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3 years ago

Explain why teens are especially weak at impulse control

Physics

1 answer:

RSB [31]3 years ago

4 0

Answer:

One explanation may be that as a group, teenagers react more impulsively to threatening situations than do children or adults, likely because their brains have to work harder to rein in their behavior, a research team reported here yesterday at the Society for Neuroscience meeting

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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

4 years ago

Zak, helping his mother rearrange the furniture in their living room, moves a 60 kg sofa 5.9 m with a constant force of 40 N. Wh

Len [333]

Answer:236 J

Explanation:

Given

Mass of sofa $m=60 kg$

Displacement $s=5.9 m$

Force $F=40 N$

neglecting Friction

Work done is given by

$W=F\cdot s$

$W=F\cdot s\cos 0$

$W=40\cdot 5.9$

$W=236 J$

4 0

3 years ago

Read 2 more answers

This whole worksheet. I am not good at this and I am completely lost

Ket [755]

1.) potential energy
2.)potential and kinetic
3.)The roller coaster car has the most kinetic energy at point X i know this because the car is moving and kinetic energy has the power to move or change things therefore point X is when the roller coaster car has the most energy.
4.)potential energy
5.)kinetic energy
6.) potential and kinetic energy

3 0

3 years ago

What is the Potential Energy of ball 1??

Dvinal [7]

GPE= height x mass x gravitational field strength
5 x 10 x 9,8=490J

7 0

3 years ago

A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta

-BARSIC- [3]

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

= 107 cos 13.5°= 104 N

Power = $\dfrac{work\ done}{time}$

69.4 = $\dfrac{W}{33}$

W = 2290.2 J

Work done= Force x displacement

displacement= $\dfrac{2290.2}{104}$

= 22.02 m

6 0

3 years ago