The corona is to the sun as ___ to earth


A. The atmosphere is

B. The oceans are

C. The seasons are

D. The moon is

Answer:

a

[tex]D =  1162.7 \  m [/tex]

b

[tex]\beta =- 65.55^o[/tex]

Explanation:

From the question we are told that

  The speed of the airplane is  [tex]u  =  92.3 \ m/s[/tex]

   The  angle is  [tex]\theta = 51.1^o[/tex]

    The altitude of the plane is  [tex]d =  532 \  m[/tex]

Generally the y-component of the airplanes velocity is  

       [tex]u_y  =  v *  sin (\theta )[/tex]

=>     [tex]u_y  =   92.3 *  sin ( 51.1 )[/tex]

=>     [tex]u_y  =  71.83  \ m/s[/tex]

Generally the displacement  traveled by the package in the vertical direction is

       [tex]d =  (u_y)t +  \frac{1}{2}(-g)t^2[/tex]

=>       [tex] -532  = 71.83 t +  \frac{1}{2}(-9.8)t^2[/tex]

Here the negative sign for the distance show that the direction is along the negative y-axis

 =>   [tex]4.9t^2 - 71.83t - 532 = 0[/tex]

Solving this using quadratic formula we obtain that

    [tex]t =  20.06 \  s[/tex]

Generally the x-component of the velocity is  

     [tex]u_x  =  u  *  cos (\theta)[/tex]

=>    [tex]u_x  =   92.3  *  cos (51.1)[/tex]

=>   [tex]u_x  =   57.96 \ m/s[/tex]

Generally the distance travel in the horizontal  direction is    

     [tex]D =  u_x  *  t[/tex]

=>   [tex]D =  57.96  *   20.06 [/tex]

=>    [tex]D =  1162.7 \  m [/tex]

Generally the angle of the velocity vector relative to the ground is mathematically represented as

       [tex]\beta  =  tan ^{-1}[\frac{v_y}{v_x } ][/tex]

Here [tex]v_y[/tex] is the final  velocity of the package along the vertical  axis and this is mathematically represented as  

     [tex]v_y  =  u_y  -   gt[/tex]

=>  [tex]v_y  =  71.83  -    9.8 *  20.06[/tex]

=>  [tex]v_y  =  -130.05 \  m/s [/tex]  

and  v_x is the final  velocity of the package which is equivalent to the initial velocity [tex]u_x[/tex]

So

       [tex]\beta  =  tan ^{-1}[-130.05}{57.96 } ][/tex]

       [tex]\beta =- 65.55^o[/tex]

The negative direction show that it is moving towards the south east direction

Answer:

-2.478

0.379

11.14

24.78

Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

ω = √(10/2.2)

ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

b)

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

c)

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

a(t) = 2.48 * 4.494 * cos2.13

a(t) = 11.15 * 0.9993

a(t) = 11.14 m/s²

d)

F = -k * x(t)

F = -10 * -2.478

F = 24.78 N

Answer:

No. A:

1-tension

No. B:

4-the string

Explanation:

The upward force that acts on the book is called "tension" while the "string" is the object that exerts a force on the block that is directed to the right.

As the string is pulled, the tension exerts an upward force on the block. The frictional force acts on the block to the left. So, both the tension and friction will act on the block in order to effect its pulling on the surface of the table.

Answer:

3.176hours

Explanation:

270/85=3.126 hours DISTANCE / SPEED = TIME

Answer:

B

Explanation: Inertia is where it exerts.

Answer: Ranking :  a = c  > b > d      

Explanation:

The amount of electric flux is directly proportional to the amount of charge enclosed.

