Differences between regular and irregular

Answer:

a. 6.41 m/s

b. 120.85 m/s^2

Explanation:

The computation is shown below:

a. Pebble speed is

As we know that according to the tangential speed,

[tex]v = r \times \omega[/tex]

[tex]= \frac{0.68}{2} \times 18.84[/tex]

= 6.41 m/s

The 18.84 come from

[tex]= 2 \times 3.14 \times 3[/tex]

= 18.84

b. The pebble acceleration is

[tex]a = \frac{v^2}{r}[/tex]

[tex]= \frac{6.41^2}{0.34}[/tex]

= 120.85 m/s^2

We simply applied the above formulas so that the pebble speed and the pebble acceleration could come and the same is to be considered

Answer:

a) 5.19 W/m²

b) 1.79 J

Explanation:

For the calculation of intensity, I. We have

I = E(rms)² / (cμ), where

c = speed of light

μ = permeability of free space

I = (62.5 / √2)² / [(2.99 x 10^8) (1.26 x 10^-6)]

I = 1954 / 376.74

I = 5.19 W/m²

Therefore, the intensity, I = 5.19 W/m²

t = 14.9 s

A = 0.0231 m²

Amount if energy flowing, U = IAt

U = (5.19) (0.0231) (14.9) J

U = 1.79 J

Answer:

The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].

Explanation:

Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:

[tex]I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}[/tex]

Where:

[tex]I[/tex] - Moment of inertia of a disk, measured in kilogram-square meter.

[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed, measured in radians per second.

This relationship is simplified and final angular speed can be determined in terms of initial angular speed:

[tex]\omega_{f} = \frac{1}{2}\cdot \omega_{o}[/tex]

Given that [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the angular speed of the new system is:

[tex]\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)[/tex]

[tex]\omega_{f} = 3\,\frac{rad}{s}[/tex]

The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].

Answer:

v =   11 m/s   is her final speed

Explanation:

work done by gravity = m g Δh =   40×9.8×10   = 3920 Joules

Work done by friction = - force×distance =   - 20×100   =   - 2000 Joules

[minus sign because friction force is opposite to the direction of motion]

Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2   = 500 Joules

Now, by work energy theorem

Work done = change in kinetic energy.

Final K.E. = initial K.E. + total work =    500 + 3920 - 2000  = 2420 Joules

Now, Final K.E. = (1/2) m v^2  [final speed being v= speed at the bottom]

⇒  2420 = (1/2)×40×v^2

   ⇒  121 = v^ 2

  v =   11 m/s   is her final speed

Answer:

A. a (-321.393, 383.022) b (-76.40, -64.278)

B. (-397.991, 318.744)

C. a. resulting speed 509.9mph  b. bearing of the plane = 51.6°

Explanation:

Answer:

The bottom two lines.

Explanation:

They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.

Answer:

If the initial speed is doubled the time is also doubled

Explanation:

You have that a car with velocity v is decelerated by a constant acceleration a in a time t.

You use the following equation to establish the previous situation:

[tex]v'=v-at[/tex]     (1)

v': final speed of the car  = 0m/s

v: initial speed of the car

From the equation (1) you solve for t and obtain:

[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex]     (2)

To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:

[tex]v'=v_1-at'[/tex]     (3)

v' = 0m/s

v1: new initial speed = 2v

t': new time

You solve the equation (3) for t':

[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]

If the initial speed is doubled the time is also doubled

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              [tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             [tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             [tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          [tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex]  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         [tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]

                          [tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]

Answer:

it changes into liquid

Answer:

It changes in to liquids

Explanation:

This is because the water vapor cools down and condenses it attaches it self to dust forming water droplets. Those water droplets are water.

Answer:

a) WT = 137.5 J

b) v2 = 2.34 m/s

Explanation:

a) The total work done on the block is given by the following formula:

[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex]          (1)

Fp: force parallel to the displacement of the block = 150N

Ff: friction force

d: distance = 5.0 m

Then, you first calculate the friction force by using the following relation:

[tex]F_f=\mu_k N=\mu_k Mg[/tex]        (2)

μk: coefficient of kinetic friction = 0.25

M: mass of the block = 50kg

g: gravitational constant = 9.8 m/s^2

Next, you replace the equation (2) into the equation (1) and solve for WT:

[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]

The work done over the block is 137.5 J

b) If the block started from rest, you can use the following equation to calculate the final speed of the block:

[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex]     (3)

WT: total work = 137.5 J

v2: final speed = ?

v1: initial speed of the block = 0m/s

You solve the equation (3) for v2:

[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]

The final speed of the block is 2.34 m/s

Answer:

[tex]solution \\ distance \: travelled = 1.8 \: km \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1.8 \times 1000m \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1800 \: m \\ time \: taken = 5 \: minute \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 5 \times 60 \: seconds \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 300 \: seconds \\ average \: velocity = \frac{distance \: travelled}{time \: taken} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1800}{ 300} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6 \: {ms}^{ - 1} [/tex]

hope this helps..

