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I have one question related to Dynamics. Suppose you throw a ball at an angle in air, or release it from rest so it falls under free-fall or you throw the ball upwards, there can be only 2 forces acting (air resistance and weight) on the ball?
Is there no type of applied force there in the downward direction which opposes air resistance when you throw and release the ball in the downward direction from top?
And an applied force in the direction of the object similarly which opposes air resistance in the case it is released at an angle? And even when you throw an object upwards, there is no applied force in the upward direction, but only weight and air resistance acting on the object?
(By applied force I mean the force along the direction in which the object is travelling.)
Thank youYou are right that there are only 2 forces acting on the ball in any case i.e. the weight of the ball and air resistance. The weight ALWAYS acts downwards but air resistance acts in the OPPOSITE direction of the ball's motion.
When you release it from a point above ground, the ball moves downwards so the air resistance acts upwards and weight downwards.
When it is thrown upwards, both the weight and air resistance act downwards.
When thrown at an angle, weight acts downwards and resistance is changing direction every instant as the ball changes direction.
I did. Still I dont quite get it.
View attachment 64541 can
somebody pls help me put with this??
Hi Donenr,Hello, I would like about Q22 in the 9702/11/m/j/18 I want to ask for Why is the question B and not A?
View attachment 64548
Here is the examiner's report on the question
View attachment 64550
View attachment 64586
From O/N 2017 paper 11
The circuit diagram shows four resistors of different resistances P,Q,R and S connected to a battery. (SEE IMAGE) The voltmeter reading is zero. Which equation is correct?
A) P - Q = R - S
B) P - S = Q - R
C) PQ = RS
D) PS = QR
THANKS FOR THE HELP IN ADVANCE
View attachment 64587
Since V=0 therefore V1=V2.
Since PQ resistor system is essentially a potential divider, we have for V1:
View attachment 64588
and similarly for V2:
View attachment 64591
Since V=0 therefore V1=V2. Substituting V1 and V2 gives us:
View attachment 64592
And equating these gives us:
View attachment 64593
Hope this helps.
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