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tensa zangetsu [6.8K]

3 years ago

14

Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if

the system it is rotated about (a) one end of the rod and (b) the center of the rod?

1 answer:

sergij07 [2.7K]3 years ago

7 0

Explanation:

a) $I=\displaystyle \sum_{i}m_ir_i^2$

where $r_i$ is the distance of the mass $m_i$ from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

$I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2$

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

$\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2$

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4 years ago

A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the

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The strength of the electric field is 5 N/C

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The magnitude of the electric field produced by a single-point charge is given by:

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3 years ago

Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of t

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Answer:

speed of golf ball is 1.15 × $10^{-30}$ m/s

and % of uncertainty in speed = 2.07 × $10^{-30}$ %

Explanation:

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mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = $10^{-3}$ m

to find out

speed of the golf ball and % of speed of the golf ball

solution

we will apply here heisenberg uncertainty principle that is

uncertainty position ×uncertainty momentum ≥ $\frac{h}{4\pi }$ ......1

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here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × $10^{-34}$ Js

so put here all these value in equation 1

$10^{-3}$ × 0.0459 × ΔVx = $\frac{6.626*10^{-34}}{4\pi }$

ΔVx = 1.15 × $10^{-30}$ m/s

and

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3 0

3 years ago

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co

Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

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Length = 31 cm

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Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

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We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

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$t=6402.31\ sec$

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Hence, The time is 106.7 minute.

7 0

3 years ago