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tresset_1 [31]
3 years ago
11

Find the mean and median of the following data set: 98, 87, 92, 79, 65, 91, 80, 92, 85, 86.

Physics
1 answer:
Naddik [55]3 years ago
8 0
Mean is 95
Median is 96.5
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Answer :

The wavelength would be 410 m/s
3 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
4 years ago
A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf 2.5m off the ground. How much gravitati
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Gravitational potential energy  = 

                        (mass)  x  (gravity)  x  (height)

                =    (5.8 kg)  x  (9.8 m/s²)  x  (2.5 m)

                =           142.1 Joules         (C)
5 0
3 years ago
An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal
borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is <u>20 cm.</u>

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u>I hope this helps!</u></em>

8 0
3 years ago
Read 2 more answers
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
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