Which equations represent the relationship between wavelength and frequency for a sound wave? Check all that apply.

How to increase the current in a circuit

ivann1987 [24]

Answer:

So to increase current of the circuit what you can do is :

1. Use conductor of low resistivity, ¶.

2. Use conductor of small length.

3. Use thick wire.

4. Decrease the temperature of the circuit.

5. If operating temprature is high than use semiconductor, because it have negative temprature coefficient.

6. Minimise the circuit losses.

3 0

2 years ago

Why are electromagnets used in metal scrap yards<br> PLEASE HELP!!!!!!

Alinara [238K]

To help pick to up the metal and place it in a incinerator <span />

4 0

3 years ago

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

Archy [21]

Answer:

The voltage is $V = 0.993V_b$

Explanation:

From the question we are told that

The time that has passed is $t = \frac{\tau}{2}$

Here $\tau$ is know as the time constant

The voltage of the power source is $V_b$

Generally the voltage equation for charging a capacitor is mathematically represented as

$V = V_b [1 - e^{- \frac{t}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\tau}{2\tau} }]$

=> $V = V_b [1 - e^{- \frac{1}{2} }]$

=> $V = 0.993V_b$

5 0

3 years ago

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

julia-pushkina [17]

Answer:

$5.5\cdot 10^{-11} C$

Explanation:

The capacitance of the parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=190 mm^2 = 190 \cdot 10^{-6} m^2$ is the area of the plates

$d=1.2 mm = 0.0012 m$ is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F$

The energy stored in the capacitor is given by

$U=\frac{Q^2}{2C}$

Since we know the energy

$U=1.1 nJ = 1.1 \cdot 10^{-9} J$

we can re-arrange the formula to find the charge, Q:

$Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C$

8 0

3 years ago

The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not

Nuetrik [128]

Answer:

P(bat) = V²r/(R+r)²

Explanation:

Let the resistance of the coil be R

Internal resistance of the battery be r

Emf of the battery = V

Power dissipated in the internal resistance of the battery is normally given as P = I²r

where I is the current flowing in the circuit.

From Ohm's law,

V = I R(eq)

R(eq) = (R + r)

I = V/(R+r)

P = I²r

P = [V/(R+r)]²r

P = V²r/(R+r)²

Hope this Helps!!!

6 0

3 years ago