resistor 2 (R₂) = 2 KΩ

Resistor 3 (R₃) = 1 KΩ

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/10 + 1/2 + 1/1

Find the least common multiple (lcm) of 10, 2 and 1. The result is 10. Divide 10 by each of the denominator and multiply the result obtained by the numerator. This is illustrated below:

1/R = (1 + 5 + 10) / 10

1/R = 16/10

Invert

R = 10/16

R = 0.625 KΩ

Thus, the equivalent resistance is 0.625 KΩ.

form
bonds with each other.
There are many kinds of mixtures. Some mixtures are
chunky like a mixture of peanuts and raisins. These
mixtures are called
I
mixtures.

Answers

Answer:

Homogeneous mixtures

Explanation:

I think so because homogeneous means mixed mixtures

what happens to the work done when a force is doubled and the distance moved remain the same?​

Answers

Answer:

It is doubled

Explanation:

f2=2f1

x1=x2=x

W1=f1*x1=f1*x

W2=f2*x2=f2*x=2*(fi*x)=2*W1

A ray of light is incident from air on the surface of a block of clear ice (nice=1.31) at an angle of 45.0° with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and refracted light.

Answers

Answer:

A ray of light is incident from air on the surface of a block of clear ice (nice=1.31) at an angle of 45.0° with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and refracted light.

Which of the following answers are true: Group of answer choices The larger the slit, the better the wavelength resolution of the spectrometer A slit can select wavelengths spatially. This is for example used in a monochromator, where focused light after refraction in a prism or diffracted from a grating can be selected through a narrow slit. A slit can be used to direct light alternatingly to the sample or the blank. A slit can be used as a photodetector. The narrower the slit, the better the wavelength resolution of t

Answers

Answer:

true  b, e

Explanation:

The expression that describes the diffraction of a grating is

         d sin θ = m λ

where d is the separation between the slits, m is the order of the spectrum

let's analyze the different answers

a) False. The size (height) of the slits does not influence the resolution, the number of them per unit of length influences, the greater the number, the smaller the distance (d) this they

b) True. The spectrum is resolved on a screen, in the form that each wavelength corresponds to a fixed distance from the central maximum, for a given order

c) False. The exit slit selects a given wavelength, but does not deflect the beam from its path

d) False. The slit lets in light, but does not measure its intensity

e) True. For a continuous spectrum, the wavelength variation that passes through a slit is proportional to its width. For a discrete spectrum the width of the slit does not affect the wavelength

What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)

Answers

Answer:

When we have a current I, we will have a magnetic field perpendicular to this current.

Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".

Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.

Of course, the intensity of the magnetic field is proportional to the current, given by:

B = (μ*I)/(2*π*r)

Where:

μ is a constant, I is the current and r is the distance between to the current.

Now remember that for a resistor:

R = ρ*L/A

R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.

If we increase the area of the wire (if we use a thicker wire).

And the relation between resistance and current is:

I = V/R

Where V is the voltaje.

Now, if we use a thicker wire, then the cross-section area of the wire increases.

Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.

And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.

If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.

Rough Surface with: Ms = 0.8 HK = 0.4

(a) Find the magnitude of the

force F needed to prevent the

book from sliding down the

rough wall.

F

M = 1.5 kg

600

(b) Find the minimum force F

needed to set the book in

motion up the rough wall with

constant velocity

Answers

Answer:

a)    F = 18.375N, b) F = 24.5 N

Explanation:

This exercise can be solved using the translational equilibrium equations.

Let's start by fixing a reference system with the horizontal x axis and the vertical y axis, from the statement of the exercise I understand that the wall is vertical and the book is supported on it, therefore the applied force is in the direction towards the wall

a) In this part the force that does not allow the movement of the book is requested, therefore the static friction coefficient must be used (μ_s = 0.8)

X axis  

       F - N = 0

       N = F

Y axis

       fr - W = 0

       W = fr

where W is the weight of the book.

The friction force has the formula

       fr = μ_s N

we substitute

       mg = μ_s F

       F = [tex]\frac{mg}{\mu_s }[/tex]

let's calculate

       F = 1.5 9.8 / 0.8

       F = 18.375N

b) In this case the book is moving so the friction coefficient to use is kinetic (   μ_K = 0.6)

   

       F = [tex]\frac{mg}{\mu_K }[/tex]

       

we calculate

        F = 1.5 9.8 / 0.6

        F = 24.5 N

Light from a 560 nm monochromatic source is incident upon the surface of fused quartz (n = 1.56) at an angle of 60°. What is the angle of reflection from the surface?

