Have no idea if I got it right (I think I did), but I remember writing about how there is a lower cross sectional area, and so there would be more more water entering a smaller area which may lead to more turbulent flow and thus create drag.
I don't think its related to Cross sectional area.
Thx in that case i would have gained the mark because of my statement at the end. In this particular question, all that matters is that somewhere you reference higher drag/friction/resistive force
The time he spends falling 1.25m is the same time he spends going 10m horizontally. So this value of time would be the same in both the vertical and horizontal equations. Can you take it from here?
The time he spends falling 1.25m is the same time he spends going 10m horizontally. So this value of time would be the same in both the vertical and horizontal equations. Can you take it from here?
I knew that fact, however I'm confused in the velocity, they're asking for the Initial velocity right?
Moreover, No angles are given so i'm unsure what to resolve.
EDIT: Are we supposed to find the angle to the horizontal with tan?
Does anyone know why there's a drop in the reaction force when the student is crouching? I know there must be a resultant force downwards because the student is accelerating downwards but what causes the drop in reaction force??
Does anyone know why there's a drop in the reaction force when the student is crouching? I know there must be a resultant force downwards because the student is accelerating downwards but what causes the drop in reaction force??
-Thanks in advance
Hi, i think i got the answer to your question, I'm taking the concept from other questions, so i might not be 100% right, but when i look at it, it seems pretty logical.
Take F=ma (downwards) This can be written as mg-R=ma (as we are considering downward forces as positve, im doing this because the person is accelerating downwards on the scale) So, therefore R=mg-ma (where 'g' is acceleration due to gravity and 'a' is the person's downwards acceleration). Therefore the value of 'R' will be lower than the students actual weight (because now its equal to the "student's weight-the students mass*his/her acceleration".
When the student is standing still: take F=ma (upwards or downwards doesn't matter, lets just take downwards) So, "mg-R=ma" since 'a' is zero because he/she is standing still then, 'R=mg'. Which will be the students actual weight.
Hope I answered your question. Anyway, Good luck for your exams