For qn 1, what you could actually do was to solve the question using the eqns for both the horizontal and vertical components. Assume no air resistance, ur horizontal component eqn will be simply s=ut while the eqn for vertical component will be s=ut+1/2at^2.. Find t in terms of s and u using...
Well.. The best way that i still think is the best is to draw out all 10 cicuits.. However, you can still try to do this using some maths.
Let's not make this too complicated.. Split up the types of arrangements first.
Let resisitors in parallel be considered as 1 resistor.
Therefore, we...
hi..
Juz want to know if 2 waves were to interfere each other, in order to find the max intensity, is it possible to juz add the 2 wave intensity to find the max? if given that wave 1 intensity is
The weight of the object doesn't spread out, weight of the object will alwas act at the center of gravity of the block.. What will change with area is the pressure acting on the surface of the blocks.
The difference doesn't lies with the speed of the block, but the force on the object. When You pull the block out fast, the force on the block is much larger.
I don't really noe if these examples are correct..
Let me state an example.. Let's say I've got an adjustable giant clamp.. in...
In this case, the car must have enough PE to convert into KE at the bottom of the loop and back into PE as it moves around the loop.. Therefore:
change PE_i= change KE_f= change PE_f
simplifying it, change PE_i= change PE_f , provided there is no energy loses, if there is, then add it in...
was there any friction? or are u juz working on non-frictionless surface? You certainly nid the coefficient of friction if there is friction and the friction force acting on the chair.. before formulas can be applied to solve for the weight of the person.
How do i prove this \sum_{r=1}^n (r^5)=1/12n^2(n+1)^2(2n^2+2n-1)
when i have already proven this 2.. That \sum_{r=1}^n (r^3)=1/4n^2(n+1)^2
and \sum_{r=1}^n (r^3+3r^5)=1/2n^3(n+1)^3
The problem is that i know how to solve the first 2 but dun know how to solve the big one using...
Force on M= Tension in string=Force on m.
The only forces in the system are gravity on mass M and the tension of the string.. the weight of m is balanced by the normal force exerted back by the table.. Thus, since there is also no friction both on the table surface and the pully, the force on M...
The skier only experiences two forces..Friction and the normal force by the surface.. Weight is a force exerted by the skier on the surface.. there's a different..
This 2 force the skier experiences are indirect 'reactions' of the the components of the force, weight.
ok..I shall only help you with the first part coz the second part is all bout doing backwards..
Now,.. U have the angle of incline and the force of the object, ie, the person's weight. You can actually make everything look easier by drawing a free body diagram on the person..Draw out the only...
This is a qn on 2D motion, You can solve it using one of the four formulas in kinematics.. See what are the given quantities and what is it that u are finding.
Erm.. Not really exactly the same as what the magicians are doing..
If you were to look really carefully, magician tablecloth tricks uses glasswares that have little area of contact with the tablecloth or that the plates have large bases that increases its stability when they are made to stand...