Any clue for this question?

We will determine the amount of electric energy stored in a capacitor by discharging it through a light bulb. Light bulbs are ra

choli [55]

Answer:

C. the product of the power rating of the light bulb and the time that it remains lit.

Explanation:

The power rating of the light is bulb is defined as the energy supplied to the light bulb divided by the time the bulb is lit up. Therefore,

$P = \frac{E}{t}\\\\E = Pt$

where,

E = Energy Supplied to the bulb = Energy stored in capacitor = ?

P = Power rating of the bulb

t = time the bulb is lit up

Hence, the correct option is:

<u>C. the product of the power rating of the light bulb and the time that it remains lit.</u>

7 0

3 years ago

A steel ball of mass 3 kg moving at 2.4 m/s to the left hits a stationary, 4 kg glob of clay. The ball becomes imbedded in the c

Kitty [74]

Answer:

V = 1.02 m/s

Explanation:

Given that,

The mass of a steel ball, m₁ = 3 kg

The mass of glob of clay, m₂ = 4 kg

The speed of a steel ball, u₁ = 2.4 m/s

The speed of glob, u₂ = 0 (at rest)

Let V be the combined speed of the system. Using the conservation of energy to find the combined speed.

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{3(2.4)+4(0)}{(3+4)}\\\\V=1.02\ m/s$

So, the combined speed is 1.02 m/s in left direction.

3 0

3 years ago

A train that is 268.7 m long undergoes constant acceleration the moment the last car (end of

Soloha48 [4]

Answer:

its 667m just took the test

Explanation:

3 0

3 years ago

An engine flywheel initially rotates counterclockwise at 5.03 rotations/s. Then, during 23.5 s, its rotation rate changes to 2.6

Sveta_85 [38]

Answer:

the flywheel's average angular acceleration is -2.05 rad/s²

Explanation:

<u>Note:</u> counterclockwise is positive

clockwise is negative

Given;

initial angular velocity, $\omega _i$ = 5.03 rev/s = $5.03\frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 31.61 \ rad/s$

final angular velocity, $\omega_f$ = -2.63 rev/s = $-2.63 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = -16.53 \ rad/s$

duration of the flywheel rotation, Δt = 23.5 s

The average acceleration of the flywheel is calculated as;

$a_r = \frac{\Delta \omega}{\Delta t} = \frac{\omega_f - \omega _i}{t_2-t_1} = \frac{-16.53 \ - \ 31.61}{23.5} = -2.05 \ rad/s^2$

Therefore, the flywheel's average angular acceleration is -2.05 rad/s²

5 0

3 years ago

If a violin string vibrates at 248 Hz as its fundamental frequency, what are the frequencies of the first four harmonics?

Andre45 [30]

Answer:

The frequencies of the first four harmonics are 248 Hz, 496 Hz, 744 Hz, and 992 Hz

Explanation:

Given;

fundamental frequency, f₀ = 248 Hz

Frequency of the nth harmonic = n x fundamental frequency

First harmonic, f₁ = 1 x f₀ = 248 Hz

Second harmonic;

2f₀ = 2 x 248 Hz = 496 Hz

Third harmonic

3f₀ = 3 x 248 Hz = 744 Hz

Fourth harmonic

4f₀ = 4 x 248 Hz = 992 Hz

Therefore, the frequencies of the first four harmonics are 248 Hz, 496 Hz, 744 Hz, and 992 Hz

6 0

3 years ago