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3 years ago

A sample is brought to the laboratory and it is determined that one-eighth of the original

Physics

1 answer:

vlabodo [156]3 years ago

5 0

Answer:

Oh please dont be so dramatic

Explanation:

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6.Convert 22 °C to K. (°C = K - 273)

DIA [1.3K]

Answer:

295

Explanation:

if C = k - 273

k = C + 273

substitute 22 for C

K = 22 + 273

k = 295

3 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/

Jlenok [28]

The magnitude and direction of the velocity of the car at t = 8.00 s is v = 7,468 m / s and the vector direction = θ = 58,984°

<h3><em>Further explanation</em></h3>

In the coordinates of the two-dimensional plane (xy) there is a position vector of a point

You can write P as:

P = xi + yi

where i denotes the unit vector in the x and y directions

While the magnitude of the vector:

$\large{\boxed{\bold{P~=~\sqrt{x^2~+~y^2}}}}$

and its direction

$\large{\boxed{\bold{\theta~=~arc~tan\frac{y}{x}}}}$

Likewise with the instantaneous speed which is a derivative of the position function of time

$V =\frac{dP}{dt}$

$V ~=~\frac{dx}{dt}i~i~+\frac{dy}{dt}j$

V = vx i + vy j

The velocity of the car as a function of time is given by v = [5.00 m / s - (0.0180 m / s²) t²] i + [2.00 m / s + (0.550 m / s²) t] j so that:

vx = 5 - 0.0180t² i

vy = 2 + 0.55t j

So the instantaneous velocity at t = 8 s becomes:

vx = 5-0.0180.8²

vx = 3,848 m / s

vy = 2+ 0.55.8

vy = 6.4

So the velocity position vector becomes:

v = 3,848 i + 6.4 y

and the magnitude becomes:

$v=\sqrt{3.848^2+6.4^2}$

v = 7,468 m / s

While the velocity vector direction:

$\theta~=~=arc~tan\frac{vy}{vx}$

$\theta~=~arc~tan~\frac{6.4}{3.848}$

θ = 58,984

<h3><em>Learn more</em></h3>

Which vector should be negative

brainly.com/question/11236351

the statements that describe a vector

brainly.com/question/4417902

the resultant vector

brainly.com/question/11194474

Keywords: velocity, vector, two-dimensional coordinates,

4 0

3 years ago

Read 2 more answers

Verdadero o falso de La energía interna es la energía térmica asociada al objeto en virtud del movimiento de sus moléculas.

d1i1m1o1n [39]

Answer:

Es la energía cinética interna total de un objeto debido al movimiento aleatorio de sus ... a es la energía que posee un objeto o cuerpo en virtud del movimiento de sus ... Si el movimiento de estas moléculas y átomos es más rápido, se dice que el ... Por lo tanto, la energía térmica asociada con la conducción resulta de la .

Explanation:

5 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago