The magnitude of a negative of a vector is negative. O True O False

Answer:

C

Explanation:

the correct answer is c) Pre Schematic.

It is in the Pre-schematic stage of development that the children learn to draw human figures by sketching a circle with dangly lines for legs. It remains till the age of six. After pre schematic stage comes the schematic stage, where in children can draw complex structures.  

Children drawing human figures with a circle and lines for legs are in the Pre-schematic stage of art development, indicating an early attempt at representational drawing before achieving greater accuracy and detail in later stages.

When children draw human figures by sketching a circle with dangly lines for legs, they are demonstrating the Pre-schematic developmental stage in art. This stage is distinct from the earlier scribbling phase, signaling that children have begun to recognize and attempt to represent the human form in a more organized albeit simplified manner. However, this representation lacks the complexity and accuracy found in later stages, such as the Schematic stage, where more detailed and proportionate figures are drawn.

The Pre-schematic stage typically occurs in children aged between 4 and 7 years old. During this time, their cognitive and motor skills are developing rapidly, but they are still learning how to translate what they see and imagine onto paper with greater fidelity. Thus, the circle and lines approach is a developmentally appropriate method for them to begin exploring human figures in their art.

Answer:

Yes

Explanation:

Yes, the orbital direction of a planet affect its Synodic period as seen from earth. The synodic period of mercury is about 116 days on Earth. Whereas the sidereal period of mercury as seen from Earth is 88 days.  This difference is caused by the simultaneous motion of Earth in its orbit. Hence we can say that the  orbital direction of a planet affect its synodic period as seen from earth.

Answer:

[tex]N=1.56*10^{20}[/tex] protons

Explanation:

The total charge Q+=+25.0C is the sum of the charge of the N protons, with charge q :

[tex]Q_{+}=N*q_{p}[/tex]

q_{p}=1.6*10^{-19}C    charge of a proton

We solve to find N:

[tex]N=Q_{+}/q_{p}=25/(1.6*10^{-19})=1.56*10^{20}[/tex]

Answer:

Ping Pong ball move at velocity 160.4 m/s

Explanation:

given data

mass m1 = 7 kg

velocity v1 = 3 m/s

mas m2 = 2.45 g = 2.45 × [tex]10^{-3}[/tex] kg

same kinetic energy

to find out

How fast Ping Pong ball move (v2)

solution

we know same KE

so

[tex]\frac{1}{2}* m1* v1^{2} = \frac{1}{2}* m2* v2^{2}[/tex]   ...........1

so v2 will be

v2 = [tex]\sqrt{\frac{m1*v1^2}{m2} }[/tex]    .............2

put here value in equation 2 we get v2

v2 = [tex]\sqrt{\frac{7*3^2}{2.45*10^{-3}}}[/tex]

v2 = 160.4

so Ping Pong ball move at velocity 160.4 m/s

Begin with the kinetic energy equation, set the two equal and solve for the unknown velocity of the Ping Pong ball. Substitute the given values into the formula to finalize your answer.

This question involves the concept of kinetic energy, that is equal to 1/2 mass x velocity2 for an object in motion. Let's denote the mass of the bowling ball as m1 (7kg), its velocity as v1 (3m/s), the mass of the Ping Pong ball as m2 (0.00245kg, which is 2.45g in kg), and its velocity as v2, which we need to find. If we assume that the two balls have the same kinetic energy, then:

1/2 * m1 * v12 = 1/2 * m2 * v22

By calculating the left-hand side and then rearranging the equation to solve for v2, we will get the velocity of the Ping Pong ball:

v2 = sqrt((m1 * v12) / m2)

Now substitute the given values into the formula to find the velocity of the Ping Pong ball.

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Answer:

b)determining the electric field due to each charge and adding them together as vectors.

Explanation:

The electric Field is a vector quantity, in other words it has a magnitude and a direction. On the other hand, the electric field follows the law of superposition. The electric field produced by two elements is equal to the sum of the electric fields produced by each element when the other element is not present. in other words, the total electric field is solved determining the electric field due to each charge and adding them together as vectors.

The electric field due to two-point charges is found by determining the electric field of each charge and adding them together as vectors, considering both their magnitudes and directions.

