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FrozenT [24]
3 years ago
6

While driving north at 21 m/s during a rainstorm you notice that the rain makes an angle of 36° with the vertical. while drivin

g back home moments later at the same speed but in the opposite direction, you see that the rain is falling straight down. from these observations, determine the speed and the angle of the raindrops relative to the ground?

Physics
1 answer:
denis23 [38]3 years ago
8 0
Let the velocity of the vehicle be 
V₁ = (21, 0)
Let the velocity of a raindrop be
V₂ = (a, b)

When returning, the raindrop is vertical. 
Therefore a = 21 m/s

When driving north, a raindrop makes an angle of 36° with the vertical, therefore
tan(36°) = a/b = 21/b
b = 21/tan(36°) = 28.9 m/s

The speed of a raindrop is
√(a² + b²) = √(441 + 835.44) = 35.73 m/s
The angle is 90-36 = 54° with the ground.

Answer: 35.7 m/s at 54° relative to the ground.

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A man of mass 85 kg runs up a flight of stairs of height 4.6 m in a time period
seraphim [82]

Explanation:

a) Power = work / time = force × distance / time

P = Fd/t

P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)

P ≈ 319 W

b) P = Fd/t

0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)

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3 years ago
A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at
Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0
3 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
3 years ago
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