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2 years ago

5

Is it possible to hold friendships outside of your designated group Why or Why not?

1 answer:

zzz [600]2 years ago

7 0

Answer: Yes it is possible.

Explanation: The reason is being because, people tend to like to see different stuff. Like in the movie Pocahontas she was a totally different person. And John Smith was a different person too. SO it is possible.

You might be interested in

If the temperature of water warming on a hot plate rises 10 degrees Celsius [°C], what is the change in temperature in units of

Luba_88 [7]

Answer:

Delta_temp = 18[°F]

Explanation:

°F = 9/5*(10)

°F = 18

Note: It is important to clarify that it is only a temperature increase, that it is only a temperature increase. The question is not related to converting from 10°C to fahrenheit degrees

4 0

3 years ago

How is tidal energy obtained

yKpoI14uk [10]

Under water turbans that are placed at the above to middle of the ocean they are used to capture kinetic motion

7 0

3 years ago

The position-time equation for a cheetah chasing an antelope is:

pochemuha

Answer:

x = 1.6 + 1.7 t^2 omitting signs

a) at t = 0 x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0 V = 0

c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)

d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)

4 0

2 years ago

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

a)

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

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

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

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

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.

b)

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.

7 0

3 years ago

If Pressure = 1000 Pa and Area = 0.2 m square, then Force

Maksim231197 [3]

Here

Pressure=1000Pa

Area=0.2m Square

Force=?

Now,,

Pressure =force / Area

1000=force/0.2

force=1000/0.2

Therefore,Force=5000N

3 0

2 years ago