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3 years ago

According to Freud, adults fixated in which stage could feel constantly out of control?

1 answer:

6 0

AnalStage. According to Freud, adults fixated in this area could feel constantly out of control or could need to be in control all the time. Pleasure focus is on the anus, which occurs when a child learns to control bladder and bowel movements. Fixation in this area can be caused from struggles during potty training.

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A force of 70 N is applied to a 28 kg rock causing it to slow down from 25 m/s to 15 m/s, a change in velocity of 10 m/s. How lo

maksim [4K]

Answer: 4 s

Explanation:

Given

The applied force is 70 N

mass of the rock is 28 kg

initial velocity $u=25\ m/s$

final velocity $v=15\ m/s$

Deceleration provided by force is

$a=-\dfrac{70}{28}=-2.5\ m/s^2$

using the equation of motion

$v=u+at\\\Rightarrow 15=25-2.5t\\\Rightarrow 2.5t=10\\\Rightarrow t=4\ s$

7 0

3 years ago

Which statement is true of forces?

jasenka [17]

FORCES HAVE STRENGTH AND DIRECTION

8 0

3 years ago

a book is sitting still on your desk. are the forces acting on the book balanced or unbalanced? explain.

GrogVix [38]

They are balanced. If the forces were not balanced the book would move*. In this example, the downward force of gravity on the book is counterbalanced by the upthrust of the desk.

8 0

3 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.