A boy weighing 70 kg is standing on the sand. Calculate the pressure on if you are resting on your feet, whose surfaces add up to 0.035 m2 and if you are resting on diving fins of 0.300 × 0.42 m each

Answer:

2 oxygen 1 carbon

Explanation:

Answer:

2 oxygen, 1 carbon

Explanation:

Answer:

A) Change in potential energy is approximately 766 KJ

B) The minimum work required by the hiker is 765 KJ

C). Yes, the actual work can be greater than this

Explanation:

A) The hiker's potential energy can be calculated at the two different elevations, and then subtracted.

P.E at 1280 m = m X g X h = 60.5kg X 9.81 m/s2 X 1280 = 759, 686 J

P.E at 2570 m = m X g X h = 60.5kg X 9.81 m/s2 X 2570 = 1, 525, 307.85 J

Change in P.E = 1, 525, 307.85 J -759, 686 J = 765, 621.85 J = 766 KJ

B) Work which was done = Force X distance moved.

The force can be got from the effect of gravity on the hiker's mass = 60.5 kg X 9.81 = 593.5 Newton.

The distance moved can be obtained by subtracting the two elevations = 2570 -1280 = 1290 m

Work done = 593 X 1290 = 765, 615 J

The minimum work required by the hiker is 765 KJ

C) Yes, the actual work can be greater than this. This is because we can now put into consideration the fact that the hiker is climbing up the elevation against the force of gravity. This will mean that the hiker is actually doing more work than if he covered the distance on flat terrain,

Answer: 90 m/s

Explanation:

The formula for speed is distance/time. The distance is 180 m and the time it took for the ball to travel is 2 s.

180 m/2 s = 90 m/s

Answer:

The wavelength is  [tex]\lambda = 10.01 *10^{-15} \ m[/tex]

Explanation:

From the question we are told that

  The  speed is  [tex]v = 0.132 c[/tex]

Where c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

Generally the wavelength is mathematically represented as

    [tex]\lambda = \frac{h}{m* v }[/tex]

where m is the mass of the proton with the value  [tex]m = 1.6726 ^{-27} \ kg[/tex]

           h is the Planck's constant with value  [tex]h = 6.626 *10^{-34} \ J\cdot s[/tex]

=>    [tex]\lambda = \frac{6.626 *10^{-34}}{1.6726 *10^{-34}* 0.132*3.0*10^8 }[/tex]

=>   [tex]\lambda = 10.01 *10^{-15} \ m[/tex]

Answer:

Explanation:

Given,

F = 100 , M = 25

Now, let's find the value of A

[tex] \sf{F = MA}[/tex]

plug the values

⇒[tex] \sf{100 = 25 A}[/tex]

Swap the sides of the equation

⇒[tex] \sf{25A = 100}[/tex]

Divide both sides of the equation by 25

⇒[tex] \sf{ \frac{25A}{25} = \frac{100}{25} }[/tex]

Calculate

⇒[tex] \sf{A = 4}[/tex]

Hope I helped!

Best regards!!

Given:-

To find out:-

Calculate the acceleration, a ?

Formula applied:-

F = m × a

Solution:-

We know,

[tex] \sf{F = m × a}[/tex]

Substituting the values of mass and acceleration,we get

[tex] \sf\implies \: 100 = 25 \times a[/tex]

[tex] \sf \implies a = \cancel \dfrac{100}{25} [/tex]

[tex] \sf \implies a = 4 \: ms {}^{ - 1} [/tex]

Answer:

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Answer:

a = 264.14 m/s²

Explanation:

From the question;

Initial velocity; u will be 0 m/s since the ball will start from rest.

Final velocity; v = 43 m/s

distance covered by the motion; s = 3.5m

To get the acceleration, we will make use of Newton's third equation of motion which is;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging in the relevant values to give;

a = (43² - 0)/(2 × 3.5)

a = 264.14 m/s²

The average acceleration of the ball during the throwing motion is 265.14m/s².

In order to get the acceleration, the Newton's third law of motion will be used. This will be:

v² = u² + 2as

We'll make a to be the subject of the formula and this will be:

a = (v² - u²) / 2s

We'll plug in the value into the equation and this will be:

a = (43² - 0) / (2 × 3.5)

a = 1849 / 7

= 264.14 m/s²

Therefore, the acceleration is 265.14m/s.

