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3 years ago

What are common features of both the MMPI-2 and CPI? Select all that apply.

Physics

1 answer:

Amiraneli [1.4K]3 years ago

8 0

Answer: D

Both MMPI-2 and CPI check for serious personality disorders.

MMPI - 2

MMPI - 2 stands for Minnesota Multiphasic Personality Inventory -2. It is the most widely used psychometric test to measure adult psychopathology. MMPI-2 differentiates people with hypochondriasis from normal one. MMPI-2 is able to detect unconsciously somatizing or consciously malingering in patients.

CPI

CPI stands for california psychological inventory. It is a self report inventory created by Harrision G. This test used to test the people, which are 13 years and older. It takes 40 minutes to complete the test. CPI works on "folk-concepts" which means an ordinary people can use it to describe other people around him.

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The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is

almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

$I = I_{0}e^{\frac{t}{\tau\\ } }$

$\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }$

Apply inverse logarithmic function:

$ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}$

The final form will be:

$\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}$

Inputing values for I, IO, and t:

$\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95$

3 0

3 years ago

A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e = v_c/m + vc_e

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

5m*vs_e - 4m*(v_c/m) = 5m*vc_e

vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

vc_e = 3760 - 0.8*(90)

vc_e = 3,688 km/h

5 0

3 years ago

Read 2 more answers

A book, that has a mass of 0.5 grams, is pushed across a table with a force of 20 newtons. What is the acceleration of the book?

Bas_tet [7]

Answer:

$4\cdot 10^4 m/s^2$

Explanation:

The acceleration of an object is given by Newton's second law:

$a=\frac{F}{m}$

where

F is the net force applied on the object

m is the mass of the object

For the book in the problem, we have:

$m=0.5 g =5\cdot 10^{-4} kg$ is the mass

$F=20 N$ is the force applied

Substituting into the formula, we find the acceleration:

$a=\frac{20 N}{5\cdot 10^{-4} kg}=4\cdot 10^4 m/s^2$

6 0

2 years ago

With a mass of 109 kg, Baby Bird is the smallest monoplane ever

OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

m_1=61.4kg
m_2=109kg
v_1=1.84m/s
v_2=?

$\\ \sf\longmapsto ∆P=P$

$\\ \sf\longmapsto m_1v_1=m_2v_2$

$\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}$

$\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}$

$\\ \sf\longmapsto v_2=112.976/109$

$\\ \sf\longmapsto v_2\approx 1.3m/s$

4 0

2 years ago

Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d

Helga [31]

I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7

or theta = 64.62 deg, it is D. Think about it. I hope you see it.

5 0

2 years ago