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Artemon [7]
4 years ago
5

Why current remains same in series combination of resistors in all resistors and p.d. remains different?

Physics
1 answer:
IgorC [24]4 years ago
4 0
Current at all points of a series circuit must be the same, because there's no place in the circuit where electrons are being manufactured, and no place where they're leaking out and falling on the floor. The nimber of electrons that leaves the loop is the same number that entered it. I'm not sure what is nmeant by "p.d. remains different" .
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A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
A visible violet light emits light with a wavelength of 4.00 × 10-7 m.
RideAnS [48]

Answer:

The correct option is B. 7.5 * 10¹⁴ Hz

Explanation:

Frequency = (speed) / (wavelength)

= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)

= (3/4 x 10¹⁵) ( m / m - s )

= (0.75 x 10¹⁵) /sec

= 7.5 x 10¹⁴ Hz

= 750,000 GHz

4 0
3 years ago
Read 2 more answers
The Astronomer Carl Sagan is famous. Why?
Zolol [24]
Mainly because he was Johnny Carson's advisor and consultant
on space, astronomy, and science in general, and he appeared
on The Tonight Show Starring Johnny Carson many times.
6 0
3 years ago
Please help extra points!!!!
Allushta [10]

Answer:

If you increase either mass or velocity, the momentum of the object increases proportionally.

6 0
3 years ago
What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w
Delvig [45]
(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
f_1 =  \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where
L is the wire length
T is the tension
m is the wire mass

In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
T= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }= \frac{1}{2 \cdot 10.9 m} \sqrt{ \frac{253 N}{0.0558 kg/10.9 m} }=    10.2 Hz

b) The frequency of the nth-harmonic for a standing wave in a wire is given by
f_n = n f_1
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz

c) Similarly, the third lowest frequency (third harmonic) is given by
f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz

8 0
3 years ago
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