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Answer:

Explanation:

SI unit of

length=meter

volume =dm^3

mass =kilogram

time=second

energy= joule

temperature =kelvin

Answer:

89.6 cm

Explanation:

From the question,

Volume of the rectangular object = Mass/Density.

V = m/D.................. Equation 1

Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³

Substitute into equation 1

V = 1.278/4980

V = 2.57×10⁻⁴ m³.

But,

V = lwh............... Equation 2

Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.

make h the subject of the equation

h = V/lw........... Equation 3

Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.

Substitute into equation 3

h = 2.57×10⁻⁴/(0.047×0.061)

h = 0.896 m

h = 89.6 cm

Answer:

(a) -136.8 Ns.

(b) -1.135 N

Explanation:

(a)

Impulse: This can be defined as the change in momentum.

From the question,

I = mv-mu.................. Equation 1

Where I = impulse, m = mass of the hammer, v = final velocity, u = initial velocity.

Given: m = 18 kg, u = 7.6 m/s, v = 0 m/s (to rest)

Substitute these values into equation 1

I = 18(0)-18(7.6)

I = -136.8 Ns.

(b)

Average force = It.............. Equation 2

Where t = time.

Given: t = 8.3 ms = 0.0083 s.

Average force = -136.8(0.0083)

Average force = -1.135 N

Although they're all 'close', none of the planets orbits in the same plane as any other planet.  They're all in slightly different planes.

The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit).  But Pluto is officially not a planet, so I don't think it's a good answer.

The next greatest inclination compared to Earth's orbit is Mercury.  That one is about 7 degrees.

The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.

Answer:

[tex] \boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise} [/tex]

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ([tex] \sf \eta [/tex])

Explanation:

[tex] \boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}[/tex]

[tex] \sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v} [/tex]

Substituting values of r, ρ, σ, v & g in the equation:

[tex] \sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25} [/tex]

[tex]\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25} [/tex]

[tex]\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25} [/tex]

[tex]\sf \implies \eta = \frac{2}{9} \times 64.7196[/tex]

[tex]\sf \implies \eta = 2 \times 7.191[/tex]

[tex]\sf \implies \eta = 14.382 \: poise[/tex]

Answer:

7 degree

Explanation:

given data

width = 4.1 inches

length = 35.4 inches

solution

we consider as per fig

O is mid point of BC

so  OB  = 2.05 inches

and

AB = [tex]\sqrt{OB^2 + OA^2}[/tex]

AB = [tex]\sqrt{2.05^2 + 35.4^2}[/tex]

AB = 35.078 inches

so

[tex]sin \frac{\alpha }{2} = \frac{OB}{AB}[/tex]

[tex]sin \frac{\alpha }{2} = \frac{2.05}{35.078}[/tex]

[tex]\alpha = 7 degree[/tex]

Answer

Answer:

a. a photon of energy 5 is absorbed

Explanation:

Because when an electron in a lower energy state absorbs energy, in form of photons it moves to higher energy stage in this case 5 photons because it moved from 5 to 10

1/200 that should get your

answer

Answer:

Explanation:

Given the following data

initial temperature T1= 150 °C

final temperature T2= 250 °C

specific heat of gold c= 0.13 J/g°C

mass of gold m= 75.0 grams

we can use the expression stated below to solve for the quantity of heat

[tex]Q= mc(T2-T1)---------1[/tex]

Substituting our known data into the expression we can solve for the value of Q

[tex]Q= 75*0.13(250-150)---------1\\\\Q= 75*0.13(100)\\\\Q= 975 Joules[/tex]

The quantity of heat need to raise the temperature from 150°C to 250°C  is 975 J

Answer:

B. 980 joules

Explanation:

Answer:

  b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

               fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

               w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

               w₀² = k / m

               w² = w₀² - b² / 4m²

               b² = (w₀² -w²) 4 m²

Let's find the angular velocities

             w₀² = 5 / 0.1

             w₀² = 50

             w = 2π f

             w = 2π 1

             w = 6.2832 rad / s

we subtitute

               b² = (50 - 6.2832²) 4 0.1²

               b = √ 0.42086

                b = 0.6487 kg / s

The coefficient friction of the mass during the measurement is 0.648 kg/s.

