All categories

2 years ago

Describe sound and record

Physics

2 answers:

Iteru [2.4K]2 years ago

8 0

Audible: a sound that is loud enough to hear

A record is a database entry that may contain one or more values

oksian1 [2.3K]2 years ago

3 0

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

You might be interested in

Which of the following is a mood disorder?

gulaghasi [49]

Bipolar disorder is the answer

5 0

3 years ago

Read 2 more answers

Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.

lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, $q=3\times 10^{-9}\ C$

Distance of the given point from the point charge is, $d=3\ cm=0.03\ m$

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

$E=\frac{kq}{d^2}$

Plug in $k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m$ . Solve for 'E'.

$E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C$

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0

3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

jeka94

Answer:

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt ( 2.6^2 + 0.15^2)

d_l = sqrt ( 6.7825 )

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)

t_l - t_r = ( 0.0043 / 343 )

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157

3 0

3 years ago

If the net force acting on an object is zero, its inertia is also zero.<br> True or false

S_A_V [24]

Answer:

its false

Explanation:

5 0

3 years ago

A current of 15.0 A flows through an electric heater operating on 120 V. Compute the power consumption of the electric heater.

Rzqust [24]

Answer:

There will be 1800 W power consumption in heater

Explanation:

We have given current flowing in the heater I = 15 A

Voltage on which heater is operating V = 120 volt

We have to find the power consumption in the heater

We know that power consumption is given by P = VI

So power consumption in heater = 120 × 15 = 1800 W

So there will be 1800 W power consumption in heater

5 0

3 years ago

Describe sound and record​

Describe sound and record