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3 years ago

What are the Big Five Personality Factors?

Physics

2 answers:

Zarrin [17]3 years ago

8 0

Answer:

O-penness.

C-onscientiousness.

E-xtraversion.

A-greeableness.

N-euroticism.

(ocean)

hope this helps

Veronika [31]3 years ago

5 0

Friends and skckfkfkrmgklvkfkkaksk

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Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen

m_a_m_a [10]

Answer:

$F=2.25\times 10^{-5}\ N$

Explanation:

According to question,

Charge 1 and charge 2 are $10^{-7}\ C$

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

$F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N$

So, the electric force of repulsion is $2.25\times 10^{-5}\ N$ .

8 0

3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

at a temperature of 10 c,700 ml of hydrogen is collected. if this gas is put into a 1000 ml container what will its new temperat

erastovalidia [21]

Answer:

14.3°C

Explanation:

Find the ratio of 10°C : 700ml then use the same ratio to 1000ml.

Have a great day <3

4 0

2 years ago

Read 2 more answers

A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?

Andrew [12]

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
   (15 m/s) in the direction opposite to the direction you call positive.

Displacement = (distance between start-point and end-point)
   in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
   = (1/2) (3 m/s²) (25 s²) = 37.5 meters

Displacement =
   37.5 meters in the direction opposite to the direction you call positive.

5 0

3 years ago

It took 500 newtons of force to push a car 4 meters. How much work was done?

Ede4ka [16]

Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J

3 0

2 years ago