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3 years ago

7

Your class decides to build a circuit to make a light bulb shine. When you are finished, you have a setup with wires, a light bu

lb and a battery.
What role does the battery play in this circuit design?
A) The battery works as the system's wires.
B) The battery is the electrical power supply.
C) The battery is the circuit's main conductor.
D) The battery works as the system's electrical switch.

2 answers:

Evgesh-ka [11]3 years ago

6 0

The answer to this question would be B (the battery is the electrical power supply)

Makovka662 [10]3 years ago

4 0

Answer:

B

Explanation:

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A baseball pitcher throws a baseball horizontally at a linear speed of 49.4 m/s. Before being caught, the baseball travels a hor

RideAnS [48]

Answer:

$v = 3.61 m/s$

Explanation:

As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s

so we will have

$t = \frac{x}{v_x}$

$t = \frac{24.7}{49.4}$

$t = 0.5 s$

now in the same time ball is turned by angle

$\theta = 52.7 rad$

now we know that

$\theta = \omega t$

$52.7 = \omega (0.5)$

$\omega = 105.4 rad/s$

now the tangential speed of a point at equator is given as

$v = r\omega$

$v = 0.0343(105.4)$

$v = 3.61 m/s$

4 0

4 years ago

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum

AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, $\omega=33\frac{1}{3} rev/min$ , radius of the record $r=0.1m$ and the distance between any two successive bumps on the groove as $d=1.75mm$ .

The linear speed of the record in meters per second is:

$v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\$

#From $v$ above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

$Hits/second=v/d \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199$

Hence the bumps hit the stylus at around 199hit/s

8 0

3 years ago

A piano tuner is using a 392 Hz tuning fork to tune the wire for a G-Natural note. She hears 4 beats per second. What are the tw

inysia [295]

A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,

$f_{beat} =| f_2\pm f_1 |$

The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be

$f_{beat} =|392+4|= 396Hz$

$f_{beat} =|392-4|=388Hz$

Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz

5 0

3 years ago

_______ measures an object's change in position per unit time.

Deffense [45]

I believe your answer should be C. Speed.

6 0

3 years ago

Read 2 more answers

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$

Where,

w = width

$\lambda =$ wavelength

m is an integer, m = 1, 2, 3...

We here know that as $sin\theta$ as w are constant, then

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}$

We need to find $w_2$ , then

$w_2 = \frac{w_1}{\lambda_1}\lambda_2$

Replacing with our values:

$w_2 = \frac{4.4*10^{-6}}{487}496$

$w_2 = 4.48*10^{-6}m$

Therefore the width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

3 0

3 years ago