Nearly wet myself in the library with me headphones in!

"So that's where the GCSE ends pretty much"

This is probably my 3rd time watching this video and it still catches me off guard xD

Muhaimen Khan little red daleks are cute

@@aadhishshah4186 the TARDIS is blue tho

How do you feel now that Kimi retired?

I'm sorry, but I don't get it?

@Christopher Hodges ohhhh 😂 thanks for explaining

@Christopher Hodges to be fair I don’t think pull is strictly British slang but you probably don’t care 😂

For anyone who cant be bothered to go to 7:50 he basically says "you are going to be using this equation a lot especially probably the second year of your ANAL"

@@user-hq6sp no one cares

@@hackerpro6742 I care

😂😂😂😂😂😂

lol lol lol

Kuch Batein

good luck to you too!

🤣🤣

same g good luck :P

@@Aixzyl u2 bro

big last day

Same here, hopefully the grade boundaries are low

@@JohnDiggle21 yeah

how

@@Haris1 this guy has got me wondering man.

@@haroonqadir3474 what has he got you wondering

@@Haris1 how he ever wants to come back to this after the exam. I do realise it could maybe be a levels now. Pray for me have my gcse in a couple days

@@haroonqadir3474 I don't pray

How did you do

So imagine the diagram of the circle and how the line for the velocity is in a straight line. Due to this line being straight there is a small component of the velocity acting pushing the chair away from its circular path hence why the chair swings are being flung outward.

I believe the support force is just the reactive force of the structure onto the car so will act towards the centre due to the track being round

It's exactly the same as the support force of the track in the last example. For the bottom of a vertically circular path, the tension is solely responsible for the centripetal force, but also has to balance the weight, so T=mg+mv^2/r

dont forget the 2 graphs for the practical 2moro

Meharpal Basi I'm pretty sure you do. Its in the main text book

Yes.

Because we know that the force responsible for circular motion must be greater than the centripetal force. For example, if centripetal force = 5N, it is impossible for say 3N of upthrust to provide this, which is what multiplying by cos or sin would get you.

v=wr Where w is angular frequency (2pif)

HERE WE GOOOOOOOOOOO

See updated video, explained better

a bit late but v= ms-1, if you square that you get m^2 ^-2, dividing it by r gives ms^-2 therefore giving acceleration.

It depends what angle is used

Not a force pushing you out, just your body wanting to carry on going straight but the centripetal force pushing you inward.

No friction = no force = carries on straight into the hedge

1/f = T, so 2πR/T can be re-written as 2πfR

Yeah the upthrust force(F) is kinda confusing like what is it for when mg is there already?

A planes wing doesnt just generate lift in an upwards sense, it creates a force that is always perpendicular to it. In level flight all the force generated by the wing is matching the gravity - mg. But when the plane banks (rolls) the force generated by the wings changes direction accordingly. Also the wings lift changes depending on the application of the ailerons (up + down flight controlls). When the plane is banked the pilot will use the ailerons to increase the lift of the wing allowing for only the vertical component of lift to counter mg and the horizontal to provide the centripetal force. If the pilot didnt use the ailerons mg would be larger than the lift component and the plane would loose height.

When do I say that?

Science Shorts sorry I misunderstood when you said that “velocity is at a tangent to the object’s path.” But I see now that you mean it’s at a tangent to object’s path which is a circle. I understand now. Thankyou!

The ISS orbits the Earth every 90 minutes not 30, at an average altitude of 400 kilometres above the Earth. This was likely chosen as a safe minimum altitude that the Space Shuttle could reach, and that the ISS only required minimal occasional altitude boosts to avoid re-entry. The velocity it has is a function of its altitude. A higher altitude, less velocity. Lower altitude higher velocity. However, there is still some aerodynamic drag on the ISS due to residual atmosphere even at 400 kilometres, so it wouldn't be safe orbiting much lower, as the station is expected to remain in orbit for years, not just mere months.

When your going around a round about, there is a frictional force from your tires as your moving in a circle

Rohan kataria but how’s that acting towards the centre

@@ahmadghaffar6758 friction will have a sideways component acting in the direction of the centre of the roundabout.

same g also nice amvs

@@jacobsuresh3743 thanks 😊😊

All that is saying is that in order to fly level, there must be an upwards force equal to mg.

Science Shorts what is the force acting at the top left of the plane? Why? 14:03

@@evelynlau1929 I know this is really late but thats the lift provided by the wings.

+Cardifyz That would be true if the angle was to the vertical.

Why is it different?

Because the sin and cos would be swapped, resulting in cos(theta)/sin(theta), which is 1/tan(theta).