anonymous
  • anonymous
http://prntscr.com/8i2msx
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
Anyone there?
dan815
  • dan815
okay so to a Cathode in the CRT, would you normally connect DC current or AC current

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anonymous
  • anonymous
ok?
dan815
  • dan815
thats a question lol
anonymous
  • anonymous
well.... http://www.filedropper.com/physicspapermanekji
dan815
  • dan815
whats that
anonymous
  • anonymous
I need help with the entire paper..
dan815
  • dan815
i cant see that
dan815
  • dan815
okay so tell me
anonymous
  • anonymous
You'll Have To Download It.
anonymous
  • anonymous
http://prntscr.com/8i2tn7 there
dan815
  • dan815
okay look at question 1a
dan815
  • dan815
what happens when current is passed into the chatode
anonymous
  • anonymous
idk
dan815
  • dan815
think about it
dan815
  • dan815
what is a cathode
anonymous
  • anonymous
negative terminal
dan815
  • dan815
okay so when you complete the circuit
dan815
  • dan815
what will happen between the cathode and the anode
anonymous
  • anonymous
what u mean?
dan815
  • dan815
an electron is ejected from the cathode and goes to the anode
anonymous
  • anonymous
electron beam?
dan815
  • dan815
yes..
dan815
  • dan815
but you need a high enough voltage to displace an electron from the cathode
dan815
  • dan815
so when you are alternating the current during the lower voltages it will not displace an electron, this way you can control the time betweeen when your electrons are ejected, the frequency of your alternating current give you control on the frequency of ejection timings
dan815
  • dan815
an alternating current is a result of an alternating voltage and vice vera
anonymous
  • anonymous
question 1 b?
anonymous
  • anonymous
beacuse high resistance right?
dan815
  • dan815
what do you think is the answer to b
dan815
  • dan815
high resistance where
anonymous
  • anonymous
High resistance in the wire?
dan815
  • dan815
wire for what
dan815
  • dan815
You're going to have to be more clear if u want my help :)
dan815
  • dan815
I am not going to spoon feed you answers
anonymous
  • anonymous
Thick wires are used for lighting purposes because current is less?
anonymous
  • anonymous
higher the resistance lesser the current right?
dan815
  • dan815
the question is actually saying thick wires are used for power circuits rather than lighting purpose
dan815
  • dan815
okay lets think about this a little, why does a lightbulb light up
dan815
  • dan815
what is that fillament in the lightbulb really
dan815
  • dan815
|dw:1442671036320:dw|
anonymous
  • anonymous
i'm getting confused :(
dan815
  • dan815
the fillament in the lightbulb is basically just a resistor
dan815
  • dan815
its a very good resistor
dan815
  • dan815
what happens in a lightbulb is that you want lots of current to pass through the fillament, you want enough current to pass through that resistor to heat it up enough to make it emit enough photos, ie light
dan815
  • dan815
so you dont want needless current being lost to the wires
dan815
  • dan815
if u have thick wires your circult looks like this
dan815
  • dan815
|dw:1442671207500:dw|
dan815
  • dan815
your current will be the sum of the voltage/sum of all resistances
dan815
  • dan815
you just lowered your whole current needlessly
dan815
  • dan815
and there is less current through the fillament taking longer for it to heat up
dan815
  • dan815
now when it comes to power circuits
dan815
  • dan815
you dont want too much current going through resistors needlessly using up power in your circuit
dan815
  • dan815
also wearing down the life time of your circuit
dan815
  • dan815
so you get thicker wires to take the load off the components in power circuits
anonymous
  • anonymous
got it... q1c potential energy is mgh right?
dan815
  • dan815
yes
dan815
  • dan815
make sure your units make sense
anonymous
  • anonymous
i need to convert 200 gm to kg right?
dan815
  • dan815
yes
anonymous
  • anonymous
but they haven't given the h...
dan815
  • dan815
you know the acceleration due to earths gravity is -9.8m/s^2
anonymous
  • anonymous
yes..
dan815
  • dan815
this means you are losing a speed of 9.8 m/s every second
dan815
  • dan815
i will make that 10m/s per second to make it simpler for now, u calcuate the right answer
dan815
  • dan815
okay so suppose your accelerationg is -10m/s/s means u lose 10m/s of speed every second, you began with 20m/s so in 1 second you go from 20 m/s to 10m/s in another second you fo from 10 m/s to 0 m/s in 2 seconds your veloicty is 0m/s so your ball has come to a momentary stop up in the air, before it begins to head down
dan815
  • dan815
any questions so far?
