All categories

3 years ago

What is the unit of work? Explain whether the unit of work is a fundamental uint

Physics

1 answer:

Levart [38]3 years ago

3 0

Answer:

1)the unit of work is joule. 2)it is derived unit because it is made by two different unit force and displacement .3)

You might be interested in

True north and magnetic north are not the same. the difference between them is called:

8090 [49]

The difference between true north and magnetic north is called "declination."

4 0

3 years ago

Một người đi xe đạp trong 40 phút với vận tốc là 12km/h. Hỏi quãng đường đi được là bao nhiêu?

Karo-lina-s [1.5K]

Answer:

khoảng cách = 8km / h

Explanation:

khoảng cách = tốc độ / thời gian

khoảng cách = 12/40

khoảng cách = 8km / h

3 0

2 years ago

Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong

sveta [45]

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI

8 0

3 years ago

Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not

d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

$I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},$

where $I_0(\text{at }1000\text{ Hz})$ is the hearing threshold at 1000 Hz.

Therefore we have the following

$0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}$

$I_1$ is the threshold at 1000Hz and $I_2$ is the threshold at 100Hz.

By exponentiating we have

$10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.$

Now dividing these two equations we get

$\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.$

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.