Transverse waves travel perpendicular to neighboring wave motion like a slinky while longitudinal waves depend on the displacement of a medium which include sound waves.

8 0

3 years ago

An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has

kolezko [41]

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is $1.50\times 10^{-3}\text{ C}$ .

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, t, is given by

$I(t) = I_0e^{-t/\tau}$

Here, $I_0$ is the maximum current and τ represents time constant which is given by RC (the product of the resistance and capacitance).

The maximum current is given by

$I = \dfrac{V}{R_\text{eff}}$

V is the emf of the battery and $R_\text{eff}$ is the effective resistance.

In this question, $R_\text{eff}$ = 10.0 Ω + 25.0 Ω = 35.0 Ω

$I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}$

(b) The maximum charge is given

Q = CV

where C is the capacitance of the capacitor

$Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}$

3 0

3 years ago

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A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of

bekas [8.4K]

Answer:

The magnitude of the induced Emf is $0.003371V$

Explanation:

The width of the truck is given as 79inch but we need to convert to meter for consistency, then

The width= 79inch × 0.0254=2.0066 metres.

Now we can calculate the induced Emf using expresion below;

Then the $induced EMF= B L v$

Where B= magnetic field component

L= width

V= velocity

=(40*10^-6) × (42) × (2.0066)

=0.003371V

Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V

8 0

4 years ago