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Oduvanchick [21]

2 years ago

7

Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig

ht as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them.

1 answer:

g100num [7]2 years ago

3 0

I have attached the image showing the 6 combination of charges

Answer:

Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc

Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc

Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc

Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc

Explanation:

The electrostatic force between 2 charges is; F = k•q1•q2/r²

Where r is the distance between the 2 charges q1 and q2 and k is coulombs constant.

Thus;

F ∝ q1•q2/r²

The net force on the charge q1 is the sum of the forces on charge q2 and q3.

If the two charges are opposite, the force is an attractive force while if the two charges are similar, the force is a repulsive force.

Thus, if the combination has same type of charges, i.e;(+1,+1,+1) or (-1,-1,-1), the force will be very small.

So, now the order of forces from smallest to largest is;

Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc

Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc

Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc

Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc

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Alex73 [517]

To determine the object which could give the greatest impact we will apply the concept of momentum. The object that has the highest momentum will be the object that will impact the strongest. Our values are

Mass of Object A

$m_A=1.9 kg$

Velocity of object A

$v_A=8ms$

Mass of object B

$m_B=2 kg$

Velocity of object B

$v_B=5ms$

The general formula for momentum is the product between mass and velocity, then

$p = mv$

For each object we have then,

$p_A=m_Av_A=1.9 kg(8ms)=15.2kg \cdot m/s$

$p_B=m_Bv_B=2 kg(5ms)=10kg \cdot m/s$

Since the momentum of object A is greater than that of object B, then object A will make you feel force upon impact.

3 0

3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

Ede4ka [16]

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

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2 years ago

Is the star moving toward earth, away from earth, or is there not enough information provided to determine its motion?

ivanzaharov [21]

If a star is moving towards Earth, shift towards the blue end of the spectrum, this is called blue shift. If the star is moving away from Earth the light from that star will be red and is called red shift .

The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum

if a star is moving towards Earth, it appears to emit light that is shorter in wavelength compared to a source of light that isn't moving. Because shorter wavelengths correspond to a shift towards the blue end of the spectrum, this is called blue shift.

If the star is moving away from Earth, its light will lose energy to reach Earth, therefore the light from that star will be red and is called red shift

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8 0

2 years ago

Para cargar un camión, se suele utilizar una tabla entre el contenedor y el suelo, con el fin de subir la carga, ya sea desplaza

stiks02 [169]

Answer:

A. No

B. si

Explanation:

A. El trabajo realizado en la carga es la energía potencial ganada por la carga al elevar la carga al nivel del camión y colocar la carga dentro del camión.

El trabajo realizado para elevar la carga W = m × g × h

Dónde;

m = masa de la carga

g = aceleración debido a la gravedad

h = Nivel de altura donde se coloca la carga en el camión

Por lo tanto, el trabajo realizado depende de la masa, m, de la carga y el nivel de altura, h, donde la carga se coloca en el camión y el trabajo realizado es el mismo para todos los métodos utilizados para colocar la carga en el camión

B. La ecuación para el trabajo realizado, W, también se puede escribir de la siguiente manera;

W = Fuerza, F × Distancia, D

De lo que tenemos;

F = W/D

Por lo tanto, cuando la mesa aumenta la distancia, como una rampa o un plano inclinado, la fuerza requerida disminuirá.

4 0

3 years ago