The greater the charge enclosed, the greater the electric flux through their surface

a)

 A cube with sides 10 cm long that contains a + 2.00 micro-coulomb point charge

∅ = Q/∈₀ = (2.00 x 10⁻³) / (8.85 x 10⁻¹²) = 2.26 x 10⁸ Nm²/C

b)

A cube with sides 10 cm long that contains a + 1.00 micro-coulomb point charge

∅ = Q/∈₀ = (1.00 x 10⁻³) / (8.85 x 10⁻¹²) = 1.13 x 10⁸ Nm²/C

c)  

A cube with sides 20 cm long that contains a + 2.00 micro-coulomb point charge

∅ = Q/∈₀ = (2.00 x 10⁻³) / (8.85 x 10⁻¹²) = 2.26 x 10⁸ Nm²/C

d)  A cube with sides 20 cm long that contains a + 1.00 micro-coulomb point charge

∅ = Q/∈₀ = (1.00 x 10⁻³) / (8.85 x 10⁻¹²) = 1.13 x 10⁸ Nm²/C

Ranking :

a = c  > b > d      

Answer:

Using Gauss's Law, the total amount of electric flux through a closed surface is proportional to the charge enclosed. (The surface integral of the electric field over the closed surface equals the enclosed charge divided by the constant of proportionality, the permittivity of free space.) So, since surfaces 1 and 3 have the most positive enclosed charge, they have the most positive electric flux (they have the same). Since surfaces 2 and 4 have the least positive enclosed charge, they have the least positive electric flux. and are the same

thus LHS to RHS; a=c > b=d

Answer:

acceleration = -15.3g

Explanation:

given data

speed = 6.00 m/s.

thickness = 12

moves the entire = 12.0 cm

solution

we will use here equation that is

v² - u²  = 2 × a × s    ........................1

here v = 0 is the final velocity and u = 6.0 m/s is initial velocity and s= 0.12 m is the distance covered and a is the acceleration

so we put here value and get acceleration

a = [tex]\frac{v^2-u^2}{2s}[/tex]

a = [tex]\frac{0^2-6^2}{2\times 0.12}[/tex]

a = -150 m/s² ( negative sign means it is a deceleration )

and

acceleration in units of g  

a = [tex]\frac{-150}{9.8}[/tex]

a = -15.3 g

Answer: Malleability

Explanation:

Gold in addition to being so rare, is the most malleable element there is. A Malleable element can be shaped and reshaped without it breaking and gold is so malleable that you can hit it till it becomes so flat, its transparent.

This malleability allowed for the ancient peoples to be able to shape and stamp gold into coins and their rarity made them quite valuable. Even today, investors still shape gold into exquisite designs of coins so it not only has monetary value, but aesthetic value as well.

Answer:

(a) The average velocity is 16 m/s

(b) The acceleration is 0.4 m/s^2

(c) The final velocity is 24 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

[tex]v_f=v_o+at\qquad\qquad [1][/tex]

The distance traveled by the object is given by:

[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2][/tex]

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

[tex]\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16[/tex]

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

(c) From [2] we can solve for a:

[tex]\displaystyle a= 2\frac{x-v_ot}{t^2}[/tex]

Since vo=8 m/s, x=640 m, t=40 s:

[tex]\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4[/tex]

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

[tex]v_f=8+0.4\cdot 40[/tex]

[tex]v_f=8+16=24[/tex]

The final velocity is 24 m/s

Complete Question

The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wave of the same frequency as the propeller. Consider a plane with 2 engines driving 2 propellers. You want to tune them to turn at identical frequencies.

Required:

a. If one single-bladed propellor is turning at 575 rpm and your hear 2.0 Hz beats when you run the second propeller, what are the two possible frequencies of the second propeller in Hz and rpm?

b

Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.10 Hz . In part (A), which of the two answers was the correct one for the frequency of the second single-bladed propeller ?

c. How do you know the answer in part (B) to be correct?

Answer:

a

The two possible frequencies of the second propeller in Hz

    [tex]f_1 =  11.58\  Hz [/tex]

    [tex]f_2 =  7.58 \  Hz [/tex]

The two possible frequencies of the second propeller in rpm

      [tex]f_1 =  695 \  rpm [/tex]

      [tex]f_2 =  455 \  rpm [/tex]  

b

The  correct answer for the frequency of the second single-bladed propeller is  

        [tex]f_1 =  695 \  rpm [/tex]

c

The above answer is correct because when the beat frequency of the second propeller  increases(i.e from 2.0 Hz  to  2.10 Hz) the  frequency of the second propeller becomes much greater than that of the first propeller so looking at the two possible value of frequency of the second propeller (i.e  695 rpm and  455 rpm ) we see that it is  695 rpm that is showing that increase of the second propeller compared to the first propeller