The given parameters;

During oscillatory motion, friction is directly proportional to speed.

[tex]F_k = -vb[/tex]

where;

The angular velocity is given as;

[tex]\omega ^2 = \frac{k}{m} - \frac{b^2}{4m^2} \\\\\omega ^2 = \omega _0^2 - \frac{b^2}{4m^2}\ \ ---\ (1)[/tex]

From the equation above, we will have the following;

[tex]\omega_0^2 = \frac{k}{m} \\\\\omega_0^2 = \frac{5}{0.1} \\\\\omega_0^2 = 50[/tex]

Also, the instantaneous angular speed is calculated as;

[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1\\\\\omega = 2\pi\\\\\omega = 6.284 \ rad/s[/tex]

From equation (1), the coefficient of friction is calculated as follows;

[tex]\omega ^2 = \omega ^2_0 - \frac{b^2}{4m^2} \\\\ \frac{b^2}{4m^2} = \omega ^2_0 - \omega ^2 \\\\b^2 = 4m^2( \omega ^2_0 - \omega ^2)\\\\b= \sqrt{ 4m^2( \omega ^2_0 - \omega ^2)}\\\\b = \sqrt{ 4\times 0.1^2\times ( 50 - 6.284^2)}\\\\b = 0.648 \ \ kg/s[/tex]

Thus, the coefficient friction of the mass during the measurement is 0.648 kg/s.

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Answer:

C. All waves have the same speed.

Explanation:

Wave equation is given as;

V = fλ

where;

V is the speed of the wave

f is the frequency of the wave

λ is the wavelength

The speed of the wave depends on both wavelength and frequency

The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.

Therefore, the correct option is "C"

C. All waves have the same speed.

Answer and Explanation: Centripetal Acceleration is the change in velocity caused by a circular motion. It is calculated as:

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

v is linear speed

r is radius of the curve the object in traveling along

For its first lap:

[tex]a_{c}_{1}=\frac{v_{i}^{2}}{R}[/tex]

After a while:

[tex]a_{c}_{2}=\frac{(4v_{i})^{2}}{R}[/tex]

[tex]a_{c}_{2}=\frac{16v_{i}^{2}}{R}[/tex]

Comparing accelerations:

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=16[/tex]

[tex]a_{c}_{2}=16a_{c}_{1}[/tex]

With linear speed 4 times faster, centripetal acceleration is 16 times greater.

Answer:

a = (v² – v₀²)/ 2(s – s₀)

Explanation:

v² = v₀² + 2a (s – s₀)

We can make 'a' the subject of the above expression as follow:

v² = v₀² + 2a (s – s₀)

Subtract v₀² from both side

v² – v₀² = v₀² + 2a (s – s₀) – v₀²

v² – v₀² = v₀² – v₀² + 2a (s – s₀)

v² – v₀² = 2a (s – s₀)

Divide both side by (s – s₀)

(v² – v₀²)/ (s – s₀) = 2a

Divide both side by 2

(v² – v₀²)/ (s – s₀) ÷ 2 = a

(v² – v₀²)/ (s – s₀) × 1/2 = a

(v² – v₀²)/ 2(s – s₀) = a

a = (v² – v₀²)/ 2(s – s₀)

Answer:

A) -40° C and -40° F

B) 574.25° K and 574.25° F

Explanation:

see attachment for calculation and explanation

Answer:

Option (A) : Zero

Explanation:

Electric field Intensity inside a metallic body is always ZERO.

Answer:

The average energy delivered to a surface is 19.116 W/m².

Explanation:

Given;

maximum electric field, E₀ = 120 v/m

The average energy delivered by the wave to a surface is given by

[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2}[/tex]

where;

c is the speed of light, = 3 x 10⁸ m/s

ε₀ is the permittivity of free space = 8.85 x 10⁻¹² c²/Nm²

[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2} \\\\I_{avg} = \frac{(3*10^8)(8.85*10^{-12})( 120)^2}{2}\\\\ I_{avg} =19.116 \ W/m^2[/tex]

Therefore, the average energy delivered to a surface is 19.116 W/m².

