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viktelen [127]
3 years ago
13

A 39.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to th

e edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0
Using the law of conservation of angular momentum, we have 

<span>I1 w1 = I2 w2 </span>

<span>ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 </span>

<span>ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) </span>

<span>= 1.8995 = 1.9 rad /sec ( nearly )</span>
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A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball
DerKrebs [107]
A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
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Solution :
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F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
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F = ma
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.:. The net force of object ( i ) has greater force compared to object ( ii ) by
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3 years ago
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kipiarov [429]

Answer:

it weighs 237469812734t7162341873498273417234321476281736481273648123764812736481723648273648137468127364872364 million pounds :)

Explanation:

4 0
2 years ago
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svet-max [94.6K]
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6 0
3 years ago
If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w
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Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

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E = \frac{h c}{\lambda _{o}}+K

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\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K

K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )

K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )

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3 0
2 years ago
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from th
zzz [600]

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i  the question;

mass of sun = M

minimum and maximum distance = r1 and r2 respectively

Now, using Kepler's third law,

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average distance a = ( r1 + r2 ) / 2

we know that

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T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

T = 2π √[( r1 + r2 )³ / 8GM ]

Therefore, the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

5 0
3 years ago
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