What units are used to measure mass?

At a distance of 14.0 m from a point source, the intensity level is measured to be 80 dB. At what distance from the source will

Ivan

Answer:

$d_2=19.8m$

Explanation:

The problem can be solved easily, because the relationship between intensities $I_1$ and $I_2$ at distances $d_1$ and $d_2$ respectively can be written as:

$\frac{I_2}{I_1} =\frac{d_1^2}{d_2^2}$

Where:

$I_1=Intensity\hspace{3}1\\I_2=Intensity\hspace{3}2\\d_1=Distance\hspace{3}1\\d_2=Distance\hspace{3}2$

In this case:

$I_1=80dB\\I_2=40dB\\d_1=14m\\d_2=?$

Solving for $d_2$

$d_2=\sqrt{\frac{d_1^2 *I_1}{I_2} } =\sqrt{\frac{14^2*80}{40} } =\sqrt{392} =19.79898987m\approx19.8m$

8 0

3 years ago

Read 2 more answers

The density of a rock is 2 g/mL and its’ been displaced in water with a volume of 25 mL. Calculate the mass of the rock in grams

Ludmilka [50]

Answer:

d =

m

V

m = d×V

V =

m

d

DENSITY

Density is defined as mass per unit volume.

d =

m

V

Example:

A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?

Step 1: Calculate the volume

V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³

Step 2: Calculate the density

d =

m

V

=

433

g

200

c

m

³

= 2.16 g/cm³

MASS

d =

m

V

We can rearrange this to get the expression for the mass.

m = d×V

Example:

If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?

m = d×V = 500 mL ×

1.11

g

1

m

L

= 555 g

VOLUME

d =

m

V

We can rearrange this to get the expression for the volume.

V =

m

d

Example:

What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.

Step 1: Convert kilograms to grams.

14.83 kg ×

1000

g

1

k

g

= 14 830 g

Step 2: Calculate the volume.

V =

m

d

= 14 830 g ×

1

c

m

³

19.32

g

= 767.6 cm³

Explanation: got it from a site

4 0

4 years ago

Sally turns on her cellular telephone to speak to her friend who is located thousands of miles away. Which of the following best

Marizza181 [45]

As we know that in cellular phone all the signals are transmitted by antenna in the form of electromagnetic waves

All these electromagnetic waves are transmitted through the phones and then received as a signal to the receiver of another phone

These signals are then processed and converted into sound which is produced by the phone

So here correct answer will be

C. The cellular telephone transmits information by electromagnetic waves to a receiver which then encodes them and produces sound.

7 0

3 years ago

Read 2 more answers

2. A student drives 7.8-km trip to school and averages a speed of

Alekssandra [29.7K]

Answer:

The total time is: t=451.22 sec

The average speed is: V=34.57 m/s

Explanation:

Average speed

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

$\displaystyle V=\frac{x}{t}$

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

$\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec$

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

$\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec$

The total time is:

$t=239.26\ sec+211.96\ sec=451.22\ sec$

$\boxed{t=451.22\ sec}$

The average speed is:

$\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s$

$\boxed{\displaystyle V=34.57\ m/s}$

6 0

3 years ago

Which equation could not be used to determine straight line acceration?

uranmaximum [27]

Answer:

the answer for the question is the last option

5 0

3 years ago