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lys-0071 [83]
3 years ago
14

You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho

ld a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 9.9 kg · m2 . When you pull the weights in toward your body, the moment of inertia decreases to 2.6 kg · m 2 . What is the resulting angular speed of the platform? Answer in units of rev/s. 017 (part 2 of 2) 10.0 point
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

\omega'=19.419\ rev.s^{-1}

Explanation:

Given:

angular speed of rotation of friction-less platform, \omega=5.1\ rev.s^{-1}

moment of inertia with extended weight, I=9.9\ kg.m^2

moment of inertia with contracted weight, I'=2.6\ kg.m^2

<u>Now we use the law of conservation of angular momentum:</u>

I.\omega=I'.\omega'

9.9\times 5.1=2.6\times \omega'

\omega'=19.419\ rev.s^{-1}

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

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Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc
kaheart [24]

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

f'=f_{1}-f...(I)

f'=f_{1}+f...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

f'=462-5=457\ Hz

f'=462+5=467\ Hz

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

4 0
3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
If yall friend me ill give lots of points like this in the future
Ne4ueva [31]

Answer:

ok......................

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2 years ago
What to do if vehicle struck by lightning
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Lighting flows around the outside of a truck, and the majority of the current flows from the cars metal cage into the ground below. It's not very safe to be in a car or truck during bad weather.
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Monochromatic light of a given wavelength is incident on a metal surface. however, no photoelectrons are emitted. if electrons a
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If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
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