All categories

3 years ago

A wave has a wavelength of 10 mm and a frequency of 14 hertz. What is its speed?

Physics

1 answer:

Katena32 [7]3 years ago

3 0

0.14 meters per second .

You might be interested in

Help! Il give brainlest to who answers first

Readme [11.4K]

Answer:

1. The density of the cube is 1.03 g/mL.

2. Dish soap

Explanation:

1. Determination of the density of the cube.

From the question given above, the following data were obtained:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Thus, density can be expressed mathematically as:

Density (D) = mass (m) / volume (V)

D = m / V

With the above formula, we can obtain the density of the cube as follow:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

D = m / V

D = 21.7 / 21

D = 1.03 g/mL

Thus, the density of the cube is 1.03 g/mL.

2. Determination of the layer of density the cube will settle in.

From the question given above,

Subtance >>>>>>>> Density

Vegetable oil >>>>> 0.91 g/mL

Grape juice >>>>>> 0.97 m/L

Water >>>>>>>>>>> 1 g/mL

Dish soap >>>>>>>> 1.03 g/mL

Maple syrup >>>>>> 1.37 g/mL

Comparing the density of the cube (i.e 1.03 g/mL) with those in the table able, we can conclude that the cube will settle in the DISH SOAP layer since they both have the same density.

7 0

3 years ago

Potassium is a crucial element for the healthy operation of the human

Degger [83]

Answer:

The mass of the Potassium-40 is $m_{40}} = 2.88*10^{-6} kg$

The Dose per year in Sieverts is $Dose_s = 26.4 *10^{-10}$

Explanation:

From the question we are told that

The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

Their abundance is 93.26%, 0.012% and 6.728%

The mass of potassium contained in human body is $m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg$ per kg of the body

The mass of the first body is $m_1 = 80 \ kg$

Now the mass of potassium in this body is mathematically evaluated as

$m_p = m * m_1$

substituting value

$m_p = 80 * 0.0003$

$m_p =0.024 kg$

The amount of Potassium-40 present is mathematically evaluated as

$m_{40}} =$ 0.012% * 0.024

$m_{40}} = \frac{0.012}{100} * 0.024$

$m_{40}} = 2.88*10^{-6} kg$

The dose of energy absorbed per year is mathematically represented as

$Dose = \frac{E}{m_1}$

Where E is the energy absorbed which is given as $E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}$

Substituting value

$Dose = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}$

$Dose = 22*10^{-10} J/kg$

The Dose in Sieverts is evaluated as

$Dose_s = REB * Dose$

$Dose_s = 1.2 * 22*10^{-10}$

$Dose_s = 26.4 *10^{-10}$

3 0

3 years ago

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago