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Sergio [31]
3 years ago
10

Convert 0.700 to scientific notation

Physics
1 answer:
ELEN [110]3 years ago
7 0

Answer:

7 × 10^-1

Explanation:

Hope this helps :) mark branliest if you could!

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At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does
insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

V = V_R + V_L

The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

where

\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s is the time constant of the circuit

At time t=0,

V_L(0) = V e^0 = V = 20 V

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

V_R = V-V_L = 20 V-20 V=0

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

Since \tau is at the order of less than milliseconds.

Using eq.(1), we see that for t >> \tau, the exponential becomes zero, and therefore the potential difference across the coil is zero:

V_L = 0

Therefore, the potential difference across the resistor will be

V_R = V-V_L = 20 V- 0 = 20 V

(c) Yes

The two voltages will be equal when:

V_L = V_R (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

Substituting into (2) we find

V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V

So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

We said that the two voltages will be equal when

V_L=\frac{V}{2}

Using eq.(1), and this last equation, this means

V e^{-\frac{t}{\tau}} = \frac{V}{2}

And solving the equation for t, we find the time t at which the two voltages are equal:

e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

I_0 = 3.20 A

The problem says this current is stable: this means that we are in a situation in which t>>\tau, so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

V_L = 0

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7
Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I=\frac{P_{avg}}{A}

I=\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}

I=7.296\times 10^{-6} W/m^2

Amplitude of electric field at receiver end

E_{max}=\sqrt{2I\mu _0c}

Amplitude of induced emf

=E_{max}d

=\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75

=17.591\times 10^{-2}=0.1759 v

7 0
3 years ago
How many type of expansion physic
Yuki888 [10]

Answer:

Three types of thermal expansion are linear expansion,s superficial expansion,cubical expansion

4 0
3 years ago
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