_____ are group of tissue working together to perform a certain job.

Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.

KengaRu [80]

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

5 0

3 years ago

Identify five common business uses for electronic spreadsheets

Sloan [31]

For the answer to the question above
Forecasting how a business might do in the future.
Calculating tax.
Doing basic payrolls.
Calculating Revenues.
Producing charts.

--Going past 5--

Inventory tracking
Very (VERY) basic CRM for small businesses
I hope my answer helped you.

4 0

4 years ago

Read 2 more answers

Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

$x_{cm}$ ’= 1/92 ((24 (-1.9) +56 3.9)

$x_{cm}$ ’= 1/92 (172.8)

$x_{cm}$ ’= 1.88 cm

This distance is to the right of the central marble

3 0

3 years ago

A traveler pulls on a suitcase strap at an angle 36° above the horizontal. If of work are done by the strap while moving the sui

ollegr [7]

Answer:

The tension in the strap is 74.82 N.

Explanation:

Given that,

Angle between the horizontal and the suitcase is 36 degrees.

The distance traveled by the suitcase is 15 meters.

Let the work done by the suitcase is 908 J. We know that the work done in the vector form is given by :

$W=Fd\ \cos\theta\\\\908=F\times 15\times cos(36)\\\\F=74.82\ N$

So, the tension in the strap is 74.82 N. Hence, this is the required solution.

6 0

3 years ago

A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kpa, when the gas is allowed to expa

Katarina [22]

By idea gas equation we can say

$PV = nRT$

here P = pressure

V = volume

n = number of moles

R = Avagadro gas constant

T = temperature

So from above expression we can see that for constant number of moles we can say

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$\frac{475*4}{290} = \frac{P_2*6.5}{277}$

$6.55 = 0.023*P_2$

$P_2 = 279.1 kPa$

So finally the pressure will be 279.1 kPa

7 0

3 years ago