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Katena32 [7]
3 years ago
14

A box of books weighing 305 N is shoved

Physics
1 answer:
Anna [14]3 years ago
5 0

Answer:

3.1 s

Explanation:

from the question we are given the the following:

Weight of books (Fb) = 305 N

Push force (Fp) = 516 N

distance (s) = 5.82 m

Angle of force exerted = 32 degrees

acceleration due to gravity (g) = 9.81 m/s^{2}

coefficient of friction Uk= 0.59

time (t) = ?

mass of the book (m) = weight / (g) =  305 / 9.8 = 31 kg

lets first get the net force from the summation of vertical forces

Fnet = Fpush + Fgravity

F net = 516 + 305sin32

F net = 677.6 N

now lets get the acceleration from the summation of the horizontal forces

Fpcos32 - friction force = m x a

Fpcos32 - (Uk x F net) = m x a

516cos32 - (0.59 x 677.6) = 31 x a

37.8 = 31a

a =  1.22m/s^{2}

now that we have our acceleration we can get the time from the equation of motion

s = ut + o.5at^{2}

u ( initial velocity ) = 0 because the box was initially at rest

5.82 = (0 x t ) + (0.5 x 1.22 x t^{2})

5.82 = 0.61t^{2}

t = 3.1 s

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Answer:

Only  two diffraction orders +1 and -1 are possible

Explanation:

The diffraction net is a system with many uniformly spaced lines, which is described by the expression

      d sin θ = m λ               m = 1, 2, 3,…

Where d is the spread between the lines, λ the wavelength and m is an integer that corresponds to the order of diffraction, all wavelengths

For this case the spacing is

          d = 775 nm

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With this data we can calculate how many diffraction orders are visible

         θ = sin⁻¹ (m λ / d)

m = 1

         λ1 = 589.00 nm

         θ1 = sin⁻¹ (1  589/775)

         θ1 = 49.46 °

         λ2 = 589.59 nm

        θ2 = sin⁻¹ ((1  589.59 / 775)

        θ2 = 49.53 °

m = 2

        λ1 = 589.00 nm

       θ1 = sin⁻¹ (2  589/775)

       θ1 = sin⁻¹ (1.5)

This order is not possible because the sin cannot be greater than 1

m = -1

       λ1 = 589 nm

       θ1 = sin⁻¹ (-1 589/775)

       θ1 = -49.46

       θ= sin⁻¹ ( -1 589.59/775)        

       λ2 = 589.59 nm

       θ2 = -49.53 °

m = -2

       λ1 = 589 nm

       θ1 = sin-1 (-2 589/775)

This order does not occur because the breast cannot be greater than 1

Only  two diffraction orders +1 and -1 are possible

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