A box is being pulled to the right over a rough surface. t > fk , so the box is speeding up. Suddenly the rope breaks. What happens? The box:_________.
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
b. stops immediately.
c. continues speeding up for a short while, then slows and stops.
d. continues with the speed it had when the rope broke.

Answers

Answer 1

Answer:

a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.

Explanation:

Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.

Since there is a net force, there is an acceleration and thus an increasing velocity.

When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.

Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.

So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.

So, the answer is a.


Related Questions

two factor of a number are 5 and 6 .what is the number show working​

Answers

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT

Answers

Answer:

B

Explanation:

The seasons are measured in how far or close the earth is to the sun.

Phát biểu nào sau đây là SAI?
A. Cường độ điện trường là đại lượng
đặc trưng cho điện trường về phương
diện tác dụng lực.
B. Điện trường tĩnh là điện trường có
cường độ E không đổi tại mọi điểm.
C. Đơn vị đo cường độ điện trường là
vôn trên mét (V/m).
D. Trong môi trường đẳng hướng,
cường độ điện trường giảm  lần so với
trong chân không

Answers

Answer:

B.

Explanation:

sana makatulong sayo

A solid aluminum sphere of radius R has moment of
inertia I about an axis through its center. What is the
moment of inertia about a central axis of a solid
aluminum sphere of radius 2R?
1. 21
2. 41
3. 87
4. 161
5. 321​

Answers

Answer:

5. 32I

Explanation:

The moment of inertia of a solid sphere about its central axis is given by

I = [tex]\frac{2}{5} MR^2[/tex]            ------------------(i)

Where;

M = mass of the sphere

R = radius of the sphere.

From the question;

Case 1: The aluminum sphere has a radius R and moment of inertia I.

This means that we can substitute these values of R and I into equation (i) and get;

I = [tex]\frac{2}{5} MR^2[/tex]       --------------(ii)

M is the mass of the aluminum sphere and is given by;

M = pV

Where;

p = density of aluminum

V = Volume of the sphere = [tex]\frac{4}{3} \pi R^3[/tex]

=> M = p([tex]\frac{4}{3} \pi R^3[/tex])              --------------------(*)

Case 2: An aluminum sphere with a radius of 2R instead.

Let the moment of inertia in this case be I' and mass be M'

Substituting R = 2R, M = M' and I = I' into equation (i) gives

I' = [tex]\frac{2}{5} M'(2R)^2[/tex]       ------------------(iii)

Where;

M' = pV'

p = density of aluminum

V' = volume of the sphere = [tex]\frac{4}{3} \pi (2R)^3[/tex]

=> M' = p([tex]\frac{4}{3} \pi (2R)^3[/tex])

Rewriting gives;

M' = p([tex]\frac{4}{3} \pi (2)^3(R)^3[/tex])

M' = p([tex]\frac{4}{3} \pi8(R)^3[/tex])

M' = 8p([tex]\frac{4}{3} \pi R^3[/tex])

From equation (*), this can be written as

M' = 8M

Now substitute all necessary values into equation (ii)

I' =  [tex]\frac{2}{5} M'(2R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(2R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(2)^2(R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(4)(R)^2[/tex]

I' =  [tex]\frac{2}{5} (32M)(R)^2[/tex]

I' =  [tex]32[\frac{2}{5}MR^2][/tex]

Comparing with equation (ii)

I' =  [tex]32[I][/tex]

Therefore, the moment of inertia about a central axis of a solid

aluminum sphere of radius 2R is 32I

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground?

Answers

Explanation:

the answer is in the above image

A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)

Answers

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

What is distance ?

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

learn more about distance

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Can someone help me
Btw the last one say current

Answers

Magnitude: Magnitude generally refers to the quantity or distance.

Attraction: A force that makes things move together and stay together.

Repulsive: The feeling of being repelled, as by the thought or presence of something; distaste, repugnance, or aversion.

Current: Current is the rate of flow of electric charge. A potential difference (voltage) across an electrical component is needed to make a current flow through it.

I hope that helps :]

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?

