Evaluate (x +y)0 for x= -3 and y=5.<br> 0 1<br> 2<br> 01

A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the wate

Vilka [71]

Answer:

f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

∑ F = ma

where the acceleration is

a = $\frac{d^2 y}{dt^2 }$

B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

B = W

In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains

B’= m $\frac{d^2 y'}{dt^2 }$

the thrust is given by the Archimedes relation

B = ρ_liquid g V_liquid

the volume is

V = π r² y'

we substitute

- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

$\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0$

this differential equation has a solution of type

y = A cos (wt + Ф)

where

w² = ρ_liquid g π r² /m

angular velocity and frequency are related

w = 2π f

we substitute

4π² f² = ρ_liquid g π r² / m

f = $\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }$

calculate

f = $\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }$

f = 5.3 Hz

6 0

3 years ago

If you doubled the load resistor in a Wheatstone bridge, the load current would not be half as much. Why not?

Fofino [41]

Doubling the size of a load resistor increases the load current. Increasing the load resistance, in turn the total circuit resistance is reduced. The load current would not be half as much since when you increase the size of load resistor then load current increases.

6 0

3 years ago

Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar ( $a_{D}$ ), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities ( $v_{D}$ , $\omega_{BD}$ ), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations ( $a_{D}$ , $\alpha_{BD}$ ), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$ (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$ (2)

<h3>Accelerations</h3>

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$ (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$ (4)

If we know that $\theta = 60^{\circ}$ , $\gamma = 19.889^{\circ}$ , $r_{BD} = 10\,in$ , $\omega_{AB} = 16\,\frac{rad}{s}$ , $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$ , then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

$v_{D}+3.402\cdot \omega_{BD} = -41.569$ (1)

$9.404\cdot \omega_{BD} = -24$ (2)

$v_{D} = -32.887\,\frac{in}{s}$ , $\omega_{BD} = -2.552\,\frac{rad}{s}$

<h3>Accelerations</h3>

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$ (3)

$-9.404\cdot \alpha_{BD} = -687.264$ (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$ , $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557