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3 years ago

Research on the Big Five has indicated that approximately _____ percent of traits appear to be inherited.

Physics

2 answers:

Evgen [1.6K]3 years ago

4 0

D 50
Hopefully this helps!

Sergeu [11.5K]3 years ago

4 0

Answer:

D) 50

Explanation:

The personality of an individual is characterized by five important traits . They are as follows.

1) Openness

2) conscientiousness

3) Extraversion

4)Agreeableness

5) Neuroticism

This as per FFM or OCEAN model . Researches have shown that half of the variation in these traits result from genetics and half from environment.

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If a person weighs 600N on earth, how much does he weigh on the moon

Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0

3 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

In one experiment the electric field is measured for points at distances r from a uniform line of charge that has charge per uni

Yuki888 [10]

Answer:

(a) A. Uniform line of charge and B. Uniformly charged sphere

(b) To three digits of precision:

λ = 1.50 * 10^-10 C/m

p = 2.81 * 10^-4 C/m^3

Explanation:

8 0

3 years ago

What determines if a star turns into a red giant or a red supergiant?

mixer [17]

Answer:

Explanation:

As stars age, they evolve away from the main sequence to become red giants or supergiants. The core of a red giant is contracting, but the outer layers are expanding as a result of hydrogen fusion in a shell outside the core. The star gets larger, redder, and more luminous as it expands and cools.

3 0

3 years ago

Sometimes you can see a faint reflection in the surface of a shiny plate or cup.Why ?

bazaltina [42]

Answer:

The degree of reflection whether faint or bright you see on the surface of an object is an indication that light particles had hit the surface. Since light is a wave and as part of its characteristics can get reflected. However, the amount of light reflected by a surface is dependent on the smoothness of the surface which can be shiny or dull, it can also be dependent on the nature of the surface which can be glass, water, and so on. So, from the question, you can see a faint reflection on the surface of a shiny plate or cup because of the smoothness of the surface which reflects the lights that hit it from a particular direction at the same angle.

4 0

2 years ago