Answers

Answer:

[tex]v=6.65m/sec[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=97.6[/tex]

Coefficient of kinetic friction  [tex]\mu k=0.555[/tex]

Generally the equation for Frictional force is mathematically given by

 [tex]F=\mu mg[/tex]

 [tex]F=0.555*97.6*9.8[/tex]

 [tex]F=531.388N[/tex]

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 [tex]a_f=-\mu g[/tex]

 [tex]a_f=-0.555 *9.81[/tex]

 [tex]a_f=-54455m/sec^2[/tex]

Therefore

 [tex]v=u-at[/tex]

 [tex]v=0+5.45*1.22[/tex]

 [tex]v=6.65m/sec[/tex]

calculate the electrical potential at a point P a distance of 1 m from either two to charge of +10 micro coulomb and -5 micro coulomb which are 10 cm apart calculate also the potential energy of a +2 micro coulomb charge placed at a point p​

Answers

Answer:

a)  V = 45 10³ V, b) U = 4.59 J

Explanation:

a) The electric potential for a series of point charges is

         V = k ∑ [tex]\frac{q_i}{r_i}[/tex]

in this case point P is at a distance of 1 m from each charge, so the point is located perpendicular to the charges at its midpoint

 

         V = k ( [tex]\frac{q_1}{r} + \frac{q_2}{r}[/tex])

         V = 9 10⁹ (10 - 5/ 1) 10⁻⁶

         V = 45 10³ V

b) the potential energy is

           U = k (  [tex]\frac{q_1q}{r} + \frac{q_2q}{r} + \frac{q_1q_2}{r_2}[/tex] )

where r = 1m and r₂ is the distance between the two charges r₂ = 0.10 m

           U = 9 10⁹ (10 2 / 1 - 5 2/1 - 10 5 /0.10) 10⁻¹²

           U = 9 10⁻³  510

           U = 4.59 J

Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied

Answers

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km​

Answers

Answer:

[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]

Answer:

108 km/hr or 0.03 km/s

Explanation:

conversion factor for m/s to km/hr is 5/18

conversion factor for m/s to km/s is 1/1000

Two skaters, both of mass 75 kg, are on skates on a frictionless ice pond. One skater throws a 0.4-kg ball at 6 m/s to his friend, who catches it and throws it back at 6.0 m/s. When the first skater has caught the returned ball, what is the velocity of each of the two skaters

Answers

Answer:

v = 0.064 m/s

Explanation:

Given that,

The mass of two skaters = 75 kg

The mass of a ball = 0.4 kg

The speed of the ball = 6 m/s

The speed of skater = 6 m/s

We need to find the velocity of each of the two skaters.

Under the values given the moment with respect to the ball and which is subsequently transmitted to people it would be given by:

[tex]P=0.4(6)+0.4(6)\\\\=4.8\ kg-m/s[/tex]

We know that,

P = mv

Where

v is the velocity of each skater.

[tex]v=\dfrac{p}{m}\\\\v=\dfrac{4.8}{75}\\\\=0.064\ m/s[/tex]

So, the velocity of each of the skaters is 0.064 m/s.

Can someone write this question clearly and send it to me? Don't just say the answer. Draw and write clearly please​

Answers

Explanation:

acceleration is weight*gravity

tension is the weight In Newtons

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-high hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 430 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. Part A What spring constant should you specify

Answers

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -  

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is  

 k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

How do solar panels work with conduction, convection and radiation?

Answers

Answer:

In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.

Explanation:

Part A
Calculate the work done when a force of 6.0 N moves a book 2.0 m
Express your answer with the appropriate units.

Answers

Answer:

Work done applied = 12 newton-meter

Explanation:

Given examples:

Force applied = 6 newton

Distance of book = 2 meter

Find from the given data:

Work done

Computation:

The equation can be used to compute work.

Work done applied = Force applied x Distance of book

Work done applied = Force x Distance

Work done applied = 6 x 2

Work done applied = 12 newton-meter

A major artery with a 1.3 cm^2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area of 0.6 cm^2. By what factor is the average velocity of the blood reduced when it passes into these branches?

Answers

Answer:

When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12

Explanation:

Given;

initial area of the artery, A₁ = 1.3 cm²

Area of each smaller 18 arteries, a₂ = 0.6 cm²

Total area of the smaller 18 arteries, A₂ = 18 x 0.6 cm²

Apply flow rate equation;

Q = AV

where;

Q is the flow rate of the blood

V is the average velocity of the blood

If the flow rate is constant, then;

A₁V₁ = A₂V₂

[tex]V_2 = \frac{A_1V_1}{A_2} = \frac{1.3\times V_1}{18\times 0.6} \\\\V_2 = 0.12 \ V_1[/tex]

When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12

The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cross-sectional area. According to theory, what should be the ratio of the resistance of the second coil to the first coil and the fourth coil to the third

Answers

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Which of the following changes would double the force between two charged particles?
A. Doubling the amount of charge on each particle
B. Increasing the distance between the particles by a factor of 2
C. Decreasing the distance between the particles by a factor of 2
D. Doubling the amount of charge on one of the particles

Answers

Answer:

Doubling the amount of charge on one of the particles.

Explanation:

The force between two charges is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

Where

r is the distance between charges

or

[tex]F\propto \dfrac{1}{r^2}[/tex]

On doubling the charge on one of the particle,

F' = 2F

So, the force gets doubled. Hence, the correct option is (d).

don't answer for points you will be reported ​

Answers

Explanation:

Glasses or Contacts. You might not realize it, but if you wear glasses or contact lenses, this is light refraction at play. ...

Human Eyes. Human eyes have a lens. ...

Prism. ...

Pickle Jar. ...

Ice Crystals. ...

Glass. ...

Twinkling Stars. ...

Microscope or Telescope.

Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.

highschool physics, not college physics

Answers

Answer:

Answer:

A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.

Explanation:

Edge.

Answer:

The motion of the paper airplane  is best explained by horizontal inertia and vertical pull of gravity.

Explanation:

What is horizontal inertia and vertical pull of gravity?

Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .

While vertical pull is due to the earth .

In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.

Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.

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plsss help me
thank you very much

Answers

It’s not loading. My bad

Có bao nhiêu đặc trưng sinh lý của âm

Answers

Answer:

Ba đặc trưng sinh lí của âm là độ cao, độ to và âm sắc.

Explanation:

You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of [10] kg, where should the mass be placed to give the pendulum a period of 2.00 seconds

Answers

Answer:

The distance is 1.026 m.

Explanation:

mass of rod, M = 1.23 kg

Length, L = 1.25 m

mass, m = 10 kg

Time period, T = 2 s

Let the distance is d.

The formula of the time period is given by

[tex]T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16 = 0.64 + 10d^2\\\\d= 1.026 m[/tex]

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3m3 of water flows through the dam each second. The water is released 220 mm below the top of the reservoir. If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?

Answers

Answer:

The output electric power is 1338876 W.

Explanation:

Volume, V = 690 cubic meter

height, h = 220 mm = 0.22 m

efficiency = 90 %

time , t = 1 s

Let the mass is m.

m = volume x density  

m = 690 x 1000 = 690000 kg

The input power is

P = m g h = 690000 x 9.8 x 0.22 = 1497640 W

The electric power out put is

[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]

The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3.25 × 10-9 mm.

Answers

Answer:

Option B. 3.25×10¯¹⁰ mm.

Explanation:

Measurement (cm) = 0.00325×10⁻⁸ cm

Measurement (mm) =?

The measurement in mm can be obtained as follow:

1 cm = 10 mm

Therefore,

0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm

0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm

Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.

The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm

The number given is in standard form and can be written as 3.25*10^-11 cm.

To convert this from centimeter to millimeter, we have to multiply this value by 10.

Conversion Units1 cm - 10mm100cm = 1m1000m = 1km

So, let's 3.25*10^-11 by 10 and get our value in mm

[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]

From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]

Learn more about conversion of units here;

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How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.

Answers

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

The motion that is not repeated in regular interval of time is called....?
please help me!​

Answers

Answer:

Motion which repeats itself after regular intervals of time is known as periodic motion.
Motion which repeats itself

If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?

Answers

Answer:

if 1 light year was one millimeter then 105,700 light years = 105,700 mm, (or 105.7 meters in case you needed to simplify or something)

A projectile of mass m is fired horizontally with an initial speed of v0​ from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0​, h, and g : Are any of the answers changed if the initial angle is changed?

Answers

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Other Questions
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