Which equation is used to determine the density of a substance

One joule of work is needed to move one coulomb of charge from one point to another with no change in velocity. Which of the fol

podryga [215]

It B the current is one ampere

6 0

4 years ago

For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn

sdas [7]

Answer:

a) $v_{max} = 2\ \textup{m/s}$

b) $v_{avg} = 1\ \textup{m/s}$

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

$u(r) = 2(1-\frac{r^2}{R^2} )$

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

$v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )$

or

$v_{max} = 2\ \textup{m/s}$

Now,

$v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}$

or

$v_{avg} = \frac{2}}{2}\ \textup{m/s}$

or

$v_{avg} = 1\ \textup{m/s}$

Now, the flow rate is given as:

Q = Area of cross-section of pipe × $v_{avg}$

or

Q = $\frac{\pi D^2}{4}\times v_{avg}$

or

Q = $\frac{\pi 0.04^2}{4}\times 1$

or

Q = 1.256 × 10⁻³ m³/s

7 0

3 years ago

A 9 m3 container is filled with 300 kg of r-134a at 10°c. what is the specific enthalpy (kj/kg) of the r-134a in the container?

andreev551 [17]

Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg

8 0

3 years ago

A 26.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.10 m -long str

mart [117]

Answer:

$7.3 ms^{-1}$

Explanation:

Consider the motion of the ball attached to string.

In triangle ABD

$Cos50 = \frac{AB}{AD} \\Cos50 = \frac{AB}{L}\\AB = L Cos50$

height gained by the ball is given as

$h = BC = AC - AD \\h = L - L Cos50\\h = 1.10 - 1.10 Cos50\\h = 0.393 m$

$M$ = mass of the ball attached to string = 110 g

$V$ = speed of the ball attached to string just after collision

Using conservation of energy

Potential energy gained = Kinetic energy lost

$Mgh = (0.5) M V^{2} \\V = sqrt(2gh)\\V = sqrt(2(9.8)(0.393))\\V = 2.8 ms^{-1}$

Consider the collision between the two balls

$m$ = mass of the ball fired = 26 g

$v_{o}$ = initial velocity of ball fired before collision = ?

$v_{f}$ = final velocity of ball fired after collision = ?

using conservation of momentum

$m v_{o} = MV + m v_{f}\\26 v_{o} = (110)(2.8) + 26 v_{f}\\v_{f} = v_{o} - 11.85$

Using conservation of kinetic energy

$m v_{o}^{2} = MV^{2} + m v_{f}^{2} \\26 v_{o}^{2} = 110 (2.8)^{2} + 26 (v_{o} - 11.85)^{2} \\v_{o} = 7.3 ms^{-1}$

3 0

3 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

4 years ago