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Andru [333]
3 years ago
12

A 10 g dime and a 200 g gold coin are dropped from the Empire State Building in

Physics
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

Explanation:

The both hit the ground at the same time. There is no air resistance and that includes the factors brought up in B.

C and D are both eliminated. If you dropped an 800 kg honda Fit it would hit the ground at the same time as the two coins. When you look at the formulas used to derive time, they look like

d = vi * t + 1/2 a* t^2

vi = 0

the distance to fell is the same for all three masses.

a is the same for all 3 masses.

So when you solve for t, there is no mention of the mass of the object falling. Therefore t is dependent only on the height.

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3 years ago
A car initially traveling at 15.0 m/s accelerates at a constant rate of 4.50 m/s2 over a distance of 45.0 m. How long does it ta
anygoal [31]

Answer:

2.24 seconds

Explanation:

xf = xo + vo t + 1/2 at^2

45 = 0 + 15 t  + 1/2 (4.5) t^2

   2.25 t^2 + 15t - 45 = 0              Quadratic formula shows  t = 2.24 seconds

4 0
2 years ago
A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?
givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

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8 0
3 years ago
2. A multipurpose wiring tool can be used to
Alex_Xolod [135]

Answer:

Option C

Crimp terminals

Explanation:

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5 0
4 years ago
An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
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