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2 years ago

What do you think is the main idea?

Physics

1 answer:

Tanzania [10]2 years ago

3 0

I think the main idea is that the middle planets have a solid inner core and that they all could have life on them

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Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus

ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0

1 year ago

1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ________

Ghella [55]

Answer:

Explanation:

I'm in 17th column , a nometal, and a solid at room temperature. What am i

6 0

2 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

Delvig [45]

Answer:

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

$E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}$

$E_w=\dfrac{\lambda }{2\pi \varepsilon _or}$

Now the electric filed due to hollow metal cylinder

$E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}$

$E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}$

Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

5 0

3 years ago

What is the relation between the weight of a body and acceleraton due to gravity?

emmainna [20.7K]

Answer:

All objects on Earth, regardless of their mass, accelerate due to gravity at the same rate - that is, 9.8 m/sec2. The weight of an object can be calculated using the formula for force - F = m * a - where F equals the weight of the object and now the acceleration (a) is the acceleration of gravity (g).

6 0

2 years ago

1) The equilibrium constant Kc for the reaction N 2(g) + O 2(g) 2NO(g) at 1200 C is 1.00x 10^-5. Calculate the molar concentrati

Elina [12.6K]

Explanation:

1) N₂ + O₂ → 2 NO

Kc = [NO]² / ([N₂] [O₂])

Set up an ICE table:

$\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]$

Plug into the equilibrium equation and solve for x.

1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))

1.00×10⁻⁵ = (2x)² / (0.114 − x)²

√(1.00×10⁻⁵) = 2x / (0.114 − x)

0.00316 = 2x / (0.114 − x)

0.00361 − 0.00316x = 2x

0.00361 = 2.00316x

x = 0.00018

The volume is 1.00 L, so the concentrations at equilibrium are:

[N₂] = 0.114 − x = 0.11382

[O₂] = 0.114 − x = 0.11382

[NO] = 2x = 0.00036

2(a) Cl₂ → 2 Cl

Kc = [Cl]² / [Cl₂]

$\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]$

1.2×10⁻⁷ = (2x)² / (2 − x)

1.2×10⁻⁷ (2 − x) = 4x²

2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²

2.4×10⁻⁷ ≈ 4x²

x² ≈ 6×10⁻⁸

x ≈ 0.000245

2x ≈ 0.00049

2(b) F₂ → 2 F

Kc = [F]² / [F₂]

$\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]$

1.2×10⁻⁴ = (2x)² / (2 − x)

1.2×10⁻⁴ (2 − x) = 4x²

2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²

2.4×10⁻⁴ ≈ 4x²

x² ≈ 6×10⁻⁵

x ≈ 0.00775

2x ≈ 0.0155

F₂ dissociates more, so Cl₂ is more stable at 1000 K.

7 0

2 years ago