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3 years ago

14

A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at

150 rpm . Part A What is the speed of the 150 g ball? Express your answer to two significant figures and include the appropriate units

1 answer:

stira [4]3 years ago

6 0

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

Angular speed ,ω = 150 rpm

The speed in rad/s

$\omega =\dfrac{2\pi N}{60}$

$\omega =\dfrac{2\pi \times 150}{60}$

ω = 15.7 rad/s

The distance of center of mass from 150 g

$r=\dfrac{150\times 0+240\times 33}{150+240}\ cm$

r= 20.30 cm

The speed of the mass 150 g

v= ω r

v= 20.30 x 15.7 cm/s

v= 318.71 cm/s

v= 3.18 m/s

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Umnica [9.8K]

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So, 180 * 1/3 = 60 degrees.

We aren't done yet.

60 degrees is not the angle which you have turned from your path.

You turned 120 degrees to describe an angle of 60 degrees on the interior side.

The answer is B.

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3 years ago

An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu

never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

$N_{12_{c}}=5.02\times10^{22}\ nuclie/g$

For $N_{c_{14}}$

$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

$N_{c_{14}}=6.526\times10^{10}\ nuclei/g$

Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

$R=15.0650\ decay/min g$

100 g carbon will decay with rate

$R=100\times15.0650=1507\ decay/min$

We need to calculate the total half lives

$(\dfrac{1}{2})^{n}=\dfrac{390}{1507}$

$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

$t=1.948\times5700$

$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

5 0

4 years ago

A certain wire has a resistance of 110 Ω. What is the resistance of a second wire, made of the same material, that is 1/4 as lon

Svet_ta [14]

Answer:

New Resistance = 247.5 ohm

Explanation:

Resistance = resistivity * length / area

Since resistivity for the material is constant, resistance is directly proportional to (length/area).

This means that if (length/area) decreases or increases by any ratio, then resistance will increase or decrease by the same ratio.

So let's find the change in length/area

New length = 0.25 old length

New area = (1/9) old area (This is because area equation contains a square of the diameter. if diameter decreases by 1/3, area decreases by (1/3)^2 )

So we now get length /area:

New length / new area = ( 0.25 old length) / (1/9 of old area)

New length / new area = 9*0.25 (old length / old area)

New length / new area = 2.25 (old length / old area)

To get the new resistance, we simply multiply it by the ratio we just found.

This equals:

110 * 2.25 = 247.5 ohm

4 0

3 years ago

Read 2 more answers

Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

bearhunter [10]

Answer:

$strain = 1.4 \times 10^{-3}$

Explanation:

As we know by the formula of elasticity that

$E = \frac{stress}{strain}$

now we have

$E = 110 GPA$

$E = 110 \times 10^9 Pa$

Area = 15.2 mm x 19.1 mm

$A = 290.3 \times 10^{-6}$

now we also know that force is given as

$F = 44500 N$

here we have

stress = Force / Area

$stress = \frac{44500}{290.3 \times 10^{-6}}$

$stress = 1.53 \times 10^8 N/m^2$

now from above formula we have

$strain = \frac{stress}{E}$

$strain = \frac{1.53 \times 10^8}{110 \times 10^9}$

$strain = 1.4 \times 10^{-3}$

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3 years ago

a bicyclist travels with an averahe velocity of 15km/h north, for 20 minutes .what is his displacement ?

Lunna [17]

Explanation:

velocity = 15 km/hr = 15×1000m / 60×60s

= 4.17 m/s

time= 20 min= 20 × 60s

= 1200s

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we know that,

V = D / t

or, 4.17 = D / 1200

or, D = 4.17 × 1200

or,D = 5004 m

7 0

3 years ago