A star with the same mass and diameter as the sun rotates about a central axis with a period of about 24.0 days. Suppose that the sun runs out of nuclear fuel and collapses to form a white dwarf star with a diameter equal to that of the earth. Assume the star acts like a solid sphere and that there is no loss of mass in the process. You will need some data from the inside front cover of you text. (a) What would be the new rotation period (s) of the star? (b) What is the ratio of final to initial kinetic energies (Kf /Ki)?

Answer:

The correct answer is option (A) that is KEA > KEB .

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

[tex]W=F_d- F_f_r_id-F_gh[/tex]

[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]

The change in kinetic energy is ,

   [tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]

At the top of the inclined plane , the velocity is zero

So,

[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]

[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]

From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so

[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

For object A-

[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

For object B

[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]

[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

[tex]KE_A >KE_B[/tex]

Therefore , option A is correct .

Answer:

a. Speed = 1.6 m/s

b. Amplitude = 0.3 m

c. Speed = 1.6 m/s

Amplitude = 0.15 m

Explanation:

a.

The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:

frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz

The wavelength of the wave is the distance between consecutive crests of wave. Therefore,

Wavelength = 6.4 m

Now, the speed of the wave is given as:

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

Speed = 1.6 m/s

b.

Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:

Amplitude = (0.5)(0.6 m)

Amplitude = 0.3 m

c.

frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

Speed = 1.6 m/s

Amplitude = (0.5)(0.3 m)

Amplitude = 0.15 m

Answer:

   f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

        ∑ F = ma

where the acceleration is

         a = [tex]\frac{d^2 y}{dt^2 }[/tex]

        B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

           B = W

In this frame of reference, the variable y'  when it is oscillating is positive and negative, therefore Newton's equation remains

         B’= m [tex]\frac{d^2 y'}{dt^2 }[/tex]

the thrust is given by the Archimedes relation

         B = ρ_liquid g V_liquid

the volume is

        V = π r² y'

we substitute

          - ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

          [tex]\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0[/tex]

this differential equation has a solution of type

         y = A cos (wt + Ф)

where

         w² = ρ_liquid g π r² /m

angular velocity and frequency are related

         w = 2π f

we substitute

          4π² f² = ρ_liquid g π r² / m

          f = [tex]\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }[/tex]

calculate

         f = [tex]\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }[/tex]

         f = 5.3 Hz

Answer:

390 light-years

Explanation:

50 x 7.8 =390

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = [tex]\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx[/tex]     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

Answer:

i dont know but i should know try g o o g l e

Explanation:

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ([tex]\Delta t[/tex]), in seconds:

[tex]\Delta t = t_{A}-t_{C}[/tex] (1)

Where:

[tex]t_{C}[/tex] - Time spent by the sound in concrete, in seconds.

[tex]t_{A}[/tex] - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

[tex]\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}[/tex]

[tex]\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)[/tex]

[tex]x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }[/tex] (2)

Where:

[tex]v_{C}[/tex] - Speed of the sound in concrete, in meters per second.

[tex]v_{A}[/tex] - Speed of the sound in the air, in meters per second.

[tex]x[/tex] - Distance traveled by the sound, in meters.

If we know that [tex]\Delta t = 6.4\,s[/tex], [tex]v_{C} = 3000\,\frac{m}{s}[/tex] and [tex]v_{A} = 343\,\frac{m}{s}[/tex], then the distance travelled by the sound is:

[tex]x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }[/tex]

[tex]x = 2478.585\,m[/tex]

The impact occured at a distance of 2478.585 meters from the person.

Answer:

a. T = 39.41 N

b. t = 1.76s

c. 150.78 N

Explanation:

Given:

Mass of bucket of water, Mb = 14.6 kg

Mass of cylinder, Mc = 11.1 kg

Diameter of cylinder, D =  0.320 m, or radius, r = D/2 = 0.16m

Displacement of the bucket from the top, that is the vertical displacement , y = 11.0 m

a. The tension in the rope while the bucket is falling is:

F = mg - T = ma

Where F= The force

m= mass

g= Acceleration due to gravity

T = tension in the rope

a = acceleration

T= m(g - a)

Then, calculating the angular acceleration of the pulley system in relation to its radial acceleration

T= 1/2Ma

Merging the two final equation so as to solve for a

M(g - a) = 1/2Ma

Make a the subject of the formula

Mg - Ma = 1/2Ma

1/2Ma + Ma = Mg

a (1/2 M + M) = Mg

Divide both side by (1/2 M + M)

a = Mg ÷ (1/2 M + M)