Explanation:

From the question we are told that

    The number of engines is  n  =  2

    The  number of propellers is m =  2

    The  angular frequency  of the single-bladed propellor [tex]w =   575 rpm[/tex]

     The frequency of the beat heard at this velocity is  [tex]f =  2.0 \  Hz[/tex]

Converting the beat frequency  to rpm

            [tex]f =  2 * 60  = 120 \ rpm[/tex]

Generally the the two possible frequencies of the second propeller in  rpm is

     [tex]f_1 =  w + f[/tex]

=>   [tex]f_1 =  575  + 120[/tex]

=>   [tex]f_1 =  695 \  rpm [/tex]

And

      [tex]f_2 =  w - f[/tex]

=>   [tex]f_2 =  575  - 120[/tex]

=>   [tex]f_2 =  455 \  rpm [/tex]  

Converting the angular frequency  of the single-bladed propellor to rpm

     [tex]w =   \frac{575}{60} [/tex]

       [tex]w =  9.58 \  Hz [/tex]

Generally the the two possible frequencies of the second propeller in  

Hz is

          [tex]f_1 =  w + f[/tex]

=>   [tex]f_1 =  9.58  + 2[/tex]

=>   [tex]f_1 =  11.58\  Hz [/tex]

And

      [tex]f_2 =  w - f[/tex]

=>   [tex]f_2 =  9.58  -  2[/tex]

=>   [tex]f_2 =  7.58 \  Hz [/tex]    

Answer:

Solids can hold their shape because their molecules are tightly packed together. Atoms and molecules in liquids and gases are bouncing and floating around, free to move where they want. The molecules in a solid are stuck in a specific structure or arrangement of atoms.

Answer:

The number of pi groups are   P =  3 pi groups            

Explanation:

From the question we are told that

    The number of parameters is  n =  6  

These parameters are  : Fluid  thermal conductivity : k

                                        Kinematic viscosity: v

                                        The temperature difference ΔT  

                                         The length of the object L

                                          Gravitational constant g

                                           Thermal expansion coefficient β

      The number of the basic dimensional unit is  N = 3

                                            Mass  M  

                                            Length L

                                             Time  T

Generally the number of  Pi groups is  mathematically evaluated as

                  P =  n  -  N

=>               P =  6  -  3

=>               P =  3 pi groups                                        

Answer:

B. 1.29 x 10^5

Answer:

B. 1.29 x 10^5

Explanation:

just took the test on ed2020

Answer:

Acceleration

Explanation:

I am writing this because it must be 20 characters to submit

Complete Question

Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.

A) Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two speeds reach their separations?

A) Electrons A is moving faster than electron B.

B) Electron B is moving faster than electron A.

C) Both electrons are moving at the same (nonzero) speed in the opposite direction

.D) Both electrons are moving at the same (nonzero) speed in the same direction.

E) Both electrons are momentarily stationary.

2) What is the minimum separation[tex]r_{min}[/tex] that the electrons reach?

Answer:

1

The  correct option is  E

2

[tex]r_{min} =  \frac{kq^2}{4 mv^2}[/tex]

Explanation:

From the question we are told that

   The mass of each electron  is  m  

    The  charge of each electron  is  q

    The speed of electron A is  v

    The  speed of electron B  is  3v

Generally at their point of separation the repulsion force is equal to the force that is propelling the electrons due to this the electrons are  momentarily stationary

Generally the total initial kinetic energy of both electron is mathematically represented as  

         [tex]K_{inT} =  K_A + K_B[/tex]

=>      [tex]K_{inT}  =  \frac{1}{2}m (v)^2 + \frac{1}{2} m (3v)^2[/tex]

=>      [tex]K_{inT} =  \frac{1}{2} (mv^2 + 9v^2m)[/tex]

=>      [tex]K_{inT} =  5mv^2 [/tex]

Generally the total  final  kinetic energy of both electron is mathematically represented as

         [tex]K_{fT} =  \frac{1}{2} *m * v^2 + \frac{1}{2} *m * v^2[/tex]

Here v is the velocity due to the repulsion force

          [tex]K_{fT} =  mv^2 [/tex]

Generally the final  potential energy of the both electrons is  

         [tex]P_f  =  \frac{ k *  q^2}{r_{min}}[/tex]

Here k is the coulombs constant

So according to energy conservation law

     [tex]K_{inT} =  K_{fT}  +  P_f[/tex]

=>   [tex]5mv^2 =  mv^2 +   \frac{ k *  q^2}{r_{min}} [/tex]

=>   [tex]r_{min} =  \frac{kq^2}{4 mv^2}[/tex]

Answer:

?

