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2 years ago

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Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

is the electric force between them?

1 answer:

vesna_86 [32]2 years ago

4 0

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles 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 of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

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A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together

viktelen [127]

Answer:

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

$m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v$ (1)

Where:

$m_{A}$ , $m_{B}$ - Masses of the cars, in kilograms.

$v_{A}$ , $v_{B}$ - Initial velocities of the cars, in meters per second.

$v$ - Velocity of the resulting system, in meters per second.

If we know that $m_{A} = 2120\,kg$ , $v_{A} = 13.4\,\frac{m}{s }$ , $m_{B} = 2810\,kg$ and $v_{B} = 0\,\frac{m}{s}$ , then the velocity of the resulting system:

$v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}$

$v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}$

$v = 5.762\,\frac{m}{s}$

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ( $K$ ), in joules, is dissipated due to work done by friction ( $W_{f}$ ), in joules, that is to say:

$K = W_{f}$ (2)

$\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s$

$\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s$ (2b)

Where:

$\mu$ - Coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Travelled distance, in meters.

If we know that $v = 5.762\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $s = 1.97\,m$ , then the coefficient of friction is:

$\mu = \frac{v^{2}}{2\cdot g\cdot s}$

$\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}$

$\mu = 0.859$

The coefficient of friction between the cars and the road is 0.859.

6 0

2 years ago

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

Tomtit [17]

a) p₀ = 1.2 kg m / s

b) p_f = 1.2 kg m / s

c) $v_{2f}$ = 1.278 m/s

d)1.2482 = $v_{2f}$ cos θ

e) 0.2736 = $v_{2f}$ sin θ

g) θ = 12.36

<h3>What is momentum?</h3>

Momentum is the product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.

According to the question,

We define a system formed by the two balls, which are isolated and the forces during the collision are internal.

Hence , the momentum is conserved

a) The momentum of the system before the collision:

$p_o = m v_1_0 + 0 p_o= 0.6 2 p_o = 1.2 kg m / s$

b) The momentum of the system after the collision, as the system is isolated, the momentum is conserved so

$p_f = 1.2 kg m / s$

d. On X axis

$p_ox$ = 1.2 kg m / s

$p_{fx}$ = m v1f cos 20 + m v2f cos θ

$p_o = p_f$

1.2 = 0.6 (-0.8) cos 20+ 0.6 $v_{2f}$ cos θ

1.2482 = $v_{2f}$ cos θ

e. On Y axis

$p_{oy} = 0$

$p_{fy}$ = m $v_{1f}$ sin 20 + m $v_{2f}$ cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 $v_{2f}$ sin θ

0.2736 = $v_{2f}$ sin θ

Here,

g. we write our system of equations

0.2736 = $v_{2f}$ sin θ

1.2482 = $v_{2f}$ cos θ

0.219 = tan θ (divide)

θ = tan⁻¹ 0.21919

θ = 12.36

c. Now To find the velocity

0.2736 = $v_{2f}$ sin θ

$v_{2f}$ = 0.2736 / sin 12.36

$v_{2f}$ = 1.278 m / s

Learn more about momentum here:

brainly.com/question/4956182

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2 years ago

During a race, a dragster is 200m from the finish line when something goes wrong and it stops accelerating. It travels at a cons

GarryVolchara [31]

Answer:

135 m

Explanation:

Speed = 45 m/s, time = 3 second

The formula for the distance is given by

Distance = speed x time = 45 x 3 = 135 m

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3 years ago

Click on the reset button, and stack one 50kg. crate on top of the other, so that the total mass is 100kg. The Friction should b

Helen [10]

Answer:

490.5 N

Explanation:

Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction

F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.

F=kmg=0.5*100*9.81=490.5 N

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3 years ago

Which is the best procedure to make a permanent magnet?

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placing a magnetically hard material in a strong magnetic field

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