Learning Goal: To understand how to find the wavelength and diffraction patterns of electrons.

A rock falls off a cliff and falls for 3 secs. How high was the cliff?

SVEN [57.7K]

44.1 meters initial height of the rock.

3 0

3 years ago

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is

levacccp [35]

Answer:

-775 rad/s^2

Explanation:

Knowing the initial and final angular speed is $\omega_0 = 440.9 rad/s$ and $\omega = 0 rad/s$ , respectively. We can use the following formula for equation of motion to calculate the average angular acceleration in t = 0.569 seconds

$\Delta \omega = \alpha t$

$\omega - \omega_0 = \alpha t$

$0 - 440.9 = \alpha 0.569$

$\alpha = \frac{-440.9}{0.569} = -775 rad/s^2$

6 0

3 years ago

Which causes an air conditioner

BigorU [14]

Answer:

The wind blowing mark me brainliest

Explanation:

6 0

3 years ago

What should you do if...

Hunter-Best [27]

Answer:

3. you need to ask your available lab instructors what to do.

4. You immediately have to drop down your cloth and roll it to extinguish the fire or move to the emergency shower if available

5. You have too keep calm and report to the lab instructor but do no shout.

6. Move immediately to the eye rinse basin if available and wash your eyes gently and thoroughly

6 0

3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

2 years ago