How do you the average speed

A 3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

anyanavicka [17]

Answer:

I think the answer is

Explanation:

140N/m/20N =

7m is the answer

8 0

2 years ago

Read 2 more answers

A hollow conductor is positively charged. Asmall uncharged metal ball is lowered by a silk thread through asmall opening in the

dimulka [17.4K]

Answer:

Explanation:

According to the property of a conductor, the entire charge will reside on the outer surface of the conductor, there is no charge on the inner side of the conductor. As the uncharged metal ball touches the inner surface of the conductor, it does not attain any charge as the inner side of the conductor has no charge.

So option (c) is correct.

8 0

3 years ago

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

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