Help..... disregard my answers...plz help

kramer

Explanation:

<em>The height of the pendulum is measured from the lowest point it reaches (point 3). </em>

At 1, the kinetic energy of the pendulum is zero (because it is not moving), and it has maximum potential energy.

At 2, the pendulum has both kinetic and potential energy, and how much of each it has depends on its height—smaller the height greater the kinetic energy and lower the potential energy.

At 3, the height is zero; therefore, the pendulum has no potential energy, and has maximum kinetic energy.

At 4, the pendulum again gains potential energy as it climbs back up, Again how much of each forms of energy it has depends on its height.

At 5, the maximum height is reached again; therefore, the pendulum has maximum potential energy and no kinetic energy.

Hope this helps :)

8 0

3 years ago

The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

8090 [49]

Α $=2.4 \frac{m}{s}$

β $=1.2 \frac{m}{s^2}$

$x(t)=at$

$y(t)=3-$ β $t^2$

$Vx(t)=$ α

$Vy(t)=-2$ β $t$

$vectorV=[$ α $;-2$ β $t]$

$ax(t)=0$

$ay(t)=-2$ β $t$

$vector a [0;-2$ β $t]$

3 0

3 years ago

A wave with a speed of 9 m/s and a frequency of 0.5 Hz has a λ of what?

patriot [66]

Wave speed = (wavelength) x (frequency)

Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

<em>Wavelength = 18 m</em>

6 0

2 years ago

The atomic number of a nucleus descended during which nuclear reactions?

kolbaska11 [484]

Answer:

(A.)Nuclear fission and beta decay (electron emission)

8 0

3 years ago

Read 2 more answers

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago