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vitfil [10]
3 years ago
14

In a certain oscillating LC circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in

the inductor in 1.40 μs. What are (a) the period of oscillation in microseconds and (b) the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again?
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

A = 5.6μs

B = 178.57kHz

C = 2.8μs

Explanation:

A. It takes ¼ of the period of the circuit before the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor.

t = T/4

T = 4*t

T = 4 * 1.4 = 5.6μs

B. f = 1/T

Frequency is the inverse of period

f = 1 / 5.6*10⁻⁶

f = 178571.4286Hz

f = 178.57kHz

C. time taken for maximum energy to occur is T/2

t = 5.6 / 2 = 2.8μs

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Why is the ozone layer important?
adoni [48]

Answer:

Ozone is a gas in the atmosphere that protects everything living on the Earth from harmful ultraviolet (UV) rays from the Sun. Without the layer of ozone in the atmosphere, it would be very difficult for anything to survive on the surface.

Explanation:

Ozone layer, also called ozonosphere, region of the upper atmosphere, between roughly 15 and 35 km (9 and 22 miles) above Earth’s surface, containing relatively high concentrations of ozone molecules (O3). Approximately 90 percent of the atmosphere’s ozone occurs in the stratosphere, the region extending from 10–18 km (6–11 miles) to approximately 50 km.

7 0
3 years ago
Read 2 more answers
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u
kenny6666 [7]

Answer:

induced emf =  28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt  = Adb/dt

induced emf =  A × ΔB / Δt

induced emf =  πr² × ΔB / Δt

induced emf =  π (0..65)² × ( 0.61 - (-0.28))  / 0.13

induced emf =  0.0286538 V

so induced emf =  28.65 mV

3 0
2 years ago
Why do you choose MCB in place of a fuse?
astraxan [27]

It protects the electrical appliance and the person from electrical shocks/faults.

Hope it Helped!

3 0
3 years ago
A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms
zmey [24]

Answer:

2156J

Explanation:

Given parameters:

Height of lift  = 10m

Mass  = 22kg

Unknown:

Work done by the machine  = ?

Solution:

Work done is the force applied to move a body through a certain distance.

So;

        Work done  = Force x distance

Here;

       Work done  = mass x acceleration due to gravity x height

      Work done  = 22 x 9.8 x 10  = 2156J

5 0
3 years ago
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