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2 years ago

Which of the following characterizes Erikson's identity versus role formation stage?

Physics

2 answers:

Rainbow [258]2 years ago

7 0

I think that the best one is determining one's sense of individuality and place in society.

melamori03 [73]2 years ago

3 0

Answer:

Determining one's sense of individuality and place in society

Explanation:

This is the fifth stage of Erickson’s eight stages of psychosocial development which occurs in the age range of 12-18.

In this stage adolescent's try to figure out what kind of person they are i.e., finding one's self. In this stage people try different things to know their likes and dislikes. In order to find their likes and dislikes they try different things to understand their place in the world i.e, they will understand their role that they will occupy.

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What is a conclusion?

Vikki [24]

The end or finish of an event or process.

8 0

2 years ago

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What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude

Bas_tet [7]

Answer:

The fraction of kinetic energy to the total energy is $\frac{K}{T}=\frac{8}{9}$ .

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

$K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)$

The total energy is

$T =0.5 mw^2A^2 ..... (2)$

Divide (1) by (2)

$\frac{K}{T}=\frac{8}{9}$

5 0

2 years ago

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

Luden [163]

Answer:

$\rho _{liquid}=1995.07kg/m^{3}$

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have $T_{3}+B = W_{rock}$

$T_{3}+\rho _{Liquid}V_{rock}g$ = $W_{rock}.....(\alpha)$

When the rock is suspended in air for equilibrium we have

$T_{1}=W_{rock}....(\beta)$

When the rock is suspended in water for equilibrium we have

$T_{2}$ + $\rho _{water}V_{rock}g$ = $W_{rock}.....(\gamma)$

Using the given values of tension and solving α,β,γ simultaneously for $\rho _{Liquid}$ we get

$W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\$

Solving for density of liquid we get

$\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000$

$\rho _{liquid}=1995.07kg/m^{3}$

5 0

2 years ago

A car traveling 90 km/hr is 100 m behind a truck traveling 50 km/hr. How long will it take the car to reach the truck?

Ber [7]

v = speed of car = 90 km/h

u = speed of truck = 50 km/h

d = initial separation distance = 100 m = 0.1 km

They meet at time t such that

vt = d + ut

t(v - u) = d

t = d/(v - u) = (0.1 km) / [(90 - 50) km/h] = 0.0025 hours

8 0

3 years ago

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A rigid, insulated tank contains steam at 3 MPa, 400 Celsius. A valve on the tank is opened, allowing steam to escape. The overa

enot [183]

Answer:

x=0.8

Explanation:

Defining our values we have

$p_1=3Mpa\\T_1=400\° c\\q_{1-2}=0$

Where $p_1$ is the pressure of the super heated steam

$T_1$ is the Temperature of the super heated steam

and $q_{1-2}$ is the pressure in an adiabatic process.

State 1

$v_1 = 0.09936m^3/Kg\\s_1=6.9213kJ/kg.K$

State 2

$s_2=s_1=6.9212$

Note that here in state 2 the process is Reversible.

At the value where $T_2 = 141\°c$ we have

$v_2=(v_g)_{T_2}=0.4972m^3/kg$

Through this values we can calculate the Fraction of steam,

$x=\frac{m_1-m_2}{m_1}\\x=1-\frac{m_2}{m_1}\\x=1-\frac{v_1}{v_2}\\x=1-\frac{0.9936}{0.4972}\\x=0.8$

4 0

2 years ago