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dybincka [34]
3 years ago
9

A physics student tests the theory of projectile motion by leaping off a 225 meter tall building. She runs off the building hori

zontally at 12.5 m/s. How far away from the base of the building should she place the safety net?
Physics
1 answer:
tamaranim1 [39]3 years ago
4 0
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,

   d = V₀t + 0.5gt²

Substituting the known values,

  225 = (0 m/s)(t) + (0.5)(9.8)(t²)

Simplifying,
 
   t = 6.776 s

To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.


   range = (12.5 m/s)(6.776 s)

   range = 84.7 m

<em>Answer: 84.7 m</em>

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2 years ago
events occur at the same place in an inertial reference frame S and are sepa- rated in time by an interval of 4 s. What is the s
lana [24]

The spatial separation between the 2 events is 13.416 × 10⁸ m

In space time-interval, the invariance of line element explains that if there are two inertial reference frames S and S', the spatial separation is invariant in all inference frames.

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\mathbf{\Big [ [\Delta x]^2 -[c^2 \Delta t ^2] \Big]_{frame \ 1} = \Big [[\Delta x]^2 -[c^2 \Delta t ^2] \Big]_{frame\   2}      }

\mathbf{[\Delta x_1]^2 -[c^2 \Delta t_1 ^2]  = [\Delta x_2]^2 -[c^2 \Delta t_2 ^2] }

where;

  • Δx₁ = 0  (since it occurs at same place)
  • Δt₁ = 4 s
  • Δt₂ = 6 s

\mathbf{[0]^2 -[c^2 (4) ^2]  = [\Delta x_2]^2 -[c^2 (6) ^2] }

\mathbf{ [\Delta x_2]^2    =36(c^2)  -  16(c^2)]}

\mathbf{ [\Delta x_2]^2    =20(c^2)}

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here;

  • c= speed of light = 3 × 10⁸

\mathbf{ \Delta x_2 = \sqrt{20} \times 3 \times 10^8}

\mathbf{ \Delta x_2 =13.416 \times 10^8 \ m}

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5 0
2 years ago
A 60-kg uniform board 2.4 m long is supported by a pivot 80 cm from the left end and by a scale at the right end (see the Fig. b
yKpoI14uk [10]

Answer:

a) x = 0.61 m

b) x = 1.4 m

Explanation:

Given data;

M = 60 kg,

m = 40 kg,

F considered to be a  reading

Torque of N about the given pivot point = F*(2.4 - 0.80) = 1.6 F (counterclockwise)

Torque of Mg about the given pivot Point = Mg*(1.2 - 0.80)

= 0.4*60*9.8 = 235.2 N-m (clockwise)

a) F = 100 N

note 1.6F = 160 < 235.2, so

mass m should be placed at left of pivot,

its torque = mg*(0.80 - x)

                = 40*9.8*(0.80 - x) = 392(0.80 -x)

160 + 392(0.8 - x) = 235.2

x = 0.61 m

b) N = 300 N

note 1.6F = 480 > 235.2, so

mass m should be placed at right of pivot,

its torque = mg*(x - 0.80) = 40*9.8*(x - 0.80) = 392(x -0.80)

480 = 392(x - 0.8) + 235.2

x = 1.4 m

3 0
3 years ago
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