What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 square meters?

Answer:

Acceleration

Explanation:

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

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:

Yes

Explanation:

Beacause u have 25 kg,and falls off from a cliff hanger cause its too bit heavy

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: from the vertical, one should aim  86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² +  1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ  into equ 2

365.76(cos∅) ×  5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ =  3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

Answer:

15 seconds

Explanation:

Given that the initial speed of a cannon ball is 0.20 km/s. If the ball is to strike a target that is at a horizontal distance of 3.0 km from the cannon. The minimum time of flight for the ball can be calculated by using the formula for speed

speed = distance / time

Where

speed = 0.2 km/s

distance = 3 km

Substitute the two parameters into the formula

0.2 = 3 / t

make t the subject of the formula

t = 3/0.2

t = 15 s

Therefore, the minimum time of flight for the ball is 15 seconds

Answer:

3.176hours

Explanation:

270/85=3.126 hours DISTANCE / SPEED = TIME

Complete question

The diagram for this question is shown on the first uploaded image

Answer:

 [tex]T_a   =  44.8 \  N [/tex] ,   [tex]  F =  86.03 \ N  [/tex]

Explanation:

From the question we are told that

    The magnitude of  [tex]T_b  =  80 N[/tex]

From the diagram we can see that

     [tex]Ta * sin (38) +  F *  sin (29) =   T_b *  cos  (30) \  \cdots (1)[/tex]

    [tex]Ta * sin (38) +  F *  sin (29) = 80 *  cos  (30) \  \cdots (1)[/tex]

   [tex]0.616 Ta +  0.485F  = 69.3 \  \cdots (1)[/tex]

=>  [tex]T_a   = \frac{69.3 - 0.485F}{0.616}[/tex]

Also

    [tex]T_a  *  cos(38) + T_b *  sin (30)= F *  cos (29) \  \cdots  (2)[/tex]

=>   [tex]T_a  *  cos(38) + 80*  sin (30)= F *  cos (29) \  \cdots  (2)[/tex]

=>   [tex]0.788 T_a + 40 =  0.875 F \  \cdots  (2)[/tex]

=>  [tex]0.788 [\frac{69.3 - 0.485F}{0.616}]+ 40 =  0.875 F[/tex]

=>  [tex]  88.65  -  0.6204 F + 40 =  0.875 F[/tex]

=>  [tex]  88.65  + 40 =  0.875 F+0.6204 F [/tex]

=>  [tex]  128.65  =  1.4954 F [/tex]

=>  [tex]  F =  86.03 \ N  [/tex]

substituting  this obtained value  for F in the above equation

       [tex]T_a   = \frac{69.3 - 0.485(86.03)}{0.616}[/tex]

      [tex]T_a   =  44.8 \  N [/tex]

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

Learn more about interference of sound waves here:

https://brainly.com/question/14006922

https://brainly.com/question/12096166

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:

a) 25.76 m/s

b) 30°

Explanation:

See attachment

Answer:

a = 16.5 x 10⁻⁶ m = 16.5 μm

Explanation:

Here we will use the diffraction equation:

y = mλL/a

where,

y = distance between two consecutive dark fringes = 5.34 cm = 0.0534 m

m = order of diffraction = 1

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

L = Distance between hair and screen = 1.4 m

a = thickness of hair = ?

0.0534 m = (1)(6.33 x 10⁻⁷ m)(1.4 m)/(a)

a = (6.33 x 10⁻⁷ m)(1.4 m)/(0.0534 m)

a = 16.5 x 10⁻⁶ m = 16.5 μm

Answer:

21

Explanation:

21

Answer:

speed is equal to distance divided by time

(for the chart i'm assuming that the time is measure in seconds but if not just change the s with whatever time unit you are using)

A. 0.224 m/s

B. 0.230 m/s

C. 0.258 m/s

D. 0.265 m/s

E. 0.301 m/s     (fastest speed)

F. 0.217 m/s     (slowest speed)

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:

(b) 0.0176

(c) -0.0124

(d) 209

(e) Also comes to rest

(a) 2.38

(b) 5.95

Explanation:

(a) Your answer is correct.  Angular momentum is conserved, so as the lighter beetle moves clockwise, the heavier turntable will move counterclockwise at a slower speed.

(b/c) Initial angular momentum = final angular momentum

L₀ = L

I₁ ω₁,₀ + I₂ ω₂,₀ = I₁ ω₁ + I₂ ω₂

0 = mr² ω₁ + ½ Mr² ω₂

0 = 2m ω₁ + M ω₂

The beetle's angular velocity relative to the turntable is 0.03 rad/s, so ω₁ = ω₂ + 0.03.  Plugging in:

0 = 2 (30 g) (ω₂ + 0.03 rad/s) + (85 g) ω₂

0 = 60ω₂ + 1.8 rad/s + 85ω₂

145ω₂ = -1.8 rad/s

ω₂ = -0.0124 rad/s

ω₁ = ω₂ + 0.03

ω₁ = 0.0176 rad/s

Relative to a stationary observer, the beetle moves 0.0176 rad/s clockwise and the turntable moves -0.0124 rad/s counterclockwise.

(d) Angular distance = angular velocity × time

2π rad = (0.03 rad/s) t

t = 209 s

(e) Angular momentum is conserved.  Since both the beetle and the turntable were originally at rest, the turntable will again come to rest when the beetle stops.

(a) Angular momentum is conserved.

L₀ = L

I₀ ω₀ + I₂ ω₀ = I ω + I₂ ω

(I₀ + I₂) ω₀ = (I + I₂) ω

(M (R/2)² + ½ (3M) (R)²) ω₀ = (M (R)² + ½ (3M) (R)²) ω

(¼ MR² + ³/₂ MR²) ω₀ = (MR² + ³/₂ MR²) ω

(¼ + ³/₂) ω₀ = (1 + ³/₂) ω

(1 + 6) ω₀ = (4 + 6) ω

7ω₀ = 10ω

ω = 0.7ω₀

ω = 0.7 (3.40 rad/s)

ω = 2.38 rad/s

(b) Angular momentum is conserved.

L₀ = L

I₀ ω₀ + I₂ ω₀ = I ω

(I₀ + I₂) ω₀ = I ω

(M (R/2)² + ½ (3M) (R)²) (3.40 rad/s) = M (R)² ω

(¼ MR² + ³/₂ MR²) (3.40 rad/s) = MR² ω

(¼ + ³/₂) (3.40 rad/s) = ω

ω = 5.95 rad/s

Notice we could also have used our answer from part a and I₀ = MR².

(I₀ + I₂) ω₀ = I ω

(M (R)² + ½ (3M) (R)²) (2.38 rad/s) = M (R)² ω

(MR² + ³/₂ MR²) (2.38 rad/s) = MR² ω

(1 + ³/₂) (2.38 rad/s) = ω

ω = 5.95 rad/s

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:

it's uhh 0.05 my g yup that's the answer

Explanation:

5

I'm pretty sure your answer should be

C. Riding uphill, the boy's work is converted into potential energy, riding on level ground, his work is converted into kinetic energy.

Answer:  The thermal energy transfer is When a fluid, such as air or a liquid, is heated and then travels away from the source, it carries the thermal energy along.

Explanation: heat transfer is called convection.  hopefully this was helpful.

Answer:

B. 1.29 x 10^5

Answer:

B. 1.29 x 10^5

Explanation:

just took the test on ed2020

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:

True

Explanation:

Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. Three ways electrons can be transferred are conduction, friction, and polarization. In each case, the total charge remains the same.

I tried, hope this helps :)

* I might be wrong though

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:

it is cool and warm

Explanation:

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: 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:

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:

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.