HELP ME ASAP!!!!100 points!!!!!How do you get your fingers in the Labia?

Physics

2 answers:

Korvikt [17]3 years ago

5 0

genitals. And p,ease put appropiate questions.

Nina [5.8K]3 years ago

4 0

You literally just put your fingers in your genitals? is this for a sex ed course...?

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Define and explain the Police Power

Allushta [10]

Answer:

Police powers are the fundamental ability of a government to enact laws to coerce its citizenry for the public good, although the term eludes an exact definition. The term does not directly relate to the common connotation of police as officers charged with maintaining public order, but rather to broad governmental regulatory power. Berman v. Parker, a 1954 U.S. Supreme Court case, stated that “public safety, public health, morality, peace and quiet, law and order. . . are some of the more conspicuous examples of the traditional application of the police power”; while recognizing that “an attempt to define police powers reach or trace its outer limits is fruitless.”

6 0

2 years ago

What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

svetoff [14.1K]

Base on my research, within 2 hours you have a number of atoms which remain.
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0

So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0

It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms

N0= 5.55*10*8/0.2057 = 2.698*10^9 atoms

Therefore, 2.698*10^9 atoms is the number of N0

4 0

3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir

tia_tia [17]

Answer:

7560 Joules

Explanation:

$m_1$ = Mass of first car = $1.5\times 10^5\ kg$

$m_2$ = Mass of second car = $2\times 10^5\ kg$

$u_1$ = Initial Velocity of first car = 0.3 m/s

$u_2$ = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s$

Energy lost is

$\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J$