3

Answer:

#0.28%#

Explanation:

The layer chromosphere is #2000# km thick.

The radius of the sun is #0.7 * 10^6# km.

To calculate this you have to just do same as we do in calculation of our exam's marks percentage.

Just imagine,

Your exam is of total marks of #0.7 * 10^6#(not possible but you have #color(white)0000000000000000000000000000000#to just imagine this )
out of this you got #2000# marks.

To calculate the percentage you have to do something like this:

#=2000/(0.7 * 10^6) * 100#

Right?

Just do same calculation for this question,
Total radius of sun is #0.7 * 10^6# km.
out of this #2000# km is chromosphere.

#=2000/(0.7 * 10^6) * 100#

#=0.002857*100#

#=0.28%#

I have added image for your help by which you can know where is chromosphere of the sun.

www.haikudeck.com

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More great answers

1

Answer:

It was Nazi Germany's last attempt to win the war. It was the largest battle on the Western Front.

Explanation:

The Battle of the Bulge was the result of a last ditch effort by the Germans to turn the tide of battle back in their favor in the West.

The German plan was to drive through the Ardennes and capture the port of Antwerp vital to the Allied war effort. If successful the American and British forces would have to bring supplies all the way from Normandy crippling the Allied war effort.

The American forces defending the area were unprepared for the massive assault of the German forces. The Germans made rapid advances and threatened to achieve a decisive breakthrough.

The stubborn defense of Bastogne by the 101 airborne, the rapid response of Patton's armored units, and the break in the weather allowing the impact of American air superiority combined to stop the German offensive.

The Battle left a large bulge in the Allied lines, but did not break the Allied forces as hoped. The German offensive cost the Germans armored units and elite infantry. The Germans would not be able to launch another offensive after the Battle of the Bulge.

2

Answer:

When the location is angled towards the sun, it gets longer days and higher temperatures. When angled away, days are shorter and cooler.

Explanation:

The earth rotates on an axis but the North-South axis is not "vertical" but tilted slightly. As the earth orbits the sun, the axis doesn't change.

June-July
The Northern Hemisphere faces the sun while the Southern Hemisphere is facing away from the sun. Temperatures north of the equator will be higher and the days will be longer.

September-October
The Northern and Southern Hemispheres face the sun equally.

December-January
The Northern Hemisphere faces away from the sun while the Southern Hemisphere faces toward the sun. Temperatures north of the equator will be lower and days will be shorter.

March-April
The Northern and Southern Hemispheres face the sun equally.

nasa.gov

More info: NASA SpacePlace

1

Answer:

# y = sqrt(pi)/2 e^(x^2) erf(x) + Ae^(x^2) #

Where #erf(x)# is the Error Function :

# erf(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt #

Explanation:

# dy/dx = 1 + 2xy #
# :. dy/dx - 2xy = 1 # ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

# I = e^(int P(x) dx)#
# \ \ = e^(int \ -2x \ dx)#
# \ \ = e^(-x^2) #

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

# dy/dx - 2xy = 1 #
# :. e^(-x^2)dy/dx - 2xye^(-x^2) = 1*e^(-x^2) #
# :. d/dx(ye^(-x^2)) = e^(-x^2) #

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

# ye^(-x^2) = int \ e^(-x^2) \ dx#

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

# erf(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt #

Which gives us:

# ye^(-x^2) = sqrt(pi)/2erf(x) + A#
# y = sqrt(pi)/2 e^(x^2) erf(x) + Ae^(x^2) #

1

Answer:

#-16#

Explanation:

When evaluating an expression with #color(blue)"mixed operations"# there is a particular order that must be followed.

Following the order as set out in the acronym PEMDAS

[Parenthesis(brackets) , Exponents(powers), Multiplication, Division, Addition, Subtraction ]

#rArr8+6xx(4-20)/2^2#

#=8+6xx(-16)/4larrcolor(red)"brackets/powers"#

When multiplication/division are in the calculation, as here, we evaluate from left to right.

#rArr8+(6xx(-16))/4#

#=8+(-96)/4larrcolor(red)" multiplication"#

#=8+(-24)larrcolor(red)" division"#

#=8-24=-16#

2

Answer:

#P = P(t -2tsqrt(1/(4t^2+1)), t^2 +sqrt(1/(4t^2+1))) #

So the parametric equations of P are:

# P_x(t) = t -2tsqrt(1/(4t^2+1)) #
# p_y(t) = t^2 +sqrt(1/(4t^2+1)) #

Explanation:

enter image source here

Our parabola #y=x^2# is parametrised by #x=t# and #y=t^2#, and #A(t,t^2)# is a general point on the parabola, and we give the point #P# the coordinates (to be determined) #P(alpha, beta)#

Differentiating wrt #x# we have:

#dy/dx=2x = 2t #

So the gradient of the tangent at #A# is given by #dy/dx=t#, We are told that #AP# is perpendicular to to parabola, in other words, it is perpendicular to the tangent at A. Hence the gradient of AP is #-1/(2t)# as the product of their gradients are -1.

We can now form the equation of the normal #AP# using the point-slope form of the straight line, #y-y_1=m(x-x_1)#:

# y-t^2=-1/(2t)(x-t) #
# :. 2ty-2t^3=-(x-t) #
# :. 2ty-2t^3=+t-x #

#P(alpha, beta)# lies on this line, giving:

# :. 2tbeta-2t^3 = t -alpha#

And we are also told that #AP# has length 1, so by Pythagoras:

# (t-alpha)^2+(t^2-beta)^2=1^2 #

We can combine these equations to eliminate #t -alpha#:

# (2tbeta-2t^3)^2+(t^2-beta)^2=1 #
# :. 4t^2beta^2-8t^4beta+4t^6 + t^4-2t^2beta+beta^2 = 1 #
# :. (4t^2+1)beta^2-2t^2(4t^2+1)beta +t^4(4t^2+1)=1 #
# :. beta^2-2t^2beta +t^4=1/(4t^2+1) #
# :. beta^2-2t^2beta +t^4-1/(4t^2+1) =0#

We can solve this quadratic in #beta# by completing the square to get:

# (beta-t^2)^2-t^4 +t^4-1/(4t^2+1) =0#
# :. (beta-t^2)^2 = 1/(4t^2+1)#
# :. beta-t^2 = +-sqrt(1/(4t^2+1))#
# :. beta = t^2 +-sqrt(1/(4t^2+1))#

We have found two solutions because there is one point #AP# distance #1# unit extending inwards,and one point extending outwards. Intuitively we choose #beta=t^2 +sqrt(1/(4t^2+1))#.

Now from before we have:

#2tbeta-2t^3 = t -alpha# :
# :. alpha = t -2tbeta+2t^3 #

Substituting our valu of #beta#, from above, gives us:

# alpha = t -2t(t^2 +sqrt(1/(4t^2+1)))+2t^3#
# alpha = t -2t^3 -2tsqrt(1/(4t^2+1))+2t^3#
# alpha = t -2tsqrt(1/(4t^2+1))#

And so the general coordinates of #P(alpha,beta)# in term of #t# are:

#P = P(t -2tsqrt(1/(4t^2+1)), t^2 +sqrt(1/(4t^2+1))) #

So the parametric equations of P are:

# P_x(t) = t -2tsqrt(1/(4t^2+1)) #
# p_y(t) = t^2 +sqrt(1/(4t^2+1)) #

The loci traced out by #P# as #A# moves along the parabola is shown in the following graph in green:

enter image source here

1

Answer:

The major caesura in 'And afterwards, remember, do not grieve.' is after 'remember', but a lesser one after 'afterwards'.

Explanation:

Caesura is a pause in the rhythm of a line of poetry. Rhythmic devices like caesuras, line breaks and punctuation subtly modulate the sense of the actual words used. Consider the slight shift of emphasis if the quoted words were written:
"And afterwards remember,
do not grieve." (Light caesura after 'afterwards', followed by heavily emphatic 'do not grieve'.

1

Answer:

the static coefficient of friction is #0.4264# (4dp)
the kinetic coefficient of friction is #0.3340# (4dp)

Explanation:

enter image source here

For our diagram, #m=8kg#, #theta=pi/12#

If we apply Newton's Second Law up perpendicular to the plane we get:

#R-mgcostheta=0#
#:. R=8gcos(pi/12) \ \ N#

Initially it takes #12N# to start the object moving, so #D=12#. If we Apply Newton's Second Law down parallel to the plane we get:

# D+mgsin theta -F = 0 #
# :. F = 12+8gsin (pi/12) \ \ N#

And the friction is related to the Reaction (Normal) Force by

# F = mu R => 12+8gsin (pi/12) = mu (8gcos(pi/12)) #
# :. mu = (12+8gsin (pi/12))/(8gcos(pi/12)) #
# :. mu = 0.426409 ... #

