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zheka24 [161]
3 years ago
12

Estimate the number of Ping-Pong balls that can be packed into an average size room (without crushing them). Given that Ping-Pon

g ball has a radius of about 1.5 cm and assume that a typical room has dimensions12 ft ×18 ft ×9 ft.
Physics
1 answer:
scoray [572]3 years ago
7 0

Answer: 3,893,845.918 Ping-Pong balls

Explanation:

The volume of an average room is:

V_{room}=(length)(width)(height) (1)

V_{room}=(12 ft)(18 ft)(9 ft)=1944 ft^{3} (2)

Now let’s transform this V_{room} to units of cm^{3}, knowing 1 ft=30.48 cm:

V_{room}=1944 ft^{3}\frac{{(30.48 cm)}^{3}}{1ft^{3}}=55,047,949.78 cm^{3} (3)

On the other hand, we have Ping-Pong balls with a radius r=1.5 cm, and their volume is given by:

V_{balls}=\frac{4}{3} \pi r^{3} (4)

V_{balls}=\frac{4}{3} \pi (1.5 cm)^{3} (5)

V_{balls}=14.137 cm^{3} (6)

Now, the number n of Ping-Pong balls that can be packed into the room is:

n=\frac{V_{room}}{V_{balls}} (7)

n=\frac{55,047,949.78 cm^{3}}{14.137 cm^{3}}  (8)

n=3,893,845.918 This is the number of Ping-Pong balls that can be packed into an average size room

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Degger [83]

Answer:

1

  The mass of the Potassium-40 is  m_{40}} = 2.88*10^{-6} kg

2

  The Dose per year in Sieverts is   Dose_s = 26.4 *10^{-10}

Explanation:

From the question we are told that

   The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

    Their abundance is 93.26%, 0.012% and 6.728%

   The mass of potassium contained in human body is  m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg per kg of the body

    The mass of the first body is  m_1 = 80 \ kg

Now the mass of  potassium  in this body is mathematically evaluated as

       m_p =  m * m_1

substituting value

       m_p =  80  * 0.0003

      m_p  =0.024 kg

The amount of Potassium-40 present  is mathematically evaluated as

      m_{40}} =0.012% * 0.024

      m_{40}} = \frac{0.012}{100}  * 0.024

      m_{40}} = 2.88*10^{-6} kg

The dose of energy absorbed per year is mathematically represented as

          Dose  = \frac{E}{m_1}

Where E is the energy absorbed which is given as E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}

    Substituting value

            Dose  = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}

            Dose  = 22*10^{-10} J/kg

The Dose in Sieverts is evaluated as

       Dose_s = REB * Dose

       Dose_s = 1.2 * 22*10^{-10}

       Dose_s = 26.4 *10^{-10}

             

3 0
3 years ago
Consider a neutron star of radius 10 km that spins with a period of 0.8 seconds. Imagine a person is standing at the equator of
Arada [10]

Answer:

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Explanation:

Given that,

The radius of the neutron star, r = 10 Km

                                                     = 10,000 m

The time period of the neutron star, T = 0.8 s

The centripetal acceleration is given by the formula,

                                  a = v²/r

The linear velocity is given by the relation,

                                    v = rω

The time taken to complete one complete rotation is given by the relation

                                   T = 2π /ω

Where,

                                    ω = 2π / T

Substituting v and ω into the equation for centripetal acceleration. It becomes

                                    a = 4π²r/T²

Substituting the given values in the above equation

                                      a = 4π² x 10000 / 0.8²

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Hence, the centripetal acceleration of this person is, a = 616850.28 m/s²

8 0
3 years ago
Suppose you harvested a large crop of grain this year. The grain is now in a pile and takes up an area of 50 m2 and is 2 m high.
Simora [160]
I'm pretty sure that the silo has the shape of a cylinder.
So its volume is

          (area of the base) x (height)  =  (15 m²) x (3m) = 45 m³ .


I'm going to assume that the pile has taken the shape of a cone
on the ground, so its volume is

                 (1/3) x (area of the base) x (height)

              = (1/3) x (50m²) x (2m) = 33-1/3 m³  .

The whole answer to the question rests on the exact shape of
the pile of grain on the ground.  It could really help if somebody
could take a picture of the pile, so we could study the picture
here in Chicago, and estimate how closely the shape of the
pile resembles a cone.

If my assumption is valid, and the volume of grain in the pile
can be accurately calculated as the volume of a cone with the
same dimensions as the pile, then the grain will all fit in the silo
with no problem.  The silo will still be only 74% full, and it'll still
have room for another  11-2/3 m³  of grain.  

But if, say, the grain got wet and sticky as it was being poured
onto the pile, and the pile took the shape of a huge brick, then
its volume is  (area of the base) x (height) = 100 m³ .
If that's the shape of the pile, then only 45% of it will fit into the
silo.  The silo will be full and there'll still be 55 m³ of grain left
out on the ground to rot.

With the information that they sent to us up here in Chicago,
we simply don't know.

4 0
3 years ago
What is the definition of efficent
nika2105 [10]

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8 0
3 years ago
Where are bar magnets the strongest?
Alecsey [184]

Answer:

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Explanation:

3 0
3 years ago
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