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2 years ago

How would you describe the difference between momentum and angular momentum?

Physics

1 answer:

zheka24 [161]2 years ago

3 0

Answer:

Analogous

Explanation:

Momentum is linear

Angular momentum is spinning

momentum is based on the mass and has no geometric component.

Angular momentum is based on moment of inertia and greatly depends on how far the mass is from the center of spin.

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A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe

tekilochka [14]

Answer:

Angular momentum, $L=6.47\times 10^{-3}\ m$

Explanation:

It is given that,

Radius of the axle, $r=3.21\ mm=3.21\times 10^{-3}\ m$

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

$\tau=\dfrac{dL}{dt}$

Torque acting on the top is given by :

$\tau=F\times r$

Here, F is the tension acting on it. Torque acting on the top is given by :

$\tau=2F\times r$

$2T\times r=\dfrac{L}{t}$

$L=2T\times r \times t$

$L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32$

$L=6.47\times 10^{-3}\ m$

So, the angular momentum acquired by the top is $6.47\times 10^{-3}\ m$ . Hence, this is the required solution.

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2 years ago

How often should cardio exercises be performed? A. Never B. Three to five times a month C. Three to five times a week D. Three t

EleoNora [17]

Answer:

C. Three to five time a week.

Explanation:

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2 years ago

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You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³ / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

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2 years ago

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

7 0

2 years ago

What are the effects of pressure on liquids?

Vitek1552 [10]

Answer: Changes in pressure have very little effect on the volume of a liquid. Liquids are slightly incompressible because any increase in pressure can only slightly minimize the distance between the closely packed molecules. Hope this helps.

Explanation:

3 0

2 years ago

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