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3 years ago

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An object with charge 4.3x10-5 C pushes another object 0.31 micrometers away with a force of 7 N. What is the total charge of th

e second object?

1 answer:

ad-work [718]3 years ago

5 0

Answer:

Charge on the other particle is given as

$q_2 = 1.74 \times 10^{-18} C$

Explanation:

As we know that the force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$F = 7 N$

$r = 0.31 \mu m$

$q_1 = 4.3 \times 10^{-5} C$

now we have

$7 = \frac{9 \times 10^9 (4.3 \times 10^{-5}q_2}{(0.31\times 10^{-6})^2}$

now we have

$q_2 = 1.74 \times 10^{-18} C$

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

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Bob, Jill, Kim, and Steve measure an object's length, density, mass, and brightness, respectively. Which student must derive a u

netineya [11]

The answer is A. Bob (<span>object's length)

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3 years ago

Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?

Leto [7]

The answer is 19.3 j

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3 years ago

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A sphere of volume 1.20×10−3m3 hangs from a cable. When the sphere is completely submerged in water, the tension in the cable is

KATRIN_1 [288]

Answer:

B = 62.9 N

Explanation:

This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid

B = ρ g V

write the equilibrium equation

T + B -W = 0

B = W- T (1)

use the density to write the weight

ρ = m / V

m = ρ V

W = ρ g V

substitute in 1

B = m g -T

B = $\rho_{body}$ g V - T

To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³

calculate

B = 7850 9.8 1.20 10⁻³ - 29.4

B = 92.3 - 29.4

B = 62.9 N

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3 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

charle [14.2K]

Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

Radius of circular arc = 0.27 m

Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

Calculate the total energy of the composite system at any time after the collision.

We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

$E=mgh+Mgh$

$E=(m+M)gh$

Put the value in to the formula

$E=(30\times10^{-3}+5.29)\times 9.8\times0.15$

$E=7.8\ J$

Hence, The total energy of the composite system is 7.8 J.

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4 years ago