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3 years ago

15

For a sine wave depicting simple harmonic motion, the smaller the amplitude of the wave, the smaller the displacement of the pen

dulum from the equilibrium position. The shorter the period, the ? the pendulum’s rod.

2 answers:

SOVA2 [1]3 years ago

7 0

Shorter the pendulum rod, as the period of a pendulum is solely dependent on the length of the pendulum.

stepan [7]3 years ago

6 0

The correct answer is

The shorter the period, the shorter the pendulum’s rod

In fact, the period of a simple pendulum is given by:

$T=2 \pi \sqrt{\frac{L}{g}}$

where T is the period, L is the length of the pendulum's rod, g is the gravitational acceleration. We immediately see from the formula that T and L are proportional: therefore, if the period T is shorter, the length of the pendulum's rod L is shorter as well.

You might be interested in

What kind of light results when the total spectrum of refracted light is recombined? violet green white red?

ElenaW [278]

Hello
The final light will be white. In fact, each color of the visible spectrum is an electromagnetic wave with its own specific frequency and wavelength. White, instead, does not have a specific frequency: it is the sum of all the different wavelengths of the visible spectrum. Therefore, when recombining the spectrum of the refracted light all the different frequencies recombine together, and their sum gives white light.

(edited)

4 0

3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago

2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically

Leya [2.2K]

Work=f.d
Work=100*50 = 500
Power = work/time = 500/4
=125 watt

7 0

3 years ago

Which data set represents constant acceleration?

zlopas [31]

B

the second chart shows a constant acceleration

7 0

2 years ago