http://prntscr.com/8i2msx

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

http://prntscr.com/8i2msx

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Anyone there?
okay so to a Cathode in the CRT, would you normally connect DC current or AC current

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question