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Virty [35]
3 years ago
5

14. Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenari

os below indicate what changes (if any) take place (1 pt. ea.):
a. Total resistance of the circuit
b. Total current of the circuit
c. Power dissipated by the circuit
Physics
1 answer:
Inessa [10]3 years ago
8 0
In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.

a).  In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.

b).  In series, the total current is  V / (3R) .
In parallel, the total current is  3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.

c).  In series, the power dissipated by the circuit is 

                                   (V) · V/3R  =  V² / 3R .

In parallel, the power dissipated by the circuit is

                                   (V) · 3V/R  =  3V² / R .

Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.

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Wavelength of blue photons 495 nm, what is the frequency? and what is the energy?
inessss [21]

Answer:

1.F: About 6*10^14 Hz

2.E: About 4*10^ -19 J

Explanation:

Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz).  By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)

=6.05*10^14 /s

Energy: The energy photon contains can be calculate by this formula-- E=hf

f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.

So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20=  4.051*10^-19 J

3 0
3 years ago
A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm
Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm

So, the thickness of the film is equal to 3.1 cm.

8 0
2 years ago
Which substance might lead to potentially addiction if abuse
uranmaximum [27]
Heroin is the number 1 most addictive substance.
4 0
3 years ago
Read 2 more answers
A substance freezes at -58c therefore substance melts at ?
Yanka [14]
The freezing point is the same as the melting point.

If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
3 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
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