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just olya [345]
3 years ago
13

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

Physics
2 answers:
Rainbow [258]3 years ago
7 0

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

melamori03 [73]3 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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Calculate the velocity of an apple that falls freely from rest and drops for 3.5 seconds.
padilas [110]

Answer:

initial velocity(u)=0

time(t)=3.5sec

acceleration (a)=9.8m/s²

final velocity (v)=?

Explanation:

we have

v=u+at=0+9.8×3.5=34.3m/s

6 0
3 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'
qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s


Check the calculations again !


7 0
3 years ago
2 Points
True [87]
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8 0
3 years ago
Read 2 more answers
A box exerts 10,000 Pa of pressure on the ground. If the box weighs 1000 N, how much area is in contact with the ground?
lara [203]

Pressure = (total force) / (Area)

10,000 Pa = (1,000 N) / (Area)

Multiply each side by (Area) :

(10,000 Pa) x (Area) = 1,000 N

Divide each side by (10,000 Pa) :

Area = (1,000 N) / (10,000 Pa)

<em>Area = 0.1 m² </em>

4 0
3 years ago
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