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3 years ago

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A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d

= 35cm above charge Q1. The net force on Q2 is equal to zero.
a) Write an equation for the magnitude of charge Q2 in terms of the given variables.

b) Calculate the magnitude of Q2 in units of nanocoulombs

1 answer:

cupoosta [38]3 years ago

4 0

Answer:

Explanation:

Q1 = 35 nC = 35 x 10^-9 C

m = 3.5 micro gram = 3.5 x 10^-9 Kg

d = 35 cm = 0.35 m

(a) The electrostatic force between the two charges is balanced by the weight of another charge.

F = m g

$\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{d^{2}}=mg$

$Q_{2}=\frac{4\pi \epsilon _{0}mgd^{2}}{Q_{1}}$

(b) By substituting the values

$Q_{2}=\frac{3.5\times 10^{-9}\times 9.8\times 0.35\timees 0.35}{9\times 10^{9}\times 35\times 10^{9}}$

Q2 = 13.34 x 10^-12 C

Q2 = 0.0134 nC

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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

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