anonymous
  • anonymous
PE = mgh i can't see the value of h
dan815
  • dan815
you dont have to use mgh
dan815
  • dan815
you can use normal physics
anonymous
  • anonymous
is the ans 4J?
dan815
  • dan815
okay if u wanna use mgh
dan815
  • dan815
you can use kinetic energy and potential energy
dan815
  • dan815
at the bottom there is only kinetic energy 1/2 mv^2 at the highest point there is only potential enrgy mgh
dan815
  • dan815
1/2mv^2=mgh solve for h
anonymous
  • anonymous
PE = KE=Mv^2/2?
dan815
  • dan815
okay, but i think you should really learn to solve it the other way too, energy theorem stuff comes after being able to solve the basic mechanics way
dan815
  • dan815
both are very important
anonymous
  • anonymous
energy theorem?
anonymous
  • anonymous
so m is 0.2 right?
dan815
  • dan815
yeah well you see from the equation the mass irrelevant
dan815
  • dan815
and it make sense, right if you were to drop any mass at some height, it will fall down with the same velocity
dan815
  • dan815
theoretically, but in reality, air resistance wont really allow that
dan815
  • dan815
anyways so 1/2mv^2=mgh mass cancels on both sides
anonymous
  • anonymous
2000?
dan815
  • dan815
wait lemme see
dan815
  • dan815
oh thats wrong
dan815
  • dan815
|dw:1442672220046:dw|
anonymous
  • anonymous
wait its 20m
dan815
  • dan815
yes
dan815
  • dan815
approximately remember we are using 10 m/s^2 for gravity instead of 9.8m/s^2
dan815
  • dan815
if u have a calculator use the right one for your answer
anonymous
  • anonymous
PE is 40J?
dan815
  • dan815
0.2kg*10m/s^2*20m=40j
dan815
  • dan815
yes
anonymous
  • anonymous
q1d?
dan815
  • dan815
okay so lets say some unsuspecting person happened to be touching the contact while he was pluggig it in
dan815
  • dan815
more current always wants to pass through the biggest voltage drop
dan815
  • dan815
if the ground is not in before your connection is made the current wants to pass through the person and u get the brunt of the current going through you
dan815
  • dan815
but if the prong is longer then that is always going to be connected inside before the other 2 go in
anonymous
  • anonymous
It's q related to sound
dan815
  • dan815
oh that one
anonymous
  • anonymous
like wavelength and all
dan815
  • dan815
okay well they gave u the distance and time for that so v=d/t
anonymous
  • anonymous
3/12?
dan815
  • dan815
yes
anonymous
  • anonymous
0.25 ms^-1?
dan815
  • dan815
1/4 m/s if u reduce
anonymous
  • anonymous
q2a?
anonymous
  • anonymous
yeah i reduced it to 1/4 then got 0.25 as d ans
dan815
  • dan815
ok u can do the rest, just work on all of them and post your solutions
dan815
  • dan815
i have to do my assignment
anonymous
  • anonymous
i'll try but....can u explain q2a to me?
anonymous
  • anonymous
wait its 2 m right?
anonymous
  • anonymous
moment - force times distance
anonymous
  • anonymous
=*
anonymous
  • anonymous
0.44
anonymous
  • anonymous
@dan815
dan815
  • dan815
whaat -.-
dan815
  • dan815
i got my own hw too
anonymous
  • anonymous
the ans to q2a is 0.447 m right?
dan815
  • dan815
no
anonymous
  • anonymous
whaat-.-
dan815
  • dan815
moment = force * distance 960Nm=430N * x
anonymous
  • anonymous
wait moment = force*distance so 960 = 430 times d 2.32?
dan815
  • dan815
yes
anonymous
  • anonymous
ok thanks ..
dan815
  • dan815
lol ..
anonymous
  • anonymous
i'll tag u if i need help
dan815
  • dan815
-.-
dan815
  • dan815
i wont come
anonymous
  • anonymous
:(
dan815
  • dan815
hahahah
dan815
  • dan815
just use youtube or google man
dan815
  • dan815
thats how we all learnt when we got stuck
anonymous
  • anonymous
thanks..
anonymous
  • anonymous
@hartnn

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