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4 years ago

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The drawing shows three objects rotating about a vertical axis. The mass of each object is given in terms of m0, and its perpend

icular distance from the axis is specified in terms of r0. Rank the three objects according to their moments of inertia, largest to smallest.

1 answer:

photoshop1234 [79]4 years ago

7 0

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

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A curent-carrying 2.0-cm long segment of wire is inside a long solenoid adius - 40 cm - 800 turns/m, current - 50 mA). The wire

klasskru [66]

Answer:

The value is $F_{max} = 12 \muN$

Explanation:

From the question we are told that

The length is $l = 2.0 \ cm = 0.02 \ m$

The radius of the the solenoid is $r =40 \ cm$

The number of turns per meter is $N = 800 \ turns / m$

The current through the solenoid is $I = 50 \ mA = 50*10^{-3} \ A$

The current through the segment is $I_s = 12 \ A$

Generally the magnetic force is mathematically represented as

$F = B * I_s l sin(\theta)$

At maximum $\theta = 90^o$

So

$F_{max} = B * I_s l$

Here B is the magnetic field is mathematically represented as

$B = \mu_o * N * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$B = 4\pi * 10^{-7} * 800 *50*10^{-3}$

=> $B = 5.0272 *10^{-5} \ T$

So

$F_{max} = 5.0272 *10^{-5} * 12 * 0.02$

$F_{max} = 1.2 *10^{-5}$

$F_{max} = 12 \muN$

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3 years ago

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Norma-Jean [14]

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3 years ago

A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

$\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h$

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

$V=\dfrac{x}{t}$

Where x is the distance, t is the time and v is the speed. using the data that we have:

$V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h$

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

$V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km$

7 0

3 years ago

A skier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angleof

victus00 [196]

Answer:

Explanation:

When the skier reaches the bottom of the slope , height lost by it

h = 50 sin32 m

= 26.5 m

potential energy lost

= mgh

Gain of kinetic energy

= 1/2 mv²

mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

Let μ be the coefficient of kinetic friction required.

friction force acting

= μmg

work done by friction in displacement of d (40 m ) on horizontal surface

- μmg x d

This negative work will be equal to positive kinetic energy of the skier on horizontal surface .

= μmg x d = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

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4 years ago

A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (un

Charra [1.4K]

Answer:

1.9 s

Explanation:

We are given that

Length of cable=l=15 m

We have to find the time you have to move out of the way.

We know that

Time period,T= $2\pi\sqrt{\frac{l}{g}}$

Where g= $9.8m/s^2$

By using the formula

$T=2\pi\sqrt{\frac{15}{9.8}}$

$T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s$

Time you have to move out

$t=\frac{T}{4}=\frac{7.77}{4}=1.9 s$

Hence,time you have to move out of the way=7.77 s

6 0

3 years ago