The correct way to find the electric field due to two-point charges is option (b), determining the electric field due to each charge and adding them together as vectors. The electric field is a vector quantity, meaning it has both magnitude and direction. Thus, when two or more electric fields exist in the same region, you add them as vectors, taking into account both their magnitudes and directions. For example, if you have two positive charges, the fields they generate will be in different directions. Therefore, you would add these fields together as vectors, resulting in the net electric field in that region.

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Answer:

-4.01 × 10⁸ N/C.

Explanation:

Given : Charge Q = -7.67 × 10⁻⁶ C

Charge q = + 5.03×10⁻⁶ C

Distance between the charges = 0.0338 m

Let d be the distance from the charge Q and charge q to the mid point.

Midway distance =  d = 0.0169 m

Electric field due to the negative charge = E₁ = k Q / d²

Electric field due to the positive charge = E₂ = k q / d²

Here k = 9 × 10⁹ N m²/C² is the Coulomb's constant.

E₁ = (9 × 10⁹)(7.67 × 10⁻⁶) / (0.0169)² = 2.42 × 10⁸ N/C

E₂ = (9 × 10⁹)( 5.03×10⁻⁶) /(0.0169)² = 1.585 × 10⁸N/C

Assume that the negative charge is  towards the left of the mid point and the positive charge is towards the right of the mid point,

The electric field due to the negative charge is inwards towards the charge(left) and the electric field due to positive charge is outwards and so towards the negative X direction.

Total electric field = E = E₁ + E₂ = -2.42 × 10⁸ - 1.585 × 10⁸

                                                     = - 4.01 × 10⁸ N/C

Direction towards the left (towards the negative charge).

Answer:

The first overtone frequency is 100 Hz.

Solution:

According to the question:

Length of pipe, l = 1.75 m

Speed of sound in air, v_{sa} = 350 m/s

Frequency of first overtone, [tex]f_{1}[/tex] is given by:

[tex]f_{1} = \frac{v_{sa}}{2l}[/tex]

[tex]f_{1} = \frac{350}{2\times 1.75} = 100 Hz[/tex]

Since, the frequency, as clear from the formula depends only on the speed

and the length. It is independent of the air temperature.

Thus there will be no effect of air temperature on the frequency.

Answer:

option A is correct

25% free lead and 75% lead oxide

Explanation:

we have given free lead and lead oxide %

so here we know lead oxide material usually composed in the range of for free lead and lead oxide as

lead oxide material contain free lead range is 22 % to 38 %

so here we have only option 1 which contain in free lead in the range of 22 % to 38 % i.e 25 % free lead

so    

option A is correct

25% free lead and 75% lead oxide

Final answer:

The speed of the block is 4.08 m/s, the magnitude of the block’s acceleration is 25.41 m/s^2, and the direction of the block’s acceleration is toward the wall.

Explanation:

(a) To find the speed of the block, we can use the principle of conservation of mechanical energy. The potential energy stored in the spring when it is compressed is converted into the kinetic energy of the block when it is released. The potential energy stored in the spring is given by:

PE = 0.5 * k * x^2

where k is the force constant of the spring and x is the compression of the spring. Plugging in the values, we get:

PE = 0.5 * 130.0 N/m * 0.80 m * 0.80 m = 41.60 J

The kinetic energy of the block when it is released is given by:

KE = 0.5 * m * v^2

where m is the mass of the block and v is its speed. Equating the potential and kinetic energies, we have:

PE = KE

41.60 J = 0.5 * 3.00 kg * v^2

Solving for v, we get:

v = √(41.60 J / (0.5 * 3.00 kg)) = 4.08 m/s

(b) The magnitude of the block's acceleration can be calculated using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force applied to the block minus the force of friction. The force applied to the block is given by F = 88.0 N. The force of friction can be calculated using the equation:

f = μk * m * g

where μk is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the values, we get:

f = 0.400 * 3.00 kg * 9.8 m/s^2 = 11.76 N

The net force is therefore:

net force = F - f = 88.0 N - 11.76 N = 76.24 N

Using Newton's second law, we have:

76.24 N = 3.00 kg * a

Solving for a, we get:

a = 76.24 N / 3.00 kg = 25.41 m/s^2

(c) The direction of the block's acceleration can be determined by considering the net force acting on the block. In this case, the applied force and the force of friction are in opposite directions, resulting in a net force in the direction of the applied force. Therefore, the direction of the block's acceleration is toward the wall.

