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3 years ago

Which term describes the second personality seen in a person with dissociative identity disorder (DID)?

Physics

2 answers:

sertanlavr [38]3 years ago

7 0

Answer:

C. Alter Ego

Explanation:

Dissociative Identity Disorder (DID), also known as Multiple Personality disorder is basically a mental disorder. The person suffering from DID tends to show at least two different types of personality states. Psychiatrist Paulette Gillig coined the term <em>Ego</em> state referring to the normal personality state of a person and "<em>Alter Ego</em>" state referring to the other personality state of the person. Multiple personalities occur alternatively. During one personality state the person has no memory of the other personality state.

Snezhnost [94]3 years ago

6 0

They have and alter ego

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A vector is any quantity that has no units of measurement

djyliett [7]

False. Velocity is a vector and is measured in m/s (in SI, anyway).

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3 years ago

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Which of the following would produce the most power?

Fantom [35]

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

But Energy = mgh

Substituting into the equation, we have

$Power = \frac {mgh}{time}$

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

$Power = \frac {10*9.8*10}{5} = 490 Watts$

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

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3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

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3 years ago

Follow me sped donkeys.

Artist 52 [7]

Okayyyyyyyyyy....................

5 0

3 years ago

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A merry-go-round accelerating uniformly from rest achieves itsoperating speed of 2.5rpm in five revolution.

g100num [7]

Answer:

(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²

(B) Time of motion 240.2 s

Explanation:

Given;

final angular speed, ωf = 2.5 RPM

angular distance, θ = 5 rev = 5 x 2π = 10π rad

initial angular speed, ωi = 0

final angular speed in rad/s;

$\omega_f = \frac{2.5 \ rev}{min} \times \ \frac{2\pi}{1 \ rev} \times \ \frac{1 \min}{60 s} = 0.2618 \ rad/s \\$

(A) the magnitude of its angular acceleration(rad/s^2);

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2\times 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = \frac{0.0685}{20\pi} \\\\\alpha = 1.09\times 10^{-3} \ rad/s^2$

(B) Time of motion;

$\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09\times 10^{-3} t\\\\t = \frac{0.2618}{1.09\times 10^{-3}} \\\\t = 240.2 \ s$

7 0

3 years ago