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fiasKO [112]
3 years ago
6

What is a example of a real life scientific inquiry problem

Physics
1 answer:
Papessa [141]3 years ago
8 0
What is an example of how you can use scientific inquiry to solve a real life problem.
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A homeowner uses a snowblower to clear his driveway. knowing that the snow is discharged at an average angle of 40° with the hor
Nesterboy [21]

given that snow is projected at an angle of 40 degree

It range is given as a = 19 ft

a = 19 * 0.3048 = 5.8 m

now we can use the formula of horizontal range

R = \frac{v_o^2 sin2\theta}{g}

5.8 =\frac{ v_o^2 sin(2*40)}{9.8}

5.8 = \frac{v_o^2 * sin80}{9.8}

v_o^2 = 57.7

v_o = 7.6 m/s

<u>so its initial speed must be 7.6 m/s</u>

8 0
3 years ago
Greatest to least order
3241004551 [841]

Answer:

Explanation:

FBEDAC

8 0
2 years ago
A red ladybug appears in white light, in red light, and in blue light.
kirza4 [7]
<span>A red ladybug appears red in white light,red in red light, and black in blue light.</span>
3 0
3 years ago
Read 2 more answers
A bird carries a 25 g oyster to a height of 11m what is the gravitational potential energy of the oyster
garik1379 [7]

Given data:

Mass of oyster (m) = 25 g,

                              = 0.025 kg            (<em> since 1 Kg = 1000 g)</em>

Height of oyster (h) = 11 m,

determine gravitational potential energy (GPE) = ?

Gravitational potential energy is energy of an object possesses because of its position on the earth. The amount of gravitational potential energy of an object on earth depends on its<em> mass</em> and<em> height </em>from the earth.  

Mathematically,

        GPE = m. g. h <em>Joules</em>

<em>         Where, </em>

<em>               </em>GPE =<em> </em>gravitational potential energy,

<em>         </em>          m = mass of object in Kg,

                  g = gravitational force in m/s²,     g = 9.8 m/s²

                  h = height of the object in meters,

     Now substituting all the values in the equation GPE

           GPE = 0.025 Kg × 9.8 m/s² × 11 m

                   = 2.695 J

Gravitational potential energy of the oyster is 2.695 J


7 0
3 years ago
A monochromatic light of wavelength 506 nm from a distant source is incident on a single slit 0.075 mm wide. Diffraction pattern
sladkih [1.3K]

Answer:

first minimum distance from the center = 0.417 m m

Explanation:

given,

wavelength (λ) = 506 nm

d = 0.075 mm  = 0.000075 m

tanθ = x /2

d sinθ = m λ

m=1 ( 1 st minimum  )  

0.000075 × sin θ = 506 × 10⁻⁹

sin θ =\dfrac{506 × 10⁻⁹}{0.000075}

sin θ = 0.00675

θ = sin⁻¹(0.00675)

θ = 0.00675

x =  0.000417 m

hence, the first minimum distance from the center = 0.417 m m

5 0
3 years ago
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