I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
A) Rubber stops charges from flowing. This protects people by stopping electricity from flowing.
Explanation:
The statement that best describes the point of wrapping rubber around the copper wire is that the rubber stops charges from flowing. This prevents people from getting electrical shocks by stopping the flow of electricity.
- A rubber is an insulator.
- Insulators are substances that prevents the flow of electricity.
- The lack free mobile electrons or ions that makes them conductors.
- When they are wrapped round a conductor such as copper wire, they will halt the flow of charges.
- Copper is a conductor of both heat and electricity. It has free mobile electrons.
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Answer:
Part a)
Part b)
Part c)
Part d)
Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as
for uniform cylinder we will have
so we have
now we have
now we have
Part b)
speed of the bucket can be found using kinematics
so we have
Part c)
now in order to find the time of fall we can use another equation
Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance
