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Brrunno [24]
3 years ago
15

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

surface with a tangential speed of 5 m/s. The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.250 m. What is the instantaneous speed of the mass at the moment the string reaches a length of 0.250 m?
Physics
1 answer:
drek231 [11]3 years ago
8 0

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

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We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

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a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26}  } \\t=0.125mm

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Answer:

Al's mass is 102.92  kg  

Explanation:

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F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

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m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

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where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

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2 years ago
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Answer:

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Explanation:

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m_{1} = mass of first body = 2.7 kg

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b)

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m_{2} = mass of second body = 1.35 kg

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Suppose you are talking about a Gold Coin. If you heated it so it melted, the gold would retain its value, but the fact that it is a coin and valuable as such, means that it has lost that part of its value.

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