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3 years ago

Is distance traveled by a moving train qualitative discrete continuous or na?

1 answer:

8 0

Continuous. Discrete values are values like 1, 2, 3, 4, etc. - they're values that are distinct, and typically there's some idea of a next and a previous value. When we're counting whole numbers, there's a definitive answer to which number comes after, and which number comes before. With continuous values, there's no real "next" or "last" value.

Motion is measured with continuous values; a train might move 300 yards in 1 minute, but we can look at smaller and smaller chunks of time to keep getting shorter and shorter distances. There is no "next" distance the train moves after those 300 yards - it just doesn't make sense for there to be.

It's also measured quantitatively, not qualitatively. This just means that we can use numerical values to measure it, rather than other descriptors like color, smell, or taste.

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Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c

kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

$\lambda = \frac{c}{f}$

Substituting,

$\lambda = \frac{344}{1250}$

$\lambda = 0.2752m$

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

$sin\theta = \frac{m \lambda}{W}$

Where,

W is the width

m is the integer

$\lambda$ the wavelenght

Re-arrange the expression,

$\theta = sin^{-1} \frac{m\lambda}{W}$

For m=1,

$\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°$

For m=2,

$\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°$

For m=3,

$\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°$

The angle of diffraction is directly proportional to the size of the wavelength.

5 0

3 years ago

Though some self-report error may exist, surveys/interviews are useful in that they allow researchers to collect large amounts o

Leokris [45]

Definitely true, surveys and interviews aren’t flawless but you can collect lots of data from them

4 0

3 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

3 years ago

A quartz crystal vibrates with a frequency of 93621 Hz. What is the period of the crystals motion? T=1/93621 does not give me th

hram777 [196]

Answer:0.0000107s

Explanation:

f=1/T

T=1/f

f=93621 1/s

t=1/93621

T=0.00001068 s

3 0

3 years ago

A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘

hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution, $T_f$ = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

$\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC$

$\Delta T_f=i\times K_f\times m$

$Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}$

$5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})$

On solving we get:

$\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol$ ....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of $Ca(NO_3)_2$ in the mixture:

$\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%$

The mass percentage of calcium nitrate is 31.23%.

5 0

3 years ago