All categories

2 years ago

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

1 answer:

5 0

Identity vs. Role confusion is the 5th stage of Erik Erikson's theory of psychosocial development.

During this stage, adolescents try to explore their independence. They begin to get interested in social interaction and relationships. This is also the time they begin to feel confused and insecure about themselves trying to figure out how to fit in the group. This is the reason why they engage themselves in different roles and activities.

You might be interested in

A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0

2 years ago

26. An ice-skater who weighs 200 N is gliding across the ice. If the force of friction is 4 N. what is the

Scrat [10]

Answer:

0.02

Explanation:

coefficient of kinetic friction = μ

force of friction = Ff

Normal Force = FN, but

FN = -W

Ff = -μFN

so μ = Ff/FN

= 4N/200N

= 0.02.

7 0

2 years ago

g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0

Margarita [4]

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

$v(t) =\frac{d(f(t))}{dt}$

$v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2$

Hence the velocity of the particle at time t is $v(t) = 0.04t^3 - 0.09t^2$

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

$v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05$

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second

6 0

1 year ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

= 3.1536 x 10⁷ +0.840

= 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

= 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

= 86400.0023 s

The time period of rotation is given by the formula,

ωₐ = 2π / Tₐ

Substituting the values,

ωₐ = 2π / 365.046306

= 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

= 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

= 2π /86400

= 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration

α = (ωₓ - ωₐ) / T₁

= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²

3 0

2 years ago

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e) y = 0 + ½ g t²

t = √ (2y / g)

t = √(2 125 / 9.8)

t = 5.05 s

6 0

1 year ago