a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of

Answer:

The induced current in the resistor is I = BLv/R

Explanation:

The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by

ε = BLv.

Now, the current I in the resistor is given by

I = ε/R where ε = induced emf in circuit and R = resistance of resistor.

So, the current I = ε/R.

substituting the value of ε the induced emf, we have

I = ε/R

I = BLv/R

So, the induced current through the resistor is given by I = BLv/R

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]

[tex]0.8 = e^{-\frac{88}{R} }[/tex]

[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]

[tex]ln(0.8) = -\frac{88}{R}[/tex]

[tex]R = -\frac{88}{ln(0.8)}[/tex]

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

Answer:

Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.

Explanation:

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.

              Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m

Explanation: A speed of wave on a string under a tension force can be calculated as:

[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]

[tex]F_{T}[/tex] is tension force (N)

μ is linear density (kg/m)

Determining velocity:

[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{0.00874 }[/tex]

[tex]|v| =[/tex] 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]

Δx = 11.5×[tex]10^{-6}[/tex]

With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex]  m.

Doubling Tension:

[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{2.0.00874 }[/tex]

[tex]|v| = \sqrt{0.01568}[/tex]

|v| = 0.1252 m/s

Displacement for same time:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]

[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]

With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m

Answer:

The radius is  [tex]r = 3.1905 \ m[/tex]

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  [tex]d = 2.70 \ m[/tex]

        The refractive index of the liquid is  [tex]n_i = 1.31[/tex]

Now the critical value is mathematically represented as

         [tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]

substituting values

         [tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]

         [tex]\theta = 49.76^o[/tex]

Using SOHCAHTOA rule we have that

         [tex]tan \theta = \frac{ r}{d}[/tex]

=>     [tex]r = d * tan \theta[/tex]

substituting values  

        [tex]r = 2.7 * tan (49.76)[/tex]

        [tex]r = 3.1905 \ m[/tex]

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

Answer:

Pls seeattached file

Explanation:

A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.

Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.

When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.

According to the question,

R = R₀ (1 + α ΔT)

(1 + 0.0135)R₀ = R₀(1 + α ΔT)

ΔT = (1 + 0.0135) / α

= 0.0135 / 0.0004

= 33.75 °C.

ΔT = [(1 - 0.0135) -1]/0.004

= -33.75 °C

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Answer:

b

Explanation:

Answer:

Current, I = 1.23 A

Explanation:

Given that,

Inductance, L = 35 mH

Resistance, R = 12 ohms

Potential difference, V = 18 V

We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :

[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)

Here,

[tex]I_o[/tex] is final current

[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]

[tex]\tau[/tex] is time constant

[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]

So, equation (1) becomes :

[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]

So, after 5 ms the current in the circuit is 1.23 A.

Answer:   2

Explanation:

1/2=1/1 +1/x

x=-2

magnification= 2/1

magnification=2

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

Answer:

Explanation:

Bohr's atomic model was replaced by the  modern atomic model because of its limitations, which included :

(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )

(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).

(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.

Answer:

a) Connect a series resistance of 8,47 ohms

b)14,16 [A]

c) r = 10,96 ohms

Explanation:

My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower

In America

P = V*I

1700 (w) = 120*I

I = 1700/120 [A]

I = 14,16 [A]        current needed for the blower

In Europe

120 (v)  (the drop of voltage I need) when a current of 14,16 passes through to series  resistor is

V = I*R          120 = 14,16* R         R = 8,47 ohms

c) P = I*r²

1700 (w) = 14,16 (A) * r²

r² = 120,06

r = 10,96 ohms

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

The correct answer is C. The inner planets are also called terrestrial planets.

Explanation:

Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".

Answer:

The answer is c.) The inner planets are also called terrestrial planets.

Explanation:

Answer:

-0.24diopters

Explanation:

The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)

So 1/do +1/di = 1/f

1/infinity + 1/-4.2 = 1/f

1/f = 1/-4.2 = -0.24diopters

Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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Answer:

Explanation:

Convert all lengths to meters:

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)

Therefore:

[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].

On the other hand:

[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].

Because the cotangent of [tex]\theta[/tex] is very close to zero,

[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].

[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].

[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].

Answer:

a)   I = I₀/2, b)  I = I₀/2 cos² θ

Explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

          i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

          I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

                  I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

                I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0          I = Io / 2

θ = 90º      I = 0

Answer:

Explanation:

We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.

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Answer:

B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction

Explanation:

In an electromagnetic wave the electric and magnetic fields are in phase

         c = E / B

         B = E / c

let's calculate

          B = 4.00 / 3 10⁸

          B = 1.33 10⁻⁸ T

To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.

 If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction

Answer:

The  thickness is   [tex]t = 1.415 *10^{-7 } \ m[/tex]

Explanation:

From the question we are told that

    The wavelength is  [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]

     The  index of refraction of glass is  [tex]n_g = 1.71[/tex]

     The index of refraction of the coating is  [tex]n= 1.46[/tex]

Generally the condition for destructive interference is  

         [tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]

Here m is the order of the interference pattern and given from the question that we are considering minimizing  reflection  m = 0

t = thickness of the coating

substituting values

         [tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]

    =>    [tex]t = 1.415 *10^{-7 } \ m[/tex]

Answer:

42000N

Explanation:

First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.

1) linear thermal expansion coef brass 19e-6 /K

∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm

Second part involves linear elasticity.

for brass, young's modulus is 15e6 psi or 100 GPa

cross-sectional area of rod is π(0.008)² = 0.0002 m²

F = EA∆L/L

F = (100e9)(0.0002)(0.00387) / (1.85)

F = 42000 or 42 kN

Answer:

displacements are 0.776m, 0.114m

Explanation:

We were given mass of 26-g rifle bullet , then we can convert to Kg since

Momentum is conserved here.

The initial momentum before impact = (Mi * Vi)

Where Mi= initial given mass

Vi=initial velocity given

= 0.026 * 220 = 5.72 kgm/s

The final momentum after impact is (Mf * Vf )

Mf= final mass

5.72=( 3.82* Vf )

= 5.72/ 3.82

= 1.497 m/s

the speed of the pendulum bob with bullet afterwards= 1.497 m/s

the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.

M = 3.82 kg the mass of the bob containing the bullet

E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J

When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before

E(total) = Mgh + 0 = Mgh = 4.280J

solving for h and substituting,

h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m

Since the height is found,we the angle of the pendulum at the top of the swing can also be determined

A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees

Since A is known, the displacement along the horizontal axis can be calculated as

x = 2.7* sin(A) = 0.776m

therefore, displacement is 0.776m, 0.114m

the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

            v₂ = 9 v

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

[tex]x - x_o = u_xt[/tex]

[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex]  ---- (1)

the equation of the motion y is :

[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]

[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]

[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]

[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]

[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]

[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]

By using the quadratic formula, we have;

[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]

where;

a = 4.9,   b = -5.14     c = 1

[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]

[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]

[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

[tex]\mathtt{x = u_x(0.258)}[/tex]

[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]

[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]

[tex]\mathbf{x = 1.581 \ m}[/tex]

Thus, the child is 1.581 m far from the fence