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4 years ago

What came first the chicken or the egg? (this is just a point give away)

Physics

2 answers:

Molodets [167]4 years ago

8 0

The chicken comes first

sweet-ann [11.9K]4 years ago

8 0

chicken. hop i helpd

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Project managers and their teams must keep in mind the effects of any project on the interests and needs of the entire system or

amm1812

Answer:True

Explanation:

True

Even though it is easier to concentrate on the urgent and sometimes specific issues of a given project, project managers and all other personnel must bear in mind the influence of any project on the wants and needs of the entire system or organization.

A Project manager must perform following duties

Planning Project resources
Lead the team
Time management
Budget
Documentation

6 0

3 years ago

Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite d

Leto [7]

Answer:

(a) $v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) Kinetic Energy of planet with mass m₁, is KE₁ = 1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ = 2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

$F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}$

Energy is given by

$Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ = m₁·v₁/m₂

At equilibrium, we have;

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$ which gives

$2G{m_{1}m_{2}}= d m_{1} v^2_1+ dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)$

multiplying both sides by m₂/m₁, we have

$2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)$

Such that v₁ = $\sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }$

$v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

Similarly, with v₁ = m₂·v₂/m₁, we have

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow 2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)$

From which we have;

$2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2$ and

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

The relative velocity = v₁ + v₂ = $v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

v₁ + v₂ = $(m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) The kinetic energy KE = $\frac{1}{2}mv^2$

$KE_1= \frac{1}{2} m_{1} v^2_1 \, \, \, KE_2= \frac{1}{2} m_{2} v^2_2$

Just before they collide, d = r₁ + r₂ = 3×10⁶+5×10⁶ = 8×10⁶ m

$v_1 = 8\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ = 10333.696 m/s

$v_2 = 2\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ =2583.424 m/s

KE₁ = 0.5×2.0×10²⁴× 10333.696² = 1.068×10³² J

KE₂ = 0.5×8.0×10²⁴× 2583.424² = 2.6696×10³¹ J.

7 0

3 years ago

What does the term electron orbital describe? What does the term electron orbital describe? An electron orbital describes a thre

Yuliya22 [10]

Answer:

An electron orbital describes a three-dimensional space where an electron can be found 90% of the time.

Explanation:

According to Heisenberg's theory we cannot observe the position and velocity of an electron in an orbit, but if they were around the nucleus (in orbit), it would be possible to know its velocity and position, which would be contrary to the principle of Heisenberg So we can say that no electron revolves around a certain orbit around the nucleus, so we can only predict if the electron will be in the right position at the right time.

From there we find two definitions for electron orbital let's see:

Orbital is considered the region of space, where each electron spends most of its time.

Orbital is considered the region of space that is most likely to find an electron.

3 0

3 years ago

A railroad car with a mass of 1.94 ✕ 104 kg moving at 3.28 m/s joins with two railroad cars already joined together, each with t

kap26 [50]

The total cost is going to be 536 kg

7 0

4 years ago

Convert 1gcm-3 into kgm-3?

Rom4ik [11]

When you have to convert a quantity you need to multiply by a fraction whose value is 1 but has different units of measurements.

In your case you have: $\frac{1 g}{cm^{3} }$

You will have two fractions: one that transforms grams into kilograms and one <span>that transforms</span> cm³ into m³. You need to position the quantities in such a way you can eliminate the original ones, therefore:

$\frac{1 g}{1 cm^{3} } \frac{10^{6} cm^{3} }{1 m^{3} } \frac{1 kg}{10^{3} g}$

Now, you can make the calculations and you get:

$\frac{1g}{cm^{3} }$ = $\frac{ 10^{3} kg }{m^{3} }$

Therefore your answer is: 1000kg/m³

3 0

3 years ago