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Nana76 [90]
3 years ago
14

What are two examples of population distribution?

Physics
1 answer:
Alex777 [14]3 years ago
7 0
Three basic types of population distribution within a regional range are (from top to bottom) uniform, random, and clumped.
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it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m
Jet001 [13]
We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values, 
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!
7 0
3 years ago
Read 2 more answers
A swimmer is swimming to the left with a speed of 1.0 m/s when she starts to speed up with constant acceleration. The swimmer re
kolezko [41]

Answer:

Correct answer: t = 2.86 seconds

Explanation:

We first use this formula

V² - V₀² = 2 a d    

where V is the final velocity (speed), V₀ the initial velocity (speed),

a the acceleration and d the distance.

We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

a = 0.525 m/s²

then we use this formula

V = V₀ + a t  => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds

t = 2.86 seconds

God is with you!!!

6 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
The base of a right prism is a rhombus with diagonals of 6 and 8. If the altitude of the prism is 12, what is the total surface
Agata [3.3K]

Answer:

total surface area is 432

Explanation:

Given data

base  = 6

diagonals = 8

altitude = 12

to find out

total surface area

solution

we know total surface area of prism is

total surface area = lateral surface area + 2base area  ..............1

so

first we calculate base perimeter i.e = 2 length + 2 width

so perimeter = 2(8) + 2(6) = 25

and area  = length * width = 8*6 = 48

so lateral surface area is perimeter * height i.e

lateral surface area = 28* 12

lateral surface area = 336

put this value in equation 1 we get

total surface area = lateral surface area + 2base area

total surface area = 336 + 2(48)

total surface area is 432

3 0
3 years ago
Read 2 more answers
A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in
Kruka [31]

Answer:

1020g

Explanation:

Volume of can=1100cm^3=1100\times 10^{-6}m^3

1cm^3=10^{-6}m^3

Mass of can=80g=\frac{80}{1000}=0.08kg

1Kg=1000g

Density of lead=11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3

By using 1g/cm^3=10^3kg/m^3

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can  and lead inside it they are floating in the water.

Buoyancy force =F_b=Weight of can+weight of lead

\rho_wV_cg=m_cg+m_lg

Where \rho_w=10^3kg/m^3=Density of water

m_c=Mass of can

m_l=Mass of lead

V_c=Volume of can

Substitute the values then we get

1000\times 1100\times 10^{-6}=0.08+m_l

1.1-0.08=m_l

m_l=1.02 kg=1.02\times 1000=1020g

1 kg=1000g

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0
3 years ago
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