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2 years ago

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10. In the 16 Personalities Test, what traits are ranked as the primary function according to type dynamics? O A. Judging and pr

ospecting B. Thinking and feeling C. Sociability and shyness O D. Introversion and extroversion

1 answer:

san4es73 [151]2 years ago

7 0

Of the four types of personalities, the traits that are ranked as the primary function according to type dynamics is Judging and prospecting.

<h3>What are personality?</h3>

Personalities refers to the traits characters and attributes possessed inherently by individuals.

The four groups of personalities using the 16 Personalities Tests are;

Judging and prospecting
Thinking and feeling
Sociability and shyness
Introversion and extroversion

Of the four, the traits that are ranked as the primary function according to type dynamics is Judging and prospecting.

Learn more about Personalities test at: brainly.com/question/12414191

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Moving force of air flows through areas of high pressure to areas of low pressure

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The temperature of 2.0 g of helium is increased at constant volume by ΔT. What mass of oxygen can have its temperature increased

gavmur [86]

Answer:

m = 9.6 g

Explanation:

Thermal energy given to helium gas at constant volume is given as

$Q = nC_v \Delta T$

so here we have

$C_v = \frac{3}{2}R$

$n = moles$

$n = \frac{2}{4} = 0.5$

so we have

$Q = \frac{3}{2}R(0.5)\Delta T$

now we know that

for oxygen gas we have

$C_v = \frac{5}{2}R$

for same amount of heat we have

$Q = nC_v \Delta T'$

$\frac{3}{2}R(0.5)\Delta T = \frac{m}{32} (\frac{5R}{2}) \Delta T$

$m = \frac{0.75 \times 32}{2.5}$

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3 years ago

Which question should Trudy write

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3 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d)

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

So

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

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3 years ago

a 4kg block is attatched to a vertical sspring constant 800n/m. the spring stretches 5cm down. how much elastic potential energy

kow [346]

The Potential energy stored in the system is 1 J

<u>Explanation:</u>

Given-

Mass, m = 4 kg

Spring constant, k = 800 N/m

Distance, x = 5cm = 0.05m

Potential energy, U = ?

We know,

Change in potential energy is equal to the work done.

So,

$U = \frac{1}{2} k (x)^2\\\\$

By plugging in the values we get,

$U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J\\$

Therefore, Potential energy stored in the system is 1 J

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3 years ago