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3 years ago

Which of the following is a form of potential energy

Physics

2 answers:

Strike441 [17]3 years ago

4 0

I found that it is either nuclear or electrical, I am not really sure which one it would be out of the two.

Sorry I didn't fully answer your question, Good Luck, Hope I helped:)

Zepler [3.9K]3 years ago

3 0

the ancwer is B sound hope it helps

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Analyze how friction can be both a positive and negative aspect on our lives

denpristay [2]

Positive - Friction allows us to create heat in a desperate situation, like being lost in the woods. If we didn't have friction, we would probably freeze to death.
Negative - Friction can also cause unwanted fires, such as forest fires. If friction didn't exist, we wouldn't have these.

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3 years ago

What is the difference between mass and density?

Fantom [35]

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4 years ago

Read 2 more answers

A heater is being designed that uses a coil of 14-gauge nichrome wire to generate 300 W using a voltage of V = 110 V . How long

Mazyrski [523]

Answer:

The length of the wire is 83.2 m.

Explanation:

Given that,

Power of heater, P = 300 W

Voltage, V = 110 V

The resistivity of nichrome wire is, $\rho=100\times 10^{-8}\ \Omega-m$

The electric power of a wire is given by :

$P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{300}\\\\R=40.34\ \Omega$

Area of cross section of the wire is :

$A=\pi r^2\\\\A=\pi (0.000815)^2\\\\A=2.08\times 10^{-6}\ m^2$

Resistance of a material is given by :

$R=\rho \dfrac{L}{A}\\\\L=\dfrac{RA}{\rho}\\\\L=\dfrac{40\times 2.08\times 10^{-6}}{10^{-6}}\\\\L=83.2\ m$

So, the length of the wire is 83.2 m. Hence, this is the required solution.

5 0

3 years ago

11. A fundamental property of light is that it: 15

myrzilka [38]

Answer:

Curves around objects

Explanation:

Diffraction is a property of light described by bending of light around an object. This ability of light to bend around edges has facilitated optical effects of light where there is interference of light waves. Other properties of light are: reflection, refraction, polarization, scattering of light, and interference of light.

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3 years ago

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

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