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NemiM [27]
2 years ago
7

HOLA, NECESITO AYUDA!

Physics
1 answer:
givi [52]2 years ago
6 0

The electrostatic force between the two ions is 2.9\cdot 10^{-10} N

Explanation:

The electrostatic force between two charged particle is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, the ion of sodium has a charge of

q_1 = +e = +1.6\cdot 10^{-19} C

while the ion of chlorine has a charge of

q_2 = -e = -1.6\cdot 10^{-19}C

And the distance between the two ions is

r=282 pm = 282\cdot 10^{-12} m

Substituting, we find the electrostatic force between the two ions:

F=(8.99\cdot 10^9) \frac{(1.6\cdot 10^{-19})(-1.6\cdot 10^{-19})}{(282\cdot 10^{-12})^2}=-2.9\cdot 10^{-10} N

where the negative sign simply means that the force is attractive, since the two ions have opposite charge.

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat
Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water\left ( m\right )=250gm

Initial temperature\left ( T_i\right )=20^{\circ}C

Final temperature \left ( T_f\right )=100^{\circ}C

Specific heat of water \left ( c \right )=4190 J/kg-k

heat of vaporization\left ( L\right )=22.6\times 10^5 J/kg

Heat required for process\left ( Q\right )=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc\left ( T_f-T_i\right )+mL

Q=0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5

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Q=6.338\times 10^5J=633.8 KJ

4 0
3 years ago
A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=
True [87]

Answer:

(a) average velocity = 17.6 m/s

(b) when t = 0, v = 0

    when t = 4, v = 19.2 m/s

    when t = 8, v = 28.8 m/s

(c) after starting from rest, the car will be at rest again in 20 s

Explanation:

Given;

x(t)=bt²−ct³, substitute the given values and the equation will become;

x(t)=3t²−0.1t³

(a)average velocity = total distance / total time

total distance, x(t) = 3t²−0.1t³

x(8) = 3t²−0.1t³

X(8) = 3(8)² - 0.1(8)³

X(8) = 140.8 m

total time = 8 s

average velocity = 140.8 / 8

average velocity = 17.6 m/s

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dx / dt = 6t - 0.3t²

when t = 0

v = 0

when t = 4 s

v = 6(4) - 0.3(4²) = 19.2 m/s

when t = 8 s

v = 6(8) - 0.3(8²) = 28.8 m/s

(c) the velocity is zero at dx / dt = 0

6t - 0.3t² = 0

t(6 - 0.3t) = 0

t = 0    or  6 - 0.3t = 0

t = 0     or   0.3t = 6

t = 0      or   t = 6 / 0.3

t= 0       or    t =  20 s

After starting from rest, the car will be at rest again in 20 s

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