Answer:d-6.7 km

Explanation:

Given

Bicyclist pedals at a speed of 5 km/h

so his speed in meter per second

[tex]5\times \frac{5}{18}=\frac{25}{18} m/s[/tex]

In 80 minutes he would travel

Distance traveled[tex]=\frac{25}{18}\times 80\times 60=6666.667 m\approx 6.67 km[/tex]

Answer:

a) 1.29*10^6 N/C

b) 0.703 *10^6 m/s

Explanation:

This is a parallel plates capacitor. In a parallel plates capacitor the electric field depends on the charge of the disks, its area and the vacuum permisivity (Assuming there is no dielectric) and can be found using the expression:

[tex]E = \frac{Q}{A*e_0} =\frac{11*10^{-9}C}{(\frac{1}{4}\pi*(0.035m)^2)*8.85*10^{-12}C^2/Nm^2} = 1.29 *10^6 N/C[/tex]

For the second part, we use conservation of energy. The change in kinetic energy must be equal to the change in potential energy. The potential energy is given by:

[tex]PE = V*q[/tex]

V is the electric potential or voltage, q is the charge of the proton. The electric potential is equal to:

[tex]V = -E*d[/tex]

Where d is the distance to the positive disk. Then:

[tex]\frac{1}{2}mv_1^2 +V_1q = \frac{1}{2}mv_2^2 +V_2q\\\frac{1}{2}m(v_1^2 - v_2^2)=(V_2-V_1)q = (r_1-r_2)Eq|r_2 = 0m, v_2=0m/s\\v_1 = \sqrt{2\frac{(0.002m)*1.29*10^6 N/C*1.6*10^{-19}C}{1.67*10^{-27}kg}}= 0.703 *10^6 m/s[/tex]

Answer:

Answered

Explanation:

The Kinetic energy of a bullet does not remain constant. It dissipates as the bullet travel through the air. It is partly converted into heat, sound, air resistance etc.

Whereas the momentum of a travelling body remains conserved and hence constant. therefore a bullet with more momentum will cause more damage. that is momentum is more important than kinetic energy.

Answer:0.249 s

Explanation:

Given

Ball is tossed down with a velocity of 6.8 m/s downward

height from ground=2 m

therefore time to reach ground is

[tex]s=ut+\frac{gt^2}{2}[/tex]

[tex]2=6.8\times t+\frac{9.81\times t^2}{2}[/tex]

[tex]9.81t^2+13.6t-4=0[/tex]

[tex]t=\frac{-13.6\pm \sqrt{13.6^2+4\times 4\times 9.81}}{2\times 9.81}[/tex]

[tex]t=\frac{-13.6+18.49}{19.62}=0.249 s[/tex]

Answer:

1.57 kW

Explanation:

The rate of heat loss is given by:

q = Gm * Cp * (tfin - ti)

Where

q: rate of heat loss

Gm: mass flow

Cp: specific heat at constant pressure

The Cp of air is:

Cp = 1 kJ/(kg*K)

The mass flow is the volumetric flow divided by the specific volume

Gm = Gv / v

The volumetric flow is the air speed multiplied by the cruss section of the duct.

Gv = s * h * w (I name speed s because I have already used v)

The specific volume is obtained from the gas state equation:

p * v = R * T

60 C is 333 K

The gas constant for air is 287 J/(kg*K)

Then:

v = (R * T)/p

v = (287 * 333) / 100000 = 0.955 m^3/kg

Then, the mass flow is

Gm = s * h * w / v

And rthe heat loss is of:

q = s * h * w * Cp * (tfin - ti) / v

q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)

The magnitude of the minimum deceleration needed to avoid the accident is –0.55 m/s²

To solve the question given above, we'll begin by calculating the distance travelled during the reaction time. This can be obtained as follow:

Speed = 18 m/s

Time = 0.45 s

Speed = distance / time

18 = distance / 0.45

Cross multiply

Distance = 18 × 0.45

Thus, the engineer travelled a distance of 8.1 m during the reaction time.

