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1 year ago

Please help me with this rotation and linear motion problem

Physics

1 answer:

kirill115 [55]1 year ago

5 0

The initial angular velocity of the wheel would be 15.5 rad/s

<h3>What is Velocity?</h3>

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.

The relation between the linear velocity and the angular velocity

V= ωr

as given the initial linear velocity is 8.22 m/s

the radius of the wheel is 0.530 m

By substituting the value of the linear velocity and the radius

8.22 = 0.530ω

ω = 8.22/0.530

ω = 15.5 rad/s

Thus, the initial angular velocity of the wheel comes out to be 15.5 rad/s

Learn more about Velocity from here

brainly.com/question/18084516

#SPJ1

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3 years ago

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Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc

Vanyuwa [196]

Answer:

$1\cdot 10^{-8} N$ to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

$q_1=2nC=2\cdot 10^{-9}C$ is the magnitude of charge 1

$q_2=5nC =5\cdot 10^{-9}C$ is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

$F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N$

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

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3 years ago

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Answer:

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3 years ago

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There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center)

Aneli [31]

Answer:

q₁ =± 1.30 10⁻⁶ C and q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

q₁ + q₂ = 2 q

Let's replace

F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

F1 = k q₁ q₂ / r²

F2 = ½ k (q₁ + q₂) / r²

F1 r² / k = (q₁ q₂)

F2 r² / k = (q₁ + q₂)/2

q₁ = 2F2 r² / k - q₂

We substitute in the other equation

F1 r² / k = (2F2 r² / k - q₂) q₂

0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

F1 r² / K = 0.0765 0.440² / 8.99 10⁹

F1 r² / K = 1.65 10⁻¹²

(2F2 r² / k) =2 0.10 0.44² / 8.99 10⁹

(2F2 r2 / k) = 4.30 10⁻¹²

q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

F1 = k q₁ q₂ / r²

q₁ = F1 r² / (k q₂)

q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

q₁ = 1.30 10⁻⁶ C

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3 years ago