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hahahaha really different, answer is 88 m, im getting around -7000

hmmm, i tried that and still got answer wrong, these were my steps 100t-86.6t=600, therefore t=44.78 then i plugged t=44.78 into y=1/2gt^2+vsinθt +H (with v=100 and θ =30) since we have the time of flight the value of y will be zero leading to the only unknown being H. Thanks

question asks the projectiles acceleration, at the peak would be 0 right, or am I tripping

also for mod 5, should question 1 be A, as in position Y the vertical acceleration will be 0 along with the horizontal acceleration thanks

hey jazz you've helped me with a lot of questions and i just wanna thank you for your helping hand in this community also thanks for the practice questions

my bad that was the answer for a different question, yes the answer is 100 kN Thanks

can comeone explain pls, the answer is 100kN

thanks for your time anyway man

thanks a lot, appreciate it

could you explain what you did to get that answer, would be really helpful Thanks

I assumed they used g=10, yeh the answer is 351.4

sorry forgot to mention this, but the answer on the exam is 351.4 N little bit confused by that

Thanks

oh ok, thanks for that

Thats what i was thinking, but isnt that the direction of the force on the wire, not the torque?. or is it the same thing and im missing something

This question is from a HSC test, and I understand the working out for the magnitude of the torque, I dont know why the direction of the torque is into the page. help would be appreciated Thanks

thanks a lot