Explanation:

Answer:

a) 25.76 m/s

b) 30°

Explanation:

See attachment

Answer:

cause without gravity, the earth will start to move away from the orbit and crash into the sun like a raining meteor of babies diaper falling on the ground of smelly dunken doughnuts

Explanation:

lol

Answer:

I believe it's A) "An exothermic change in which water freezes I could be wrong though just giving my opinion

Explanation:

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

[tex]\rho (r) \ \alpha \ \delta (r -R)[/tex]

[tex]\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)[/tex]

[tex]\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)[/tex]

To find the constant k, we  examine the total charge Q which is:

[tex]Q = \int \rho (r) \ dV = \int \sigma \times dA[/tex]

[tex]Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2[/tex]

∴

[tex]\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2[/tex]

[tex]\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]

[tex](2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]

Thus;

[tex]k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2[/tex]

[tex]k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2[/tex]

[tex]k * R^2= \sigma \times R^2[/tex]

[tex]k = R^2 --- (2)[/tex]

Hence, from equation (1), if k = [tex]\sigma[/tex]

[tex]\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}[/tex]

[tex]\mathbf{\rho (r) =0 \ \ at \ \ r<R \ \ or \ \ r>R}[/tex]

To verify the units:

[tex]\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}[/tex]

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

[tex]Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr[/tex]

[tex]Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr[/tex]

[tex]Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr[/tex]

[tex]Q = 4 \pi \sigma *R^2[/tex]    since  [tex]( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )[/tex]

[tex]\mathbf{Q = 4 \pi R^2 \sigma }[/tex]

Answer:

v> √ gr

Explanation:

Let's analyze this problem from Newton's second law

At the lowest point

            N - W = m a

at the highest point

            W = m a’

in both cases there is a net force towards the center of the circle that we can call the centripetal force, which is responsible for changing the direction and magnitude of the acceleration.

When the bucket is in the highest part, the centripetal force is equal to the weight of the water, but since it carries a horizontal speed, until it starts to fall, it is moving and therefore they follow the bottom of the tube. This implies that there is a minimum speed for this to occur

             v> √ gr

This principle is applied in many things, for example the roller coaster, centrifuges, simulators of the effect of acceleration on people.

Answer:

Representative Democracy.

Explanation:

It is simple and easy because you choose a representative to make choices for the good of your people. It is much simpler then all the rest.

Answer:

it is reduced to half of that strength

Answer:

it is cool and warm

Explanation:

Answer:

v = 7.42 m / s

Explanation:

For this exercise we will use the conservation of mechanical energy.

Starting point. Before jumping from the tree

          Em₀ = U = m g h

Final point . Part of the trajectory at 25º

           Em_{f} = K + U = ½ m v2 + m g y

as they indicate that there is no air resistance, mechanical energy is conserved

           Em₀ = Em_{f}

           m g h = ½ m v² + m g y

           v² = 2g (h - y)

Let's use trigonometry to find the height that has descended, how the angle is measured with respect to the vertical

           cos 25 = y / L

           y = L cos 25

we substitute

           v² = 2 g (h - L cos 25)

for this case h = L = 30 m

          v2 = 2g L (1- cos25)

let's calculate

           v² = 2 9.8 30 (1 -cos 25)

           v = √55.09

            v = 7.42 m / s

Answer:

Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Answer:

a. approximately [tex]1.1\; \rm m[/tex] (first minimum.)

b. approximately [tex]2.2\; \rm m[/tex] (first maximum.)

c. approximately [tex]3.4\; \rm m[/tex] (second minimum.)

d. approximately [tex]4.7\; \rm m[/tex] (second maximum.)

Explanation:

Let [tex]d[/tex] represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let [tex]\theta[/tex] represent the angle between:

The distance between the microphone and point [tex]P_0[/tex] would thus be [tex]9.4\, \tan(\theta)[/tex] meters.

