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2 years ago

Essay about why people should not join a gang 300 word

Physics

1 answer:

Salsk061 [2.6K]2 years ago

6 0

Answer:

I don't know if it's fair for me to write an entire essay for you

but if you would like I can list some reasons you can incorporate into an essay.

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-If you are joining a gang you are most likely going to have a greater chance of being targeted, even if it seems like you would be protected.

-Most of the time when someone joins a gang a child is being separated from a parent and will face most aggression.

-This can also cause academic failure in students, from lack of education and probably a pretty bad background.

-For teens when joining a gang they will do things that aren't suited for their age (I would list things but I'm not trying to get myself reported if you know what I mean)

-You will also have a great risk of being imprisoned and facing bad consequences, and these consequences can even be having the death penalty.

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I hope this is good enough, sorry I didn't give you a full essay. I think this should help with that though.

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A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the

storchak [24]

Answer:

a) V1 = 4V - V2/3 and V2 = 4V - 3V1

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V represent volume of the box containing the two compartments

V1 represents compartment of the left compartment

V2 represents compartment of the right compartment

Momentum of the compartments before impact:

3000V1 + 1000V2

Momentum of the compartments after impact:

V(3000 + 1000) = 4000V

a) To obtain the volume of each compartment, that is, V1 and V2, we say:

Momentum before impact = Momentum after impact

3000V1 + 1000V2 = 4000V

∴ V1 = 4000V - 1000V2/3000 = 4V - V2/3

Also, V2 = 4000V - 3000V1/1000 = 4V - 3V1

b) Change in entropy,Δe = 4000V1 - 1000V2

By substituting the V1 and V2, we have:

4000(4V - V2)/3 - 1000(4V - 3V1)

16000V - 4000V2/3 - 4000V + 3000V1

16000V - 4000V2 - 12000V + 9000V1

∴ Δe = 4000V - 4000V2 + 9000V1

6 0

2 years ago

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

2 years ago

a) Calculate the magnitude of displacement of the car in 40 seconds. b) During which part of the journey was the car acceleratin

Zielflug [23.3K]

Answer:

a) 600 meters

b) between 0 and 10 seconds, and between 30 and 40 seconds.

c) the average of the magnitude of the velocity function is 15 m/s

Explanation:

a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:

$Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m$

b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.

Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.

c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

$Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx$

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

$Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}$