Answers

Answer:

The angle of reflection is "60°".

Explanation:

The given values are:

Light from monochromatic source,

= 560 nm

Angle of incidence,

= 60°

Surface of fused quartz (n),

= 1.56

Whenever a light ray was indeed occurring at a flat surface, it should be the law or concept of reflection which contains this same rays of light, the reflected ray as well as the "normal" ray at either the mirror surface.

According to the above law,

⇒  [tex]Angle \ of \ incidence=Angle \ of \ reflection[/tex]

then,

⇒  [tex]Angle \ of \ reflection=60^{\circ}[/tex]

What is a cyclic behavior

Answers

Answer:

Uzi

Explanation:

When somebody comes to me I don’t have any

A 2.3 kg , 20-cm-diameter turntable rotates at 110 rpm on frictionless bearings. Two 460 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. Part A What is the turntable's angular velocity, in rpm, just after this event

Answers

Answer:

The correct solution is "64 RPM".

Explanation:

The given values are:

Mass,

M = 2.3 kg

Diameter,

D = 20 cm

i.e.,

   = 0.2 m

Rotates at,

N = 110 rpm

Mass of block,

m = 460 g

i.e.,

   = 0.46 kg

According to angular momentum's conservation,

⇒  [tex]I_1\omega_1=I_2\omega_2[/tex]

then,

⇒  [tex]I_1=\frac{1}{2}MR_2[/tex]

On substituting the values, we get

⇒      [tex]=\frac{1}{2}\times 2.3\times (0.1)^2[/tex]

⇒      [tex]=\frac{1}{2}\times 0.023[/tex]

⇒      [tex]=0.0115 \ kg \ m^2[/tex]

Now,

⇒  [tex]I_2=I_1+2mR^2[/tex]

        [tex]=0.0115+2\times 0.46\times (0.1)^2[/tex]

        [tex]=0.0115+0.0092[/tex]

        [tex]=0.02 \ kg \ m^2[/tex]

then,

⇒  [tex]0.0115\times 110=0.02\omega_2[/tex]

⇒              [tex]1.265=0.02\omega_2[/tex]

⇒                  [tex]\omega_2=\frac{1.265}{0.02}[/tex]

⇒                       [tex]=63.25 \ or \ 64 \ RPM[/tex]

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 8.00 m. A person is standing 12.0 m away from the wall, equidistant from the loudspeakers. When the person moves 3.00 m parallel to the wall, she experiences destructive interference for the second time. What is the frequency of the sound

Answers

Answer: [tex]278\ Hz[/tex]

Explanation:

Given

Distance between two speakers is 8 m

Man is standing 12 away from the wall

When the person moves 3 parallel to the wall

the parallel distances from the speaker become 4+3, 4-3

Now, the difference of distances from the speaker is

[tex]\Delta d=\sqrt{12^2+(4+3)^2}-\sqrt{12^2-(4-3)^2}\\\Delta d=1.85\ m[/tex]

Condition for destructive interference is

[tex]\Delta d=(2n-1)\dfrac{\lambda }{2}=(2n-1)\dfrac{\nu }{2f}\\\\\Rightarrow f=(2n-1)\dfrac{v}{2\Delta d}[/tex]

for second destructive interference; n=2

[tex]\Rightarrow f=(2\times 2-1)\dfrac{343}{2\times 1.85}=278.10\approx 278\ Hz[/tex]

180 N
40 kg
140 N
Net Force =

Also how do you find the net force?

Answers

Answer:

720N

Explanation:

180+(40×10)+140=720 remember we can only add with same units ;1kg=10N therefore 40 kg=(40×10)N=400N

The net force would be the summation of all the forces in addition to the weight force of the 40 kg weight, thus the net force of all the forces would come out to be 712.4 Newtons.

What is Newton's second law?

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

F = ma

As given in the problem we have to find the net force,

Let us assume the acceleration due to gravity would be 9.81 m/s².

The force generated by the 40-kilogram weight =40 ×9.81 Newtons.

The force generated by the 40-kilogram weight = 392.4 Newtons

Net force = 180 + 140 +392.4

         

                =712.4 Newtons

Thus, the net force of all the forces would come out to be 712.4 Newtons.

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What is the
mass
density
of an object of
equal to 100 grams and volume of 20 cubic
centimeter​

Answers

Answer:

[tex]d=5\ g/cm^3[/tex]

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3[/tex]

So, the density of the object is equal to [tex]5\ g/cm^3[/tex].

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