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bixtya [17]
2 years ago
9

What is newton first law​

Physics
1 answer:
andrew11 [14]2 years ago
5 0

Answer:

Newtons first law states that:

<em>If</em><em> </em><em>a</em><em> </em><em>body</em><em> </em><em>i</em><em>s</em><em> </em><em>in</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em>,</em><em> </em><em>it</em><em> </em><em>remains</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>or</em><em> </em><em>at</em><em> </em><em>motion</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>straight</em><em> </em><em>line</em><em> </em><em>with</em><em> </em><em>constant</em><em> </em><em>speed</em><em> </em><em>until</em><em> </em><em>and</em><em> </em><em>unless</em><em> </em><em>and</em><em> </em><em>external</em><em> </em><em>unbalanced</em><em> </em><em>force</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>it</em><em>.</em>

<em>'</em><em>This</em><em> </em><em>law</em><em> </em><em>i</em><em>s</em><em> </em><em>also</em><em> </em><em>known</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>Inertia</em><em>.</em><em>'</em>

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slava [35]

Answer:

 θ =  3.19  arc second

Explanation:

The resolution of a telescope is given by the rayleigh criterion, which establishes that two objects are separated if the principal maximum of diffraction of one of them coincides with the first minimum of diffraction of the second object, based on this the solution is given by the first diffraction minimum, the a slit is

        a sin θ = m λ

with m = 1

in the case of circular apertures the equation must be found in polar coordinates, therefore a numerical constant is introduced

        a sin θ = 1.22 λ

Angles are measured in radians and in these experiments they are small

        sin θ = θ

       θ= 1.22  λ  / a

in this case a = 6.09 in, the wavelength is wrong = 550 10⁻⁹ m which is the maximum resolution of the human eye

l

et's reduce the magnitudes to the SI system

        d = 6.09‘  2.54 10⁻-2 m / 1 inch = 15.4686 10-2 m

let's calculate

       θ = 1.22 550 10-9 / 15.468 10-2

       θ = 15.5 10⁻⁶ rad

       rad = 2.06 105 s

       θ = 15.5 10⁻⁶ rad  2.06 105s/ 1 rad

       θ =  3.19   s

     

3 0
3 years ago
Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
nignag [31]

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

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          y = 39 2

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             y _total = 80 feet

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