Next, we shall the distance between the engineer and the car. This can be obtained as follow:

Total distance = 300 m

Distance during the reaction time = 8.1 m

= 300 – 8.1

Finally, we shall determine the magnitude of the deceleration needed to avoid the accident. This can be obtained as follow:

Initial velocity (u) = 18 m/s

Final velocity (v) = 0 m/s

Distance (s) = 291.9 m

0² = 18² + (2 × a × 291.9)

0 = 324 + 583.8a

Collect like terms

0 – 324 = 583.8a

–324 = 583.8a

Divide both side by 583.8

a = –324 / 583.8

Therefore, the magnitude of the deceleration needed to avoid the accident is –0.55 m/s²

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To avoid an accident, the engineer must decelerate the train at a minimum rate of approximately -0.55 m/s^2, taking into account the reaction time which contributed an additional 8.1 m to the stopping distance.

An engineer in a locomotive sees a car stuck on the track and needs to calculate the necessary deceleration to avoid an accident. The engineer has a reaction time of 0.45 seconds, and the initial speed of the train is 18 m/s. The total stopping distance must include the distance traveled during the reaction time plus the actual braking distance.

The distance covered during the engineer's reaction time is found using the formula d = vt, where d is distance, v is velocity, and t is time. We get d = 18 m/s * 0.45 s = 8.1 m.

So, the stopping distance minus the reaction distance is 300 m - 8.1 m = 291.9 m. To find the deceleration, we can use the formula v^2 = u^2 + 2as, where v is the final velocity (which is 0), u is the initial velocity, a is the acceleration (deceleration in this case), and s is the distance. Solving for a, we get a = -u^2 / (2s). Substituting the values, a = -(18 m/s)^2 / (2 * 291.9 m), resulting in a deceleration of approximately -0.55 m/s^2.

Answer:

option A

Explanation:

the correct answer is option A

An electric field is a vector field created by charged particle.

Electric field is mathematically defined as the ratio of the vector component of the force to the charge in coulomb.

Si unit of Electric field is N/C

Electric field is created by the electric charge or by the varying magnetic field.

Electric field is responsible for the attractive force between the electron and atomic nucleus.

Answer:

Frequency, [tex]\nu=1.96\times 10^{20}\ Hz[/tex]

Explanation:

Given that,

Energy of a gamma rays, [tex]E=0.815\ MeV=0.815\times 10^6\ eV[/tex]

Since, [tex]1\ eV=1.6\times 10^{-19}\ J[/tex]

[tex]0.815\times 10^6\ eV=0.815\times 10^6\times 1.6\times 10^{-19}\ J[/tex]

[tex]E=1.304\times 10^{-13}\ J[/tex]

The energy of a wave is given by :

[tex]E=h\nu[/tex]

[tex]\nu[/tex] is the frequency of such a photon

[tex]\nu=\dfrac{E}{h}[/tex]

[tex]\nu=\dfrac{1.304\times 10^{-13}\ J}{6.63\times 10^{-34}\ Js}[/tex]

[tex]\nu=1.96\times 10^{20}\ Hz[/tex]

So, the frequency of such a photon is [tex]1.96\times 10^{20}\ Hz[/tex]. Hence, this is the required solution.

Answer:

(a) d = 15.6 m

(b) v' = - 4.3 m/s

Given:

Initial velocity, v = 9.50 m/s

Acceleration, a = [tex] - 2.30 m/s^{2][/tex] or deceleration = 2.30[tex]m/s^{2}[/tex]

Solution:

(a) For the calculation of the distance covered, we use eqn (2) of motion:

[tex]d = vt + \frac{1}{2}at^{2}[/tex]

where

d = distance covered in time t

t = 6 s (Given)

Now,

[tex]d = 9.50\times 6 - \frac{1}{2}\times 2.30times 6^{2}[/tex]

d = 15.6 m

(b) For the calculation of her final velocity, we use eqn 1 of motion:

v' = v + at

v' = 9.50 + (- 2.30)(6) = - 4.3 m/s

(c) Since, the final velocity after the body slows down comes out to be negative and the distance covered and displacement of the body are different, it does not make sense.

Answer:

(a) 14.88 ohm

(b) 1.88 A

(c)  -74.4°

Explanation:

Vo = 28 V

f = 12 kHz = 12000 Hz

R = 4 ohm

L = 30 micro henry = 30 x 10^-6 H

C = 800 nF = 800 x 10^-9 F

(a)

The inductive reactance,

XL = 2 π f L = 2 x 3.14 x 12000 x 30 x 10^-6 = 2.26  ohm

The capacitive reactance

[tex]X_{c}=\frac{1}{2\pi fC}=\frac{1}{2 \times 3.14 \times 12000 \times 800 \times 10^{-9}}[/tex]

Xc = 16.59 ohm

Let the impedance is Z.

[tex]Z=\sqrt{4^{2}+\left ( 2.26-16.59 \right )^{2}}[/tex]

Z = 14.88 ohm

(b)

The formula for the amplitude of current

[tex]I_{o}=\frac{V_{o}}{Z}=\frac{28}{14.88}[/tex]

Io = 1.88 A

(c)

Let the phase difference is Ф

[tex]tan\phi =\frac{X_{L}-X_{C}}{R}=\frac{2.26-16.59}{4}=-3.5825[/tex]

Ф = -74.4°

Answer:

783 grams

Explanation:

Here a time of 0.783 kg is given.

Some of the prefixes of the SI units are

1 gram = 10⁻³ kilogram

1 milligram = 10⁻⁶ kilogram

1 microgram = 10⁻⁹ kilogram

The number is 0.783

Here, the only solution where the number of significant figures is three is gram

[tex]1\ kilogram = 1000\ gram[/tex]

[tex]\\\Rightarrow 0.783\ kilogram=0.783\times 1000\ gram\\ =783\ gram[/tex]

So, 0.783 kg = 783 grams

Answer

Work done will be [tex]14.7\times 10^{-6}J[/tex] and it will be positive

Explanation:

We have given charge [tex]2.1\times 10^{-6}C[/tex]

We have to find work done in moving the charge from 15 volt to 8 volt

Let [tex]V_1=15V\ and\ V_2=8volt[/tex]

So potential difference [tex]V=V_1-V_2=15-8=7volt[/tex]

We know that work done [tex]W=QV[/tex], here Q is charge and V is potential difference

So work done [tex]W=QV=2.1\times 10^{-6}\times 7=14.7\times 10^{-6}J[/tex]

It will be positive work done because work is done in moving charge from higher potential to lower potential  

Answer:

[tex]v_{o}=8.85m/s[/tex]

Explanation:

To determine the muzzle velocity of the gun, we must know how long does it take the ball to strikes the ground

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

Since the ground is at y=0 and [tex]v_{oy}=0[/tex]

[tex]0=1-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t

[tex]t=0.4517s[/tex]

Now, to determine the muzzle velocity we need to find its acceleration first

[tex]x=x_{o}+v_{ox}t+\frac{1}{2}at^{2}[/tex] (1)

[tex]v=v_{ox}+at[/tex] (2)

If we analyze the final velocity is 0. From (2) we have that

[tex]v_{ox}=-at[/tex] (3)

Replacing (3) in (1)

[tex]2=-at^{2}+\frac{1}{2}at^{2}[/tex]

[tex]2=a(0.4517)^{2} (\frac{1}{2}-1)[/tex]

[tex]a=-19.60m/s^{2}[/tex]

Solving (3)

[tex]v_{ox}=-at=-(19.60m/s^{2} )(0.4517s)=8.85m/s[/tex]

The skater's speed at the bottom of the track can be determined using the law of conservation of energy.

The skater's speed at the bottom of the track can be determined by applying the law of conservation of energy. At the top of the track, the skater has only potential energy, which is converted to kinetic energy as the skater moves down the track. The potential energy at the top of the track can be calculated using the equation:

mgh = (1/2)mv2

where m is the mass of the skater, h is the height, and v is the speed of the skater. At the bottom of the track, all the potential energy is converted to kinetic energy, so we can set the potential energy equal to the kinetic energy:

mgh = (1/2)mv2

Rearranging the equation gives:

v = √(2gh)

Plugging in the values for mass (which is not given in the question) and height, you can calculate the skater's speed at the bottom of the track.

Answer:

735 J

Explanation:

mass of box, m = 15 kg

Height, h = 5 m

Work is defined as the product of force and the displacement in the  direction of force.

W = F x S

Work is a scalar quantity and its SI unit is Joule.