Inputing the given value in the formula above

g= 9.8m/s2

a = (14.6 kg) (9.8m/s2) ÷ 1/2 (11.1 kg) + 14.6 kg

a = 7.1007m/s2

Now it is easy to input the value into T= 1/2Ma

T = 1/2 (11.1 kg) (7.1007m/s2) = 39.41 N

B. Time of fall is:

Using one of the equation of motion

s = ut + 1/2 at^2

U = Initial velocity

t = time

a = acceleration

s= distance in this case displacement y

making t the subject of the formula

t = √(2s ÷ a)

u is 0 since the bucket starts from rest

so, t = √((2)(11.0 m) ÷ 7.1007m/s2)

t = 1.76s

c.  the force exerted on the cylinder by the axle = T + Mg

  = 42 N + (11.1 kg) (9.8m/s2)

= 150.78 N

Answer:

so you have a question

Explanation:

either way, have a nice day

Answer:

Acceleration, a = 0.35 m/s²

Explanation:

Given the following data;

Initial velocity, u = 3.5 m/s²

Final velocity, v = 2.1 m/s²

Time, t = 4 secs

To find the acceleration, we would use the first equation of motion;

V = u - at (the sign is negative because Lucy is slowing down).

Substituting into the formula, we have;

2.1 = 3.5 - a(4)

2.1 = 3.5 - 4a

4a = 3.5 - 2.1

4a = 1.4

a = 1.4/4

a = 0.35 m/s²

Answer: C

Explanation:

USAtestprep

This question is not complete, the complete question is;

A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.

An object is made from a uniform piece of sheet metal. The object has dimensions of α = 1.50 ft, where α is the diameter the semi-circle, b= 3.51 ft, and c = 2.20 ft. A hole with diameter d = 0.500 ft is centered at ( 1.21, 0.750 ).

Find x", y", the coordinates of the body's centroid.

Answer:

x" = 1.4857 ft

y" = 0.668 ft

Explanation:

Given the data in the question and as illustrated in the second image below;

from the image;

BC² = DC² - BD²

BC² = 2.2² - 1.5² = 4.84 - 2.25 = 2.59

BC = √2.59 = 1.61 ft

AB = 3.51 ft - 0.75 ft - 1.61 ft = 1.15 ft

so;

A₁ = [tex]\frac{1}{2}[/tex] × 1.51 ft × 1.61 ft  = 1.2075 ft²

x₁ = 0.75 + 1.15 + [tex]\frac{1}{3}[/tex](1.61 ft) = 2.44 ft

y₁ = [tex]\frac{1}{3}[/tex](1.5 ft) = 0.5 ft

A₂ = 1.15 ft × 1.5 ft = 1.725 ft²

x₂ = 0.75 ft + ( 1.15/2 )ft = 1.325 ft

y₂ = ( 1.5/2 ) ft = 0.75 ft

A₃ = [tex]\frac{\pi }{2}[/tex](0.75 ft)² = 0.88 ft²

x₃ = 0.75 - ([tex]\frac{4 }{3\pi }[/tex](0.75 ft)) = 0.43 ft

y₃ = 0.75 ft

diameter d = 0.5 ft and centered at ( 1.21, 0.750 )

A₄ = [tex]\frac{\pi }{4}[/tex]( d )² =  

x₄ = 1.21 ft

y₄ = 0.75 ft

Thus;

x" = [tex]\frac{A_1 x_1 + A_2 x_2 + A_3 x_3 - A_4x_4 }{A_1+A_2+A_3-A_4}[/tex]

so we substitute

x" = [tex]\frac{(1.2075X2.44) + (1.725 X 1.325) + (0.88X0.43) - (0.196 X 1.21 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]

x" = [tex]\frac{ (2.9463 + 2.285625 + 0.3784 - 0.23716)}{ 3.6165 }[/tex]

x" = 5.373165 / 3.6165

x" = 1.4857 ft

y" = [tex]\frac{A_1 y_1 + A_2 y_2 + A_3 y_3 - A_4y_4 }{A_1+A_2+A_3-A_4}[/tex]

so we substitute

y" = [tex]\frac{(1.2075X0.5) + (1.725 X 0.75) + (0.88X0.75) - (0.196 X 0.75 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]

y" = [tex]\frac{ (0.60375 + 1.29375 + 0.66 - 0.14112)}{ 3.6165 }[/tex]

y" = 2.41638 / 3.6165

y" = 0.668 ft

Therefore,

x" = 1.4857 ft

y" = 0.668 ft

Answer:

Potential energy

Explanation:

Potential energy is stored energy

Answer:

[tex]9.495 \times 10^3\ m[/tex]

Explanation:

From the given information:

Using the equation of Barometric formula as related to density, we have:

[tex]\rho (z) = \rho (0) e^{(-\dfrac{z}{H})} \ \ \ \ --- (1)[/tex]

Here;

[tex]p(z) =[/tex] the gas density at altitude z

[tex]\rho(0) =[/tex] the gas density  at sea level

H = height of the scale

[tex]H = \dfrac{RT}{M_ag } \ \ \ --- (2)[/tex]

Also;

R represent the gas constant

temperature (T) a= 280 K

g = gravity

[tex]M_a =[/tex] molaar mass of gas; here, the gas is Oxygen:

∴

[tex]M_a =[/tex] 15.99 g/mol

= 15.99 × 10⁻³ kg/mol

[tex]H = \dfrac{8.3144 \times 280}{15.99 \times 10^{-3} \times 9.8 }[/tex]

[tex]H =14856.43 \ m[/tex]

Now we need to figure out how far above sea level the density of oxygen drops to half of what it is at sea level.

This implies that we have to calculate z;

i.e. [tex]\rho(z) =\dfrac{\rho(0) }{(2)}[/tex]

By using the value of H and [tex]\rho(z)[/tex] from (1), we have:

[tex]\dfrac{\rho(0) }{(2)} = \rho (0) e^{(-\dfrac{z}{14856.43})}[/tex]

∴

[tex]\dfrac{1}{2} = e^{(-\dfrac{z}{14856.43})} \\ \\ e^{(-\dfrac{z}{14856.43})} =\dfrac{1}{2}[/tex]

By rearrangement and taking the logarithm of the above equation; we have:

[tex]- z = 14856.43 \times \mathtt{In}\dfrac{1}{2} \\ \\ -z = 14856.43 \times (-0.6391) \\ \\ z = 9495 \ m \\ \\ z = 9.495 \times 10^3\ m[/tex]

As a result, the oxygen density at  [tex]9.495 \times 10^3\ m[/tex] is half of what it is at sea level.

Answer:

True.

Explanation:

true

knowing your audience there General age gender education level religion language culture and gave group members is the single most important aspect of developing your speech this means the speaker dock smart the audience wasn't often without asking question or spending with any feedback

Answer:

what

you need to elaborate

Answer: Can you please write question clearly.

Explanation:

Answer:

The answer is "[tex]\bold{7.18 \times 10^3 \ m^2}[/tex]".

Explanation:

The efficiency system:

[tex]\eta =\frac{P_{req}}{P} \times 10\\\\P =\frac{P_{req}}{\eta} \times 10\\\\[/tex]

   [tex]=(\frac{2.20 \times 10^6 \ W}{30})\times 100\\\\=(\frac{220 \times 10^6 \ W}{30})\\\\=(\frac{22 \times 10^6 \ W}{3})\\\\=7.33 \times 10^6 \ W[/tex]  

Using formula:

[tex]A=\frac{P}{I}[/tex]

Effective area:

[tex]A= \frac{7.33 \times 10^6 \ W}{1020\ \frac{W}{m^2}}\\\\[/tex]

   [tex]=\frac{7.33 \times 10^6 }{1020}\ m^2 \\\\ =0.0071862 \times 10^6 \ m^2 \\\\=7.1862 \times 10^3 \ m^2 \\\\[/tex]  

Answer:

35,000 km = 35,000,000 m = 3.5 E107 m

t = S / v = 3.5 * 10E7 / 3.0 E10E8 = .117 sec

A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.

An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.

By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.

The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.

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Answer:

A: 9.8 rad/s2

B: 7.4 rad/s2

C: 8.4 rad/s2

D: 5.9 rad/s2

E: 6.5 rad/s2

I think the answer is A 9.8rad/s2

Answer:

the answer is D I tought

Answer:

so i would say 11.4 i dont have work only this link

Explanation:

https://flexbooks.ck12.org/cbook/ck-12-physics-flexbook-2.0/section/11.4/primary/lesson/wave-speed-ms-ps

Answer:

B) 10^-2 cm/s

in term of meter. it is 10^-4 m/s

Explanation:

Answer:

A)   v = 40 m / s, B)   v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

         v = vo + a t

         v = 0 + 8 5

         v = 40 m / s

B) the average velocity can be found with the relation

         v_average = vf + vo / 2

         v-average = 0+ 40/2

          v_average = 20 m / s

Answer:

The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.

Explanation:

I hope this is what you were looking for?!

Hope this helps!

Have a great day!

-Hailey!

The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.

The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.

A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.

Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.

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