Based on the assumptions and the equation from Young's double-slit experiment:

[tex]\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}[/tex].

Hence:

[tex]\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)[/tex].

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let [tex]\lambda[/tex] denote the wavelength of this wave.

[tex]\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}[/tex].

Calculate the wavelength of this wave based on its frequency and its velocity:

[tex]\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m[/tex].

Calculate [tex]\theta[/tex] for each of these path differences:

[tex]\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}[/tex].

In each of these case, the distance between the microphone and [tex]P_0[/tex] would be [tex]9.4\, \tan(\theta)[/tex]. Therefore:

Interference is the result when two or more waves combine

The distance between P₀ and

a. The first intensity minimum is approximately 1.06 m

b. The first intensity maximum is  approximately  2.165 m

c. The second intensity minimum is approximately 3.36 m

d. The second intensity minimum is approximately 4.72 m

The reasons the above values are correct are given as follows:

The known parameters are;

The distance between the two speakers = 0.94 m

Frequency of the tone produced by the two speakers = 1,630 Hz (in sync)

The line along which the microphone moves is parallel to the line between the two speakers

The distance between the parallel lines above = 9.4 m

The speed of sound in air, v₀ = 344 m/s

The interference pattern of light passing between two slits is to be applied

a. Based on the arrangement, we have;

P₀ = 9.4 × tan(θ)

Where;

θ = The angle formed formed by the line from the microphone to the midpoint of the distance between the two speakers and the perpendicular bisector to the line joining the two speakers

Based on Young's double-slit experiment, we have;

[tex]sin(\theta) = \dfrac{Path \ difference}{d}[/tex]

Where;

d = The distance between the two speakers representing the slits

The path difference for a minimum is n × λ/2

Where n = 1,  3, 5,...,(set of odd numbers)

The path difference for a maximum intensity sound is n·λ

Where n = 1, 2. 3, ..., n (n is an integer)

The wavelength, is given as follows;

[tex]\lambda = \dfrac{v}{f}[/tex]

Therefore;

[tex]\lambda = \dfrac{344}{1,630} = \dfrac{172}{815} \approx 0.211[/tex]

The wavelength, λ ≈ 0.211

Therefore, the angle, θ, for the first minima, θ, ≈arcsine(0.211/(2×0.94))

First minima, λ/2, P₀ =9.4 × tan(arcsine(0.211/(2×0.94))) ≈ 1.06

First maxima, λ, P₀ =9.4 × tan(arcsine(0.211/(94))) ≈ 2.165

Second minima, 3·λ/2, P₀ = 9.4 × tan(arcsine(3*0.211/(2×0.94))) ≈ 3.36

Second maxima, 2·λ, P₀ = 9.4 × tan(arcsine(2*0.211/(0.94))) ≈ 4.72

Therefore;

a. The distance between P₀ and the first intensity minimum is ≈ 1.06 m

b. The distance between P₀ and the first intensity maximum is ≈ 2.165 m

c. The distance between P₀ and the second intensity minimum is ≈ 3.36 m

d. The distance between P₀ and the second intensity minimum is ≈ 4.72 m

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Answer:

a. 28.28

B.14.4

C.14.4

Explanation:

A. v^2=u^ + 2as

v^2=0^2 + 2*2*200

B. v=u+at

t=v-u/a

C. V+u/2

The sprinter velocity at the end of the 200 m would be 28.28 m/s.

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

As given in the problem a sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.

By using the third equation of the motion,

v² - u² = 2×a×s

v² - 0² = 2×2×200

v² = 800

v =28.28 m/s

The sprinter velocity at the end of the 200 m would be 28.28 m/s

By using the second equation of motion

S = ut + 1/2*a*t²

u= 0 m/s , a=  m/s² s = 200 m

200  = 0 + 0.5*2*t²

t² = 200

t = √200

t = 14.14 seconds

It would take him 14.14 seconds to cover.

average velocity = initial velocity + final velocity /2

                            = 0+ 28.28/2   m/s

                             = 14.14 m/s

Thus, The sprinter velocity at the end of the 200 m would be 28.28 m/s.

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