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Alexxx [7]
2 years ago
11

19. In a 16 Personalities Test, which

Physics
1 answer:
tiny-mole [99]2 years ago
6 0

Personality test is aimed at being able to decipher the nature of human personality. The opposite of judging in a personality test is prospecting.

<h3>What is a personality test?</h3>

The term personality test is aimed at being able to decipher the nature of human personality through a  Personalities Test.

In this personality test different personality types have their respective opposites. The opposite of judging in a personality test is prospecting.

Learn more about personality test: brainly.com/question/8789865

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If a vector A has components A. 0, and Ay -0, then the magnitude of the vector is negative. Select one: True False
Dima020 [189]

Answer:

False

Explanation:

The magnitude of any vector is given by,

||A||=\sqrt{A_x^2+A_y^2}

The magnitude of anything is never negative. It can be even seen from the formula that the components are squared. A squared value can never be negative. Even if the component is negative the square will be always positive.

So, magnitude of the vector is <u>not</u> negative.

8 0
3 years ago
Read 2 more answers
Why do electrical devices have resistance​
Kamila [148]
As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.
8 0
3 years ago
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
3 years ago
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if you jump from a window that is 2 meters up you will most likely break the bones in your legs when you land however on a tramp
Gemiola [76]
<span>The de-acceleration or negative acceleration of stopping is what damages bones. The ground is rigid and therefore the change in momentum when striking the ground will be large. On the trampoline, the elasticity of the material means that the momentum changes more slowly, resulting in smaller accelerations.</span>
5 0
3 years ago
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