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3 years ago

Failure to accomplish Erikson’s psychosocial task of late adulthood leads to despair.

Physics

2 answers:

mestny [16]3 years ago

8 0

Answer:

True

Explanation:

There are eight stage of Erikson’s psychosocial task. The stages have been divided according to the age of a person and the emotions being associated with it

In the last stage i.e., the stage of late adulthood (after 65 years of age) we generally look back at our lives and judge ourselves. If we accept our life for what it is then we feel that we have our integrity. But, if we do not accept our life this will lead to a sense of unaccomplishment which leads to despair.

Y_Kistochka [10]3 years ago

6 0

False is the answer

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Dahasolnce [82]

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

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iris [78.8K]

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3 years ago

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A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car

V125BC [204]

3.375m/s is the final velocity of the car.

<h3>How do you find final velocity?</h3>

The final velocity depends on how large the acceleration is and the distance over which it acts.

Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.

According to the question,

A toy car starts from the rest and accelerates

So the acceleration = 1.50m/s²

Time = 2.25s

$x=x_{0} + vt$

$x = 0 + ( 1.50m/s^2*2.25s)$

$x = 3.375m/s$

The final velocity, of the car is 3.375 m/s.

Learn more about velocity here:brainly.com/question/18084516

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2 years ago

A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed

fgiga [73]

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

8 0

3 years ago

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

3 years ago