To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

$1atm = 2116lb/ft^2$

The ideal gas equation said us that,

PV = nRT

Here,

P = pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = Amount of substance (at this case the mass)

Then

$\frac{n}{V} = \frac{P}{RT}$

The amount of substance per volume is the density, then

$\rho = \frac{P}{RT}$

Replacing with our values,

$\rho = \frac{5.6*2116}{1716*850}$

$\rho = 0.00812slug/ft^3$

Finally the specific volume would be

$v = \frac{1}{\rho}$

$v = 123ft^3/slug$

6 0

3 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

4 0

2 years ago

Please help. I don’t understand this

skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

v (ave) = ∆x / ∆t

and under constant acceleration,

v (ave) = (v (final) + v (initial)) / 2

According to the plot, with ∆t = 4.0 s, we have v (initial) = 0 and v (final) = 10.0 m/s, so

∆x / (4.0 s) = (10.0 m/s) / 2

∆x = ((4.0 s) • (10.0 m/s)) / 2

∆x = 20.00 m

5 0

2 years ago

A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy

snow_lady [41]

Answer:

The distance required = 16.97 cm

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

e = 16.97 cm

Thus the distance required = 16.97 cm

6 0

3 years ago