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2 years ago

What is a difference between physical traits and personality traits?

Physics

2 answers:

malfutka [58]2 years ago

5 0

Answer:

A physical trait is visible to the naked eye, such as, having six fingers; and a character trait is invisible to the naked eye, such as, being charming. Both can be subjective, for instance, claiming somebody has a big nose is a subjective comment about somebody's physical appearance.

Ahat [919]2 years ago

5 0

The difference between physical traits and personality traits are that physical traits are mainly determined by a person’s genes, personality traits are determined by both a person’s genes and environment.

Answer: Option B

<u>Explanation: </u>

Physical traits are facial similarities, body language, nature of hairs and skins. Mostly the face of siblings and the children will have some similarities with their parents or other members in their family.

So, we can say that physical traits are mainly determined by person’s genes. While the personality traits determine a person’s character , hobbies, likes, dislikes, point of view etc.

Thereby, the personality traits will be coming from genes and also from the environment the person has been grown. Both genes and environment effect equally on a persons’ character and personality.

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Look at the circuit diagram. Which of these components is part of the circuit?

tangare [24]

I think the answer is c AC power source
Hope this help you?

4 0

2 years ago

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of oneâs birth. The on

Anton [14]

Answer:

Explanation:

Gravitational force between two objects having mass m₁ and m₂ at a distance R

F = G m₁ m₂ / R²

Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²

= 1.01 x 10⁻⁶ N

b )

Force between baby and Jupiter

F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²

= 1.31 x 10⁻⁶ N

c )

Ratio = 1.01 / 1.31

= .77

4 0

3 years ago

Do Small objects exert no gravitational force ? True or false

Zanzabum

Answer:

False -

F = G M1 M2 / R^2

So F depends on M1 and M2 and as long either is not zero there will be a gravitational force between them.

6 0

2 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

2 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago