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dybincka [34]
2 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]2 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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In the voltage multiplier experiment, why not use a zener diode?​
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2 years ago
The following forces act on an object: 10 N north, 7 N south, and 4 N east. What is the magnitude of the net force?
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Given data;

 Fn = 10 N

 Fs = 7 N

 Fe = 4 N

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6 0
3 years ago
A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.
ohaa [14]

Answer:

speed of a particle  = 2.31 × 10^{6} m/s

energy of proton required = 27.77 KeV

energy of electron required =  15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and  energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB    ..............1

and force due to electric field E = qE   .....................2

so without deflection force due to magnetic field  = force due to electric field  

so here qvB = qE

and V = \frac{E}{B}    ...................3

put here value

V =  \frac{0.51*10^6}{0.22}

speed of a particle  = 2.31 × 10^{6} m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × \frac{V^2}{2}

put here value

energy of proton required = 1.67 × 10^{-27} × \frac{(2.31*10^6)^2}{2}

energy of proton required = 4.45 × 10^{-15} J

energy of proton required = 4.45 × 10^{-15} J ÷ (1.602 × 10^{-19}

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × \frac{V^2}{2}

put here value

energy of electron required = 9.11 × 10^{-31} × \frac{(2.31*10^6)^2}{2}

energy of electron required =  24.305× 10^{-19} J

energy of electron required =  24.305 × 10^{-19} J ÷ (1.602 × 10^{-19}  

energy of electron required =  15.171 eV

5 0
3 years ago
Read 2 more answers
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