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3 years ago

Which is the correct formula for sulfur dichloride

Physics

2 answers:

xz_007 [3.2K]3 years ago

4 0

The formula for sulfur dichloride is SCl2

Anvisha [2.4K]3 years ago

4 0

The chemical formula for sulfur dichloride is SCl2

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Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s

PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0

2 years ago

A meter stick hurtles through space at a speed of 0. 95c relative to you, with its length perpendicular to the direction of moti

marta [7]

Answer:

Explanation:

Caty , Use the relativity formula for length. ( they teach this in H.S. ? ) it's from my Modern Physics in college, A 300 level class

L = $L_{0}$ $\sqrt{1-\frac{v^{2} }{c^{2} } }$

L = 3 $\sqrt{1-\frac{.95^{2} }{1^{2} } }$

L = 0.9367496998 meters

L = 0.94 meters approx

6 0

1 year ago

Amy crashed her bike into the fence. She was thrown over it onto the lawn. Which Newton law applies

BlackZzzverrR [31]

Answer:

Newton's First Law of Motion applies here.

Explanation:

Before crashing into the fence, Amy was moving at a certain speed on her bike. As, she crashed her bike into the fence, the collision stopped the bike suddenly. But, Amy had the same speed due to inertia of her body. Due tot his speed Amy did not stop and she was thrown over the fence onto the lawn. So, the force of inertia of Amy's body caused her to be overthrown in this case. We study about inertia in Newton's First Law of Motion, which is also known as Law of Inertia.

<u>Newton's First Law of Motion applies here.</u>

7 0

2 years ago

The potential energy of a 35 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?

Anettt [7]

From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.

We need to find its height.

We will use the formula P.E = mgh

Therefore h = P.E / mg

where P.E is the potential energy,

m is mass in kg,

g is acceleration due to gravity (9.8 m/s²)

h is the height of the object's displacement in meters.

h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters high.

6 0

2 years ago