Its average speed is (1,800 m) / (300 sec) = 6 m/s .

There's not enough information in the question to calculate the velocity with.  We would need to know the straight-line distance and direction from the place he started from to the place he ended at.

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = 360 km-m/s

The direction of the change is up /\ .

The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.

Solve the problem ?

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Answer:

404.3 J

Explanation:

Given that

Weight of the merry go round = 790 N

Radius if the merry go round = 1.6 m

Horizontal force applied = 45 N

Time taken = 4 s

To find the mass of the merry go round, we divide the weight by acceleration due to gravity. Thus,

m = F/g

m = 790 / 9.8

m = 80.6 kg

We know that the moment of inertia is given as

I = ½mr², on substitution, we have

I = ½ * 80.6 * 1.6²

I = 103.17 kgm²

Torque = Force applied * radius, so

τ = 45 * 1.6

τ = 72 Nm

To get the angular acceleration, we have,

α = τ / I

α = 72 / 103.17

α = 0.70 rad/s²

Then, the angular velocity is

ω = α * t

ω = 0.7 * 4

ω = 2.8 rad/s

Finally, to get the Kinetic Energy, we have

K.E = ½ * Iω², on substituting, we get

K.E = ½ * 103.17 * 2.8²

K.E = 404.3 J

Therefore, the kinetic energy is 404.3 J

Answer: C. Petal. Length

Explanation: Petal are unit of Corolla which are usually brightly colored. This part of a plant or flower, helps attracts insects to the plant for pollination. And also provide protection to the reproductive parts of the plant or flower.

Examples of flowers with petals is the Sun Flower, which coincidentally is the flower plant with most petals.

Answer:

a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]

Explanation:

a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

Where:

[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.

[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.

Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:

[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]

[tex]COP_{HP} = 14[/tex]

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]

The input work to deliver a determined amount of heat to the house:

[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]

If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:

[tex]W_{in} = \frac{3000\,J}{14}[/tex]

[tex]W_{in} = 214.286\,J[/tex]

b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

Where:

[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.

[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.

Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:

[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]

[tex]COP_{HP} = 7[/tex]

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]

The input work to deliver a determined amount of heat to the house:

[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]

If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:

[tex]W_{in} = \frac{3000\,J}{7}[/tex]

[tex]W_{in} = 428.571\,J[/tex]

Answer:

we see that the lights with the most extreme wavelength are blue and red

we see that the separation between the interference lines (y) increases linearly with the wavelength for which the phenomenon is best observed in the RED response 2

Explanation:

In Young's double-slit experiment, constructive interference is written by the equation

       d sin θ = m λ

where you give the gap separation, lam the length of the donda used and m the order of interference

in many he uses trigonometry to express the synth in confusing the distances on a very distant screen

so θ = y / L

in this experiment the angles are generally very small, so

     tan θ = sin θ / cos θ = sin θ

     sint θ = y / L

let's replace

      d y / L = mλ

      y = (m L / d) λ

now let's examine the effect of changing the wavelength

1 yellow lam = 600 10⁻⁹ m

2) red lam = 750 10⁻⁹m

3) blue lam = 450 10⁻⁸ nm

4) green lam = 550 10⁻⁹ nm

we see that the lights with the most extreme wavelength are blue and red

we see that the separation between the interference lines (y) increases linearly with the wavelength for which the phenomenon is best observed in the RED response 2

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

Answer:

22.66Nm²/C

Explanation:

Flux through an electric field is expressed as ϕ = EAcosθ

When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.

From the formula above where:

EA = 25N.m^2/C

θ = 25°

ϕ = 25cos 25°

ϕ = 22.66Nm²/C

Answer:

mass 20 times of an amazing and all its motion

Answer:

D

Explanation:

Answer:

It is D

Explanation: No cap

Answer: evaporation

Explanation:

Refrigerators work by causing the refrigerant circulating inside them to change from a liquid into a gas. This process, called evaporation, cools the surrounding area and produces the desired effect.

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

[tex]F=qvB[/tex]     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

[tex]F=qvB=ma[/tex]       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

Answer: 192 ft/s

Explanation:

The initial height of the object is:

576ft above the ground.

