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3 years ago

What needs to be done to this circuit so that the light bulb lights up?

Physics

2 answers:

kramer3 years ago

6 0

A) close the switch
I took the test

Pachacha [2.7K]3 years ago

5 0

The answer is the first one a

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What is the frequency of a wave having a period equal to 18 seconds <br>

Ivanshal [37]

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

or, f = \frac{1}{T}f=

= \frac{1}{18 sec}

18sec

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>

3 0

3 years ago

As a capacitor is being charged in an RC circuit. the current flowing through a resistor isa) increasingb) decreasingc) constant

Montano1993 [528]

<h2>Answer: decreasing</h2>

An RC circuit is an electrical circuit composed of resistors and capacitors, where the charging time $T$ of the circuit is proportional to the magnitude of the electrical resistance $R$ and the capacity $C$ of the capacitor.

As shown below:

$T=R.C$

In this context, the electrical resistance is the opposition to the flow of electrons when moving through a conductor.

Therefore:

<h2>When a capacitor is being charged in an RC circuit, the current flowing through a resistor <u>decreases</u>.</h2>

And the correct option is b.

4 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$