Answer:

The  ratio is  [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]

Explanation:

From the question we are told that

   The first radius is [tex]R_1[/tex]

    The  second radius is  [tex]R_2[/tex]

Generally the angular speed of the turntable is mathematically represented as

       [tex]w_1 = \frac{ v_k }{R_1 }[/tex]

Generally the angular speed of the rubber roller is mathematically represented as

        [tex]w_2 = \frac{ v_k }{R_2 }[/tex]

Where [tex]v_k[/tex] is the velocity of both turntable and  rubber roller

So

    [tex]\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }[/tex]

    [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]

Answer:

a) 800N × 20 cm = 1100N × x cm

16000= 1100x

x= 14.5

therefore it must be placed on the (50 + 14.5)cm mark

= 64.5 cm mark

b) 800N × 25 cm = (1100N × 10 cm)+(450N × x cm)

20000 = 11000 + 450x

450x = 9000

x = 20 cm

therefore it must be placed on the (50 + 20)cm mark

= 70 cm mark

a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.

b) The distance at which mass m₃(45 g) should be suspended is 70 cm.

According to the Principle of Moments, when a body is balanced or is at equilibrium, the total clockwise and anticlockwise moments about a given point are equal.

a) m₁ = 80 g

m₂ = 110 g

r₁ = 30 cm

According to the Principle of Moments,

m₁r₁ = m₂r₂

Therefore, the distance,

r₂ = m₁r₁/m₂

r₂ = 80 x 20/110

r₂ = 14.54 cm

So, the distance at which mass m₂ should be suspended is,

r' = 50 + 14.54

r' = 64.54 cm

b) m₁ = 80 g

m₂ = 110 g

m₃ = 45 g

r₁ = 25 cm

r₂ = 60 cm

According to the Principle of Moments,

m₁r₁ = m₂r₂ + m₃r₃

80 x 25 = (110 x 10) + (45 x r₃)

45 x r₃ = 2000 - 1100

r₃ = 900/45

r₃ = 20 cm

So, the distance at which mass m₃ should be suspended is,

r' = 50 + 20

r' = 70 cm.

Hence,

a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.

b) The distance at which mass m₃(45 g) should be suspended is 70 cm.

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Answer:

1999.46 mm

45.59 s

Explanation:

given that

cylindrical water tank with diameter, D = 3 m

Height of the tank above the ground, h = 2 m

Depth of the water in the tank, d = 2 m

Diameter of hole, d = 0.420 cm

We start by calculating the volume of water in the tank, which is given as

Volume = πr²h

V = (πD²)/4 * h

V = (3.142 * 3²)/4 * 2

V = 28.278/4 * 2

V = 7.07 * 2

V = 14.14 m³

If 1.0 gal of water is equal to 0.0038m³, then

1 gal is 0.0038 = A * h

the area of the tank is 7.07 m²

therefore, 0.0038 = 7.07 * h

h₁ =0.00054 m = 0.54 mm is the height of water that flow out

the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m

b)

Like we stated earlier, 1.0 gal of water is 0.0038m³

to solve this we use the formula

Q = Cd * A * √2gH

where Cd is a discharge coefficient, and is given by 0.9 for water

A is the area of the small hole

A = (πD²)/4

A = (π * 0.0042²)/4

A = 5.54*10^-5 / 4

A = 1.39*10^-5 m²

H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m

g = 9.8 m/s²

Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958

Q = 1.251*10^-5 * 6.25

Q = 7.82*10^-5 m³/s

Q = V/t

t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s

t = 45.59 s

Answer:

a). C = b/2   and C = b/4

b).  [tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]

c). m = 63.4 kg (approx.)

Explanation:

Ex. 2.4.4

The total force acting on mass m is [tex]$ F = F_{spring }= -kx $[/tex] , where x is the displacement from the equilibrium position.

The equation of motion is [tex]$ m {\overset{..}x} + kx = 0 $[/tex]

or [tex]$ {\overset{..}x}+ \frac{k}{m}x=0 $[/tex]     or   [tex]$ {\overset{..}x} + w_0^2 x = 0 $[/tex] ,    where [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex]  

The solution is [tex]$ x = A \cos (w_0t + \phi) $[/tex] ,  where A and Ф are constants.

A is amplitude of  motion

[tex]$ w_0$[/tex] is the angular frequency of motion

Ф is the phase angle.

Now, [tex]$ w_0 = 2 \pi f_0 = \sqrt{k/m} $[/tex]

or [tex]$ m = \frac{k}{4\pi f_0^2} $[/tex]

Given [tex]$ f_0 = 0.8 Hz , k = 4 N/m $[/tex]

a).  [tex]$ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$[/tex]

b). [tex]$ w_0^2 = k/m $[/tex]  

or  [tex]$ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $[/tex]

Ex. 2.4.5

a). Total force acting on the mass m is [tex]$F = F_{spring}+f $[/tex]

                                                                    [tex]$ = -kx-bv $[/tex]

    The equation of motion is [tex]$ m {\overset{..}x}= -kx-b{\overset{.}x} $[/tex]

    or [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex] ,  angular frequency of the undamped oscillation.

   γ = b/2m  is called the damping coefficient (γ=C)

   [tex]$ k = m w_0^2 = 4 \pi^2 m f_0^2 $[/tex]

   for 1 kg weight (= 9.8 N), [tex]$ f_0$[/tex] = 1.1 Hz

    k = 4 x (3.14)² x (9.8) x 1.1²  = 4.6 x 10² N/m

  For 2 kg weight (= 19.6 N), [tex]$ f_0$[/tex] = 0.8 Hz

    k = 4 x 9.8596 x 2 x 9.8 x 0.8²  = 5 x [tex]$ 10^7$[/tex] N/m

   [tex]$ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $[/tex]

or [tex]$ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $[/tex]

γ = b/2 (for 1 kg)  and   γ = b/4 (for 2 kg)

C = b/2   and C = b/4

b). [tex]$ w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $[/tex]

 For two particle problem,

   [tex]$ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $[/tex]

[tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]

where, μ is the reduced mass.

This time period is same for both the particles.

c). [tex]$ m =\frac{k}{4 \pi^2 f_0^2}$[/tex]

        [tex]$ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg $[/tex]  ( approx.)

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

Answer:

a

  [tex]v _{max } = 6.82 \ m/s[/tex]

b

 [tex]a_{max} = 37489.5 \ m/s^2[/tex]

Explanation:

From the question we are told that

   The amplitude is [tex]A = 1.24 \ mm = 1.24 * 10^{-3} \ m[/tex]

   The frequency is  [tex]f = 875 \ Hz[/tex]

Generally the maximum speed is mathematically represented as

           [tex]v _{max } = A * 2 * \pi * f[/tex]

=>        [tex]v _{max } = 1.24*10^{-3} * 2 * 3.142 * 875[/tex]

=>        [tex]v _{max } = 6.82 \ m/s[/tex]

Generally the maximum acceleration is mathematically represented as

       [tex]a_{max} = A * (2 * \pi * f)[/tex]

=>      [tex]a_{max} = 1.24*10^{-3} * (2 * 3.142 * 875 )^2[/tex]

=>       [tex]a_{max} = 37489.5 \ m/s^2[/tex]

Answer:

0.276

0.9

0.756

Explanation:

Given that

Wavelength of the light, λ = 588 nm

Distance from the slit to the screen, L = 2.7 m

Width of the slit, a = 0.0351 mm

a point on the screen, y = 1.3 cm = 0.013 m

Sinθ = y/L

Sinθ = 0.013/2.7

sinθ = 0.0081

θ = sin^-1 0.00481

θ = 0.276°

α = (π.a.sinθ)/λ

α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9

α = 5.3*10^-7 / 588*10^-9

α = 0.9 rad

I/i(m) = ((sinα)/α)²

I/I(m) = ((sin 0.9) / 0.9)²

I/I(m) = (0.783/0.9)²

I/I(m) = 0.87²

I/I(m) = 0.756

Note, our calculator has to be set in Rad instead of degree for part C, to get the answer