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3 years ago

Differentiate scalars & vectors?

2 answers:

6 0

A scalar quantity is a one dimensional measurement of a quantity, like temperature, or mass. A vector has more than one number associated with it. A simple example is velocity. It has a magnitude, called speed, as well as a direction, like North or Southwest or 10 degrees west of North.

ira [324]3 years ago

3 0

Scalar is distance and magnitude while Vector is distance with direction as well as displacement and velocity

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A wave has a length of 8 meters and a speed of 32 meters per second. What is the frequency of the wave

Naddika [18.5K]

Answer:

The frequency of wave is 4 hertz.

Explanation:

Given,

wave length = 8m

speed of sound ( v) = 32 m/s

frequency ( f) = ?

we know,

v = f × lambda

or, 32 = f × 8

or, f = 32/8

or, f = 4 Hertz.

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4 years ago

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Lilit [14]

I think the answer is b

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4 years ago

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The environment contains

Alchen [17]

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3 years ago

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Boltzmann’s constant is 1.38066×10^−23 J/K,

dmitriy555 [2]

Answer:

1.18266×10⁻²⁰ J

Explanation:

Applying,

E = (3/2)kT.................... Equation 1

Where E = kinetic energy of the gas molecule, k = Boltzmann's constant, T = Temperature

But,

PV = nRT.............. Equation 2

Where P = pressure, V = Volume, n = number of moles, R = Universal gas constant.

make T the subject of the equation

T = PV/nR............. Equation 3

Substitute equation 3 into equation 1

E = (3/2)k(PV/nR)

E = 3kPV/2nR............... Equation 4

Given: k = 1.38066×10⁻²³ J/K, V = 6.6 L = 0.0066 m³, P = 7.1 atm = (101325×7.1) N/m² = 719407.5 N/m², n = 2.9 mol, R = 8.31451 J/K.mol

Substitute into equation 4

E = (3×1.38066×10⁻²³×0.0066×719407.5)/(2×8.31451)

E = 1182.66×10⁻²³ J

E = 1.18266×10⁻²⁰ J

8 0

3 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

KATRIN_1 [288]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

3 0

3 years ago