Mike wants to build an amplifier. Which technology can he use?

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 92.2 kg hop on board for a ride

Anvisha [2.4K]

Answer:

k = 1400.4 N / m

Explanation:

When the springs are oscillating a simple harmonic motion is created where the angular velocity is

w² = k / m

w = $\sqrt{ \frac{k}{m} }$

where angular velocity, frequency and period are related

w = 2π f = 2π / T

we substitute

2π / T = \sqrt{ \frac{k}{m} }

T² = 4π² $\frac{m}{k}$

k = π² $\frac{m}{T^{2} }$

in this case the period is T = 1.14s, the combined mass of the children is

m = 92.2 kg and the constant of the two springs is

k = 4π² 92.2 / 1.14²

k = 2800.8 N / m

to find the constant of each spring let's use the equilibrium condition

F₁ + F₂ - W = 0

k x + k x = W

indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant

(k + k) x = W

2k x = W

k_eq = 2k

k = k_eq / 2

k = 2800.8 / 2

k = 1400.4 N / m

5 0

3 years ago

A gas mixture contains 320mg methane, 175 mg argon, 225 mg nitrogen (N2). The partial pressure of argon at 300K is 12.52 kPa. Wh

svet-max [94.6K]

Answer: The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 320 mg = 0.3 g (Conversion factor: 1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol$

For argon:

Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol$

For nitrogen:

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol$

To calculate the volume of the mixture, we use the equation:

PV = nRT ......(2)

We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

$12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L$

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

$P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa$

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

5 0

3 years ago

What are two reasons the period of Venus is shorter than the period of earth

Setler79 [48]

Answer: The planet, named after the Roman goddess of art and beauty, can actually be bright enough to cast shadows on a moonless night. It appears so close to the sun because its orbital radius is smaller than the Earth's, and because it also moves faster than Earth, its orbital period is shorter.

7 0

3 years ago

The atomic number of an element is 19. Which is true of this element?

bezimeni [28]

C.
Each atom has 19 proton
 the atomic number is the same as protons

4 0

3 years ago

How many grams does 15.58 L of gasoline weigh? Gasoline has a density of 0.74 g/mL

Korolek [52]

We are given volume of gasoline = 15.58 L and

Density = 0.74 g/mL.

Density is given in grams(g) per milliliter (mL). So, we need to convert given volume in mL also.

We know, 1 litre = 1000 milliliters

Therefore, 15.58 L = 15.58 * 1000 mL = 15580 mL

We could rewrite,

Volume of gasoline = 15580 mL.

Formula for density is..

Density (ρ)= $\frac{Mass}{ Volume}$

Plugging values in formula,

0.74 g/mL = $\frac{Mass}{15580 \ mL}$

Muliplying both sides by 15580.

$0.74 g/mL * 15580mL =\frac{Mass}{15580 \ mL} * 15580mL$

11529.2g = mass.

Therefore, 11529.2g grams does 15.58 L of gasoline weigh.

5 0

3 years ago