Here, Force, F = m x g = 15 x 9.8 = 147 N

So, W = 147 x 5 = 735 J

Thus, the work is 735 J.

Answer:

The height of mountain in meter will be 106.2732 m

Explanation:

We have given height of mountain = 348 ft,8 in

We know that 1 feet = 0.3048 meter

So 348 feet [tex]=348\times 0.3048=106.07meter[/tex]

And we know that 1 inch = 0.0254 meter

So 8 inch [tex]8\times 0.0254=0.2032m[/tex]

So the total height of mountain in meter = 106.07+0.2032 = 106.2732 m

The height of mountain in meter will be 106.2732 m

Answer:

g = 0.98 m

Explanation:

given data:

upper portion weight 438 N and its center of gravity is 1.17 m

upper legs weight 144 N and its center of gravity is 0.835 m

lower leg weight 88 N and its center of gravity is 0.270 m

position of center of gravity of whole body can be determine by using following relation

[tex]g = \frac{\sum ({m_i g_i)}}{\sum m_i}[/tex]

where [tex]m_i[/tex] is mass of respective part and [tex]g_i[/tex] is center of gravity

[tex]g = \frac{\sum{ 438\times 1.17 + 144\times 0.835 + 88\times 0.270}}{438+144+88}[/tex]

g = 0.98 m

Final answer:

To calculate the muzzle velocity, we use the equation for constant acceleration (v = u + at) with an initial velocity (u) of 0, the given acceleration (a) of 5.70 x 10^5 m/s^2, and the time (t) of 9.60 x 10^-4 s to find the final velocity (v), which is approximately 547 m/s.

Explanation:

The question concerns calculating the muzzle velocity of a bullet as it's accelerated from the firing chamber to the end of the barrel of a gun.

Using the formula v = u + at, where v is the final velocity, u is the initial velocity (which is zero before the gun is fired), a is the acceleration, and t is the time for which the acceleration is applied, we can find the muzzle velocity. The bullet experiences an acceleration (a) of 5.70 x 105 m/s2 for a time (t) of 9.60 x 10−4 seconds.

Plugging the values into the formula, we get:

v = 0 + (5.70 x 105 m/s2)(9.60 x 10−4 s)

v = 5.47 x 102 m/s

Therefore, the final muzzle velocity of the bullet as it leaves the barrel is approximately 547 m/s.

Answer:

The correct answer is option 'c'.

Explanation:

Gauss's law theoretically can be used to calculate the electric field by any shape of conductor

According to Gauss's law we have

[tex]\oint _s\overrightarrow{E}\cdot \widehat{ds}=\frac{q_{in}}{\epsilon _o}[/tex]

Now the integral on the right hand side of the above relation is solved easily if there is a high degree of symmetry in the electric field which is  possible in cases when the object is highly symmetric.

The cases of high symmetry include electric fields due to charged spheres, infinite line charge, point charge, infinite plane charged sheet, infinite cylindrical conductor.

But the for the case of limited height of cylinder the symmetry cannot be utilized thus the integral becomes complex to solve, thus cannot be used.  

Answer:

Xylene boils at 138.5 °C, 740.97 R and 411.65 K

Explanation:

[tex]T_{\°C}=(T_{\°F}-32)\cdot \frac{5}{9}[/tex]

We know that temperature is 281.3 °F so in °C is:

[tex]\°C=(281.3-32)\cdot \frac{5}{9}= 138.5 \°C[/tex]

[tex]T_{R}=T_{\°F}+459.67\\T_{R}=281.3\°F+459.67=740.97 R[/tex]

[tex]T_{K}=(T_{\°F}+459.67)\cdot \frac{5}{9} \\T_{K}=(281.3 \°F+459.67)\cdot \frac{5}{9} \\T_{K}=411.65K[/tex]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × [tex]10^{-9}[/tex] m

wavelength λ  = 700 nm =  700 × [tex]10^{-9}[/tex] m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = [tex]\frac{1}{900*10^3 m}[/tex]

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

[tex]\frac{1}{900*10^3}[/tex] sinθ = 1 × 380 × [tex]10^{-9}[/tex]

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

[tex]\frac{1}{900*10^3}[/tex] sinθ = 1 × 700 × [tex]10^{-9}[/tex]

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )