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Nesterboy [21]
2 years ago
13

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle conn

ected to a hose can be estimated by the formula where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units
Physics
1 answer:
suter [353]2 years ago
5 0

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × \frac{V^{2} }{2g}

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

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3 0
3 years ago
Two glasses of water have the same thermal energy. must they have the same temperature? explain.
taurus [48]
No,because they  may have more particles 
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3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
In an electricity experiment, a 1.10 g plastic ball is suspended on a 56.0 cm long string and given an electric charge. A charge
Shalnov [3]

Answer:

Tension, T = 0.0115 N                      

Explanation:

Given that,

Mass of the plastic ball, m = 1.1 g

Length of the string, l = 56 cm

A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

T\cos\theta=mg

T is tension in the string

T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{1.1\times 10^{-3}\times 9.8}{\cos(21)}\\\\T=0.0115\ N

So, the tension in the string is 0.0115 N.

8 0
3 years ago
A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. De
Len [333]

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation  a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

\int\limits^a_b {x} \, dx

<em>a=2</em>

<em>b=1</em>

<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>

<em>dx=dv</em>

Integrate the left side the standard method.

\int\limits^a_b {x} \, dx

<em>a=v</em>

<em>b=0</em>

<em>dx=dv</em>

<em>Integrating</em>

=v^2/2

Use Simpson's rule for the right site.

\int\limits^a_b {x} \, dx

<em>a=b</em>

<em>b=a</em>

<em>x=f(x)</em>

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

        =0.8376

Solve for v.  

      v=1.295

5 0
3 years ago
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