All categories

2 years ago

Riddle of the day What is so fragile that saying its name breaks it?

Physics

2 answers:

lidiya [134]2 years ago

7 0

Silence i’m pretty sure^

anyanavicka [17]2 years ago

6 0

answer:

The answer to this specific riddle is silence.

<em>-- please mark the brainliest and click the thanks button if correct :)</em>

You might be interested in

A student wants to start a small business in school. Write down six items that

Fiesta28 [93]

Answer:

packets of pen

packets of pencil

copies

books

bottles

mask

3 0

2 years ago

A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move awa

DerKrebs [107]

Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, $\lambda=53\ cm=0.53\ m$

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

$v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s$

So, the wave move with a speed of 7.95 m/s.

5 0

2 years ago

Complete the sentence to explain when waves interact.

True [87]

Answer:

Waves interact with objects and other waves.

Explanation:

8 0

2 years ago

A train rolls past a stationary observer. To him, the train is moving at a speed of 23m/s west, and a woman on the train is movi

GREYUIT [131]

Answer:

21.7 seconds.

Explanation:

Woman's velocity relative to train (23 m/s - 22.4 m/s) = 0.6 m/s

Distance woman wants to travel = 13m

To find how long she will take to move 13m relative to the train, take the distance she wants to travel divided by her velocity relative to the train.

(13m)/(0.6 m/s) = 21.6667 seconds or 21.7 seconds.

Therefore, it will take the woman 21.7 seconds to move 13m.

4 0

2 years ago

A fan blade, initially at rest, rotates with a constant acceleration of 0.029 rad/s2. What is the time interval required for it

LenKa [72]

Answer:

The time interval is $t = 21.30 \ s$

Explanation:

From the question we are told that

The constant acceleration is $\alpha = 0.029 \ rad / s^2$

The displacement is $\theta = 6.58 \ rad$

According to the second equation of motion we have that

$\theta = w_i* t + \frac{1}{2} * \alpha t^2$

given that the blade started from rest

$w_i$ which is the initial angular velocity is 0

$\theta = 0 + \frac{1}{2} * \alpha t^2$

=> $t = \sqrt{ \frac{2 * \theta }{\alpha } }$

substituting values

=> $t = \sqrt{ \frac{2 * 6.58 }{0.029 } }$

=> $t = 21.30 \ s$

6 0

3 years ago