Physics: The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters and t is in seconds. determine the angular momentum of the particle about the origin as a function of time.

Following are the equation of continuation expression: are the initial area and velocity, respectively, and are the end area and velocity. Therefore, Using the Bernoulli equation at the height of constant. Let the Human bool is the average density then: Learn more about the blood viscosity:link...

The position vector of a particle of mass 1.65 kg

Ответ:

username999990

11.01.2022

Physics

L = 41.09 Kg m2 / s The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

p = m v

Let's replace

L = m r x v

The given function is

x = 6.00 i ^ + 4.15 t j ^

We look for speed

v = dx / dt

v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

$L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]$

L = m 6 4.15 k ^

The other products give zero

Let's calculate

L = 1.65 6 4.15 k ^

L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

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Ответ:

iBrain

01.03.2022

Physics

Following are the equation of continuation expression:

$\to A_1 V_1= A_2 V_2 \\\\$

$A_1\ and \ V_1$ are the initial area and velocity, respectively, and $A_2\ and \ V_2$ are the end area and velocity.

$\to A_1 V_1= A_2 V_2 \\\\\to v_2= \frac{A_1}{A_2} \times v_1\\\\$

$= \frac{\frac{\pi}{4} d_1^2}{ \frac{\pi}{4} d_2^2 } \times v_1\\\\= (\frac{d_1}{d_2})^2 \times v_1\\\\$

$\to d_2= 0.74\ d_1 \\\\$

Therefore,

$\to v_2 = (\frac{d_1}{d_2})^2 \times v_1\\\\$

$= (\frac{d_1}{0.74 \ d_1})^2 \times 0.15 \ \frac{m}{s}\\\\= 0.2739 \ \frac{m}{s}\\\\$

Using the Bernoulli equation at the height of constant.

$\to P_1+ \frac{\rho v_1^2}{2} + \rho gh_1 = P_2 + \frac{\rho v_2^2}{2} + \rho gh_2\\\\\to P_1-P_2 =\frac{\rho}{2} (v_2^2-v_1^2)\\\\\to \Delta P =\frac{\rho}{2} (v_2^2-v_1^2)\\$

Let the Human bool is the average density then:

$\to \rho =1060 \ \frac{kg}{m^3}\\\\$

$\to \Delta P =\frac{\rho}{2} (v_2^2-v_1^2)\\\\$

$= \frac{(1060 \frac{kg}{m^3})}{2} ((0.2739 \frac{m}{s})^2 -(0.15 \frac{m}{s})^2) \\\\= \frac{(1060 \frac{kg}{m^3})}{2} (0.07502121- 0.0225) (\frac{m}{s})^2) \\= \frac{(1060 \frac{kg}{m^3})}{2} (0.05252121) (\frac{m}{s})^2) \\\\= 27.836 \ Pa\\\\$

Learn more about the blood viscosity:

link

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The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters and t is in seconds. determine the angular momentum of the particle about the origin as a function of time.

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