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PIT_PIT [208]
2 years ago
15

The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time

the bulbs (resistors) stay lit in each circuit.
Physics
1 answer:
8_murik_8 [283]2 years ago
3 0

The concept used in this is circuit analysis using the simplification of resistors and capacitors.

Explanation:

The time constant for each of the circuits in figure A, B, C, D and E. Therefore, rank the length of time the bulbs stay lit from longest to shortest by using the value of time constant for each circuit. The rank of the time constant of the circuit is C > A = E > B > DC  > A = E > B > D.

Capacitance is the central concept in electrostatics and constructed devices called capacitors are essential elements of electronic circuits.

If more charge is placed on the conductor the voltage increases proportionately. The ratio of the charge to the voltage is called the capacitance C of the conductor C= q/v.

The resistance increases if you add resistors in series and decreases if you add them in parallel. on the other hand the capacitors increases if it is added parallel and decreases if added in series. hence the circuit longest time constant takes longest time to discharge.

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A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
liq [111]

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

5 0
1 year ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
Read 2 more answers
he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th
AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x 10^{-6} T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ_{0}I/2πr

where

B = magnetic field (T)

μ_{0} = 1.26 x 10^{-6} (N/A^{2})

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ_{0}I/2πr

     = (1.26 x 10^{-6})(4.7)/(2π)(0.19)

     = 4.96 x 10^{-6} T

BV = μ_{0}I/2πr

     = (1.26 x  10^{-6})(4.7)/(2π)(0.41)

     = 2.3 x 10^{-6} T

hence,

the net magnetic field = BH - BV

                                     = 4.96 x 10^{-6} - 2.3 x 10^{-6}

                                     = 2.66 x 10^{-6} T

4 0
1 year ago
For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin
nydimaria [60]

Answer:

The width of slide is 0.092 mm.

Explanation:

Given that,

Wave length = 680 nm

Distance between slit and screen D= 5.4 m

Distance of bright band 2y = 7.9 cm

y =\dfrac{7.9}{2}

y=3.95\ cm

We need to calculate the width of slide

Using formula of width

d=\dfrac{\lambda D}{y}

Where, d = width

D =Distance between slit and screen

y = Distance of bright band

Put the value into the formula

d=\dfrac{680\times10^{-9}\times5.4}{3.95\times10^{-2}}

d=0.092\times10^{-3}\ m

d=0.092\ mm

Hence, The width of slide is 0.092 mm.

4 0
2 years ago
A Body OF Volume 36cc Floats With 3/4 of its volume submerged in water . The density Of Body is
Radda [10]

Answer:

Density of body = 0.25g/cc

Explanation:

Given:

Volume submerged in water = 3/4

Find:

Density Of Body

Computation:

Density of body = fraction of body in liquid x density of water

Density of body = [1-3/4]1

Density of body = 0.25g/cc

8 0
1 year ago
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