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Aleks04 [339]
3 years ago
15

You are helping the school nurse decide which students are most at risk for health problems. Look at these four cases. Which per

son seems most likely to abuse drugs because of peer pressure? Everyone in Margene's family smokes. She has decided not to start. Her best friends are also committed to abstaining from tobacco use. Jacob plays basketball and has to be the best player on the team. He practices all the time and eats a diet high in protein, carbohydrates, and vitamins. Rhonda lives in a housing project. Her parents work hard, and the whole family checks on her social life regularly to make sure she studies and stays active. Ramon is outgoing and likes to give parties at his house. He's easygoing, and he likes the company because his parents are away a lot.
Physics
2 answers:
Tanya [424]3 years ago
5 0
It would be Ramon because he dont have family around to help him and hes already throwing parties
Kryger [21]3 years ago
3 0
It would be D because he seems to not have much parental structure in his life and since he is easygoing, he is more susceptible to just going along with everyone else. Also, since he has parties, the people at the parties could be doing drugs or drinking alcohol and could influence him to do so as well.


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Answer:

d = 13 miles

Explanation:

Lets say the position of court house is origin in this case

her office is located at 4 miles west and 4 miles south of court house

so here we have coordinate of the office with respect to court house is given as

r_1 = (-4\hat i - 4\hat j)

now the position of her home is located at 1 miles east and 8 miles north of the court house

so the coordinates of her home is given as

r_2 = (1\hat i + 8 \hat j)

now the change in the position is given as the distance between office and home

d = r_2 - r_1

d = 5 \hat i + 12 \hat j

d = \sqrt{5^2 + 12^2} = 13 miles

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A 5kg rock and a 10kg rock are dropped from a height of 10m?
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A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
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