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3 years ago

When does the most significant drop in physical activity typically occur in people's lives?

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

Later high school years and freshman year of college

Explanation:

The transition from high school to college is an important developmental milestone that holds the potential for personal growth and behavioral change. A cohort of 2,025 students was recruited during the summer before they matriculated into college and completed Internet-based surveys about their participation in a variety of behavioral risks during the last three months of high school and throughout the first year of college. Alcohol use, marijuana use, and sex with multiple partners increased during the transition from high school to college, whereas driving after drinking, aggression, and property crimes decreased. Those from rural high schools and those who elected to live in private dormitories in college were at highest risk for heavy drinking and driving after drinking.

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Which of the following are events?

ira [324]

Answer:

G..................

3 0

3 years ago

A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?

balandron [24]

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s Result

7 0

3 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r

Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l = 4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

$F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s$

Let $\omega$ is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

$v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s$

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0

3 years ago

Questions

KIM [24]

Answer:

Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C. For comparison sake, it only takes 385 Joules of heat to raise 1 kilogram of copper 1°C.

Explanation:

6 0

3 years ago