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alisha [4.7K]
3 years ago
8

What is the mass of 1.0 l of water in grams?

Physics
1 answer:
Maksim231197 [3]3 years ago
5 0
The mass of 1.0 l of water in grams is 1,000 g ;) 
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What energy output objects work with the turbine
Angelina_Jolie [31]

Answer:

maybe alternator..generator..

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3 years ago
Which statement is true about the inner planets of our solar system
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B they formed from the dense elements as compared to the others in the solar system
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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface cha
Ivan

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  \eta  =  -\epsilon _o E

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮E \cdot dA =  \frac{-Q}{\epsilon _o}

Where E is the electric field

          dA change in unit area

           -Q is the negative charge

          \epsilon _o  is the permittivity of free space

So

          EA  =  \frac{-Q}{\epsilon _o }

           \frac{Q}{A}  =  -\epsilon _o E

          \eta  =  -\epsilon _o E

Where \eta is the charge density

   

8 0
3 years ago
If Juan upgrades from a sports car to a large truck, what will happen to the
Brrunno [24]

Answer:

The force must increase

Explanation:

According to newton's second law "force is the product of mass and acceleration".

      Force  = mass  x acceleration

Now, the mass of the sports car is lesser compared to that of the truck. Therefore, to take both automobiles to the same speed, enough force must be applied by the engine of the truck.

There must be an increase in the force in order to make both automobiles attain the same speed.

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