Once the object is moving the driving force is reduced from #12N# to #5N#. Now #D=5#, reapply Newton's Second Law down parallel to the plane and we get:

# D+mgsin theta -F = 0 #
# :. F = 5+8gsin (pi/12) \ \ N#

And the friction is related to the Reaction (Normal) Force by

# F = mu R => 5+8gsin (pi/12) = mu (8gcos(pi/12)) #
# :. mu = (5+8gsin (pi/12))/(8gcos(pi/12)) #
# :. mu = 0.333974 ... #

So the static coefficient of friction is #0.4264# (4dp)
the kinetic coefficient of friction is #0.3340# (4dp)

2

Answer:

#f(x) = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))# with radius of convergence #R=1#.

Explanation:

We have:

#f(x) = int_0^x tln(1-t)dt#

Focus on the function:

#ln(1-x)#

We know that:

# ln (1-x) = -int_0^x (dt)/(1-t)#

where the integrand function is the sum of the geometric series:

#sum_(n=0)^oo t^n = 1/(1-t)#

then we can integrate term by term, and we have:

#ln(1-x) = -int_0^x (sum_(n=0)^oo t^n)dt = -sum_(n=0)^oo int_0^x t^ndt =- sum_(n=0)^oo x^(n+1)/(n+1)#

and mutiplying by x term by term:

#xln(1-x) = - sum_(n=0)^oo x^(n+2)/(n+1)#

We can substitute this expression in the original integral and integrate again term by term:

#f(x) = int_0^x tln(1-t)dt = - int_0^x ( sum_(n=0)^oo t^(n+2)/(n+1))dt = -sum_(n=0)^oo int_0^x t^(n+2)/(n+1)dt = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))#

To determine the radius of convergence we can then use the ratio test:

#abs (a_(n+1)/a_n) = abs (frac (x^(n+4)/((n+4)(n+2))) (x^(n+3)/((n+3)(n+1)))) = abs (x) ((n+3)(n+1))/((n+4)(n+2))#

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)#

and the series is absolutely convergent for #abs(x) <1# which means the radius of convergence is #R=1#

1

Answer:

It made cotton worth growing.

Explanation:

The demand for cotton was high, but it wasn't really worth the effort to grow it. The cotton seeds had to be removed by hand, which took too long to do. So though the price for cotton was high, too much labor was required to make it profitable.

Then along came Eli Whitney's cotton gin (gin was short for engine). The machine, run by turning a crank, cleaned the seeds from the cotton as fast as 100 slaves could do by hand. Now cotton was profitable. Farmers in the deep South, where cotton grew best, now wanted to buy land and slaves, because with the demand for cotton so high, a person could now get rich growing it.

In the first half of the 19th Century, the demand for cotton grew as the North, England and France wanted it for their textile mills, which would turn the cotton into cloth. As demand for cotton increased, so did the demand for slaves to work the ever-increasing lands that were devoted to growing cotton.

As a result, "King Cotton" became our country's number one export, and slavery became even more ingrained into the fabric of the southern economy.

1

Answer:

Andrew Jackson

Explanation:

The Battle of new Orleans was fought on January 8, 1815, two weeks after the Treaty of Ghent (Belgium) was signed to end the war of 1812. But word of the treaty had not reached the U.S., as it took at least a month for news to travel across the ocean at that time.

The British wanted to capture the all-important port of New Orleans. Farmers in the West shipped their crops down the Mississippi River to New Orleans, where the crops would be put on ships that would take them to the East Coast or overseas for sale.

General Andrew Jackson was the officer in charge of defending New Orleans for the U.S. He had his men dig entrenchments to enable them to better defend their position outside the city.

General Edward Packenham was the British leader. His troops were veterans of the Napoleonic Wars. The British army was well equipped and well trained, and Packenham thought they were invincible.

The U.S. army was a patchwork collection of regular soldiers, Native Americans, free Blacks, and even a few pirates.

Packenham ordered his men to attack Jackson's position with a full frontal assault. The strategy was suicidal. The British lost some 2,000 men and Packenham was killed. The Americans lost only twelve men. It was the most lopsided victory in U.S. history. It made Jackson a household name in the U.S. He was referred to simply as "The Hero."

His fame from this battle allowed him to be elected president in 1828.