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Lelu [443]
3 years ago
14

What is the area called where the particles are spread apart?

Physics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer: Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively

Explanation:

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On Venus, the acceleration due to gravity is 8.87 m / s 2 . How far would a 17 g rock fall from rest in 6.5 s if the only force
algol [13]

Answer:

187.38 m

Explanation:

Using the equation of motion

s = ut + 1/2gt²...................... Equation 1

Where s = distance of fall, u = initial velocity of the rock, t = time taken for the rock to fall from rest, g = acceleration due to gravity of venus.

Given: u = 0 m/s ( from rest), t = 6.5 s, g = 8.87 m/s².

substituting into equation 1

s = 0(6.5) + 1/2(8.87)(6.5)²

s = 0 + 374.7575/2

s = 187.38 m.

Hence the rock will fall 187.38 m

7 0
3 years ago
True or False<br><br> The greater the speed of an object, the less kinetic energy it possesses.
Anettt [7]
That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy
3 0
3 years ago
Read 2 more answers
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
Please tell me the answer
maria [59]

Answer:

for which one

Explanation:

4 0
3 years ago
What happens when a force is applied to an object in stable equilibrium
Katarina [22]

Answer:

the object is no longer in equilibrium .

7 0
2 years ago
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