The position equation is:

p(t) = -16*t^2 + 576

in the position equation, we only can see the therm of the initial height and the term of the acceleration (that is equal to the gravitational acceleration g = 32 ft/s^2 over 2)

So we have no initial velocity, this means that at the beginning we only have potential energy:

U = m*g*h

where m is the mass of the object, g = 32m/s^2 and h = 576 ft.

Now, as the object starts to fall down, the potential energy is transformed into kinetic energy, and when the object is about to hit the ground, all the potential energy has become kinetic energy.

The kinetic energy equation is:

K = (m/2)*v^2

where v is the velocity of the object, then the maximum kinetic energy (when the object reaches the ground) is equal to the initial potential energy:

m*g*h = (m/2)*v^2

now we can solve this for v.

v = √(2*g*h) = √(2*32ft/s^2*576ft) = 192 ft/s

Answer:

Dear user,

Answer to your query is provided below

The angle for which the rope will break θ = 41.8°

Explanation:

Explanation of the same is attached in image

A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°

The tension is a kind of force which acts on linear objects when subjected to pull.

The maximum tension Tmax =2W

From the work energy principle,

T₂ = 1/2 mv²

Total energy before and after pushing off

0+mglsinθ = 1/2 mv²

v² = 2gflsinθ..............(1)

From the equilibrium of forces, we have

T= ma +mgsinθ = mv²/l +mgsinθ

2mg = mv²/l +mgsinθ

2g = v²/l +gsinθ

Substitute the value of v² ,we get the expression for θ

θ = sin⁻¹(2/3)

θ =41.81°

Hence, the angle θ for which the rope will break, is 41.81°

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Answer:

Explanation:

Given that,

The length of the beam L = 3.10m

The mass of the steam beam [tex]m_1[/tex] = 430kg

The mass of worker [tex]m_2[/tex] = 69.0kg

The distance from  the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]

and our length of beam is 3.10m

so the distance from  the fixed point to centre of gravity of beam is

[tex]\frac{3.10}{2}=1.55m[/tex]

Then the net torque is

[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]

[tex]W_s[/tex] is the weight of steel rod

[tex]=430\times9.8=4214N[/tex]

[tex]W_w[/tex] is the weight of the worker

[tex]=69\times9.8\\\\=676.2N[/tex]

Torque can now be calculated

[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]

≅ 9169Nm

Answer:

visual basics

Explanation:

not suited for programming, slower than the other languages. hard to translate to other operating systems

Answer:

Corresponding raft speed = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Explanation:

Law of conservation of momentum gives that the momentum of the diver and the raft before the dive is equal to the momentum of the diver and the raft after the dive.

And since the raft and the diver are initially at rest, the momentum of the diver after the dive is equal and opposite to the momentum experienced by the raft after the dive.

(Final momentum of the diver) + (Final momentum of the raft) = 0

Final Momentum of the diver = (mass of the diver) × (diving velocity of the diver)

Mass of the diver = 73 kg

Diving velocity of the diver = 5.54 m/s

Momentum of the diver = 73 × 5.54 = 404.42 kgm/s

Momentum of the raft = (mass of the raft) × (velocity of the raft)

Mass of the raft = 462 kg

Velocity of the raft = v

Momentum of the raft = 462 × v = (462v) kgm/s

404.42 + 462v = 0

462v = -404.42

v = (-404.42/462) = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Hope this Helps!!!

Answer:

About 6.26m/s

Explanation:

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Divide both sides by mass:

[tex]gh=\dfrac{1}{2}v^2[/tex]

Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.

[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]

Hope this helps!

The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

Given data:

The height of vertical cliff is, h = 4.0 m.

Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,

Kinetic energy = Gravitational potential energy

[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]

Here,

m is the mass of rock.

v is the speed of rock.

g is the gravitational acceleration.

Solving as,

[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]

Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.

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Answer:

The wavelength is  [tex]\lambda = 886 \ nm[/tex]

Explanation:

From the question we are told that

   The  energy band gap is  [tex]E = 1.4 eV[/tex]

Generally the energy of  a single photon of light emitted for an electron jump in a GaAS solar cell is mathematically represented as

      [tex]E = \frac{hc}{\lambda }[/tex]

Where  h is the Planck's  constant with values

     [tex]h = 4.1357 * 10^{-15} eV[/tex]

and  c is  the speed of light with values  [tex]c = 3*10^{8} \ m/s[/tex]

So  

     [tex]\lambda = \frac{hc}{E}[/tex]

substituting values

    [tex]\lambda = \frac{4.1357 *10^{-15} * 3.0 *10^{8}}{1.4}[/tex]

  [tex]\lambda = 886 \ nm[/tex]