Pytanie: a few physics questions?

( Wstecz )

Answer #1:

What do you think you are doing?
You test psychics with math problems?
Get a better hobby!

Peace,
Miss Zoe.

OOPs, My Bad. I was in the wrong section. But really, What are you doing? You do need better hobby. There you just got some unrequested genuine psychic advise.

Answer #2:

they were taking breather!
1) At a snow cross event snow mobiles launch from a horizontal jump, landing on the 20 degree slope below. If a racer leaves the jump @ 80km/hr, how far down the slope will he land?
♥ thus the horizontal speed will remain const, that is
u=80km/hr = 80*1000/3600 = (200/9) m/s;
{The vertical component of velocity will increase with time, that is v(t) = g*t};
♠ the distance in horizontal direction is x=u*t, hence time is t=x/u;
the distance in vertical direction is y= -0.5*g*t^2;
thus y=-0.5*g*t^2 = -0.5*g*(x/u)^2; or;
y= -(0.5*g/u^2)*x^2 is parabola sloped downwards;
♣ the ramp underneath is a line also sloped downwards, its equation being
y=-tan(b)*x, where b=20° to horizontal;
♦ the parabola and the line meet, that is y=y; or;
-(0.5*g/u^2)*x^2 =-tan(b)*x, hence
the distance traveled in horizontal direction