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2 years ago

PLEASE HELP!!!!

Physics

2 answers:

Snezhnost [94]2 years ago

4 0

Answer:

the cognitive abilities of an individual to learn from experience to reason well, and to cope effectively with the demands of daily living

Explanation:

Yes that can be understood by a simple example

Suppose two people are in exam

A difficult and logical question came
The person can use his cognitive abilities faster can solve that

mestny [16]2 years ago

3 0

$\large\longrightarrow$ Intelligence is the <u>cognitive abilities of an individual to learn from experience to reason well, and to cope effectively with the demands of daily living.</u>

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If a 4N weight is hung on a spring, and it extends by 0.2m, what is the spring constant (k)?

Naddika [18.5K]

Answer: 200 N/m

Explanation:

The Gravitational spring energy(Us) is equal to 1/2kx^2. So we have x as .2 m and Us as 4 N. So 4 N = 1/2 * k * .2^2. So now we solve for K and get 200 N/m.

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3 years ago

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Which of the following is an example of a noninfectious disease?

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D lung cancer is not infectious

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3 years ago

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The average speed between earth and the sun is 1.50 x10^8 km. Calculate the average speed of the Earth in its orbit in kilometer

cluponka [151]

Answer:

The average speed of the earth in its orbit is $29.86km/s$

Explanation:

The average distance between the Earth and the Sun is $1.50x10^{8} km$ .

The average speed of the earth in its orbit can be found by the next equation :

$v = \frac{2 \pi r}{T}$ (1)

Where r is the radius and T is the period.

In this case, the orbit of the Earth can be considered as a circle

( $r = 1.50x10^{8}km$ ) instead of an ellipse.

It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.

Notice that to express the period in terms of seconds, the following is needed:

$365.25d . \frac{86400s}{1d}$ ⇒ $31557600s$

Then, equation 1 can be used:

$v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}$

$v = 29.86km/s$

7 0

3 years ago

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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

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