2 hours of tv had 40 commercials __ __commercials per hour (Ratio/Rate)

J u l i a n has partially filled the p r i s m below with 16 unit cubes. How many more unit cubes will he need to completely fil

Wittaler [7]

add it

Step-by-step explanation:

11+20+25+36

8 0

3 years ago

is the product of a fraction that is greater than one and any whole number less than or greater than the whole number?Explain yo

katovenus [111]

Greater because the product of the fraction is greater than 1

4 0

3 years ago

Solve each problem by writing and solving an equation

Viktor [21]

Answer:

The anwser would be 51 i think

Step-by-step explanation:

if im right ls makr brainliest

3 0

3 years ago

Circle towns limit form a perfectly circular shape that has a population of 20000 in the population density of 480 people per sq

Alex

Answer:

$D =\frac{P}{\pi X^2}$

And solving for the radius we got:

$X = \sqrt{\frac{P}{\pi D}}$

And replacing the data given we got:

$X = \sqrt{\frac{480 \frac{people}{km^2}}{\pi *20000 people}}= 0.0874Km$

And this value converted to meters is $X = 87.40 m$

Step-by-step explanation:

For this case we know the population size $P = 20000$ and we also know the population density $D = 480 \frac{people}{km^2}$

We can assume that the area is a circle. We also know that the formula for the population density is given by:

$D= \frac{P}{A}$

Where P represent the number of people and A the area. Since we are assuming a circle then the area is given by:

$A = \pi X^2$

With X the radius of the circle

And then the populationd density become:

$D =\frac{P}{\pi X^2}$

And solving for the radius we got:

$X = \sqrt{\frac{P}{\pi D}}$

And replacing the data given we got:

$X = \sqrt{\frac{480 \frac{people}{km^2}}{\pi *20000 people}}= 0.0874Km$

And this value converted to meters is $X = 87.40 m$

6 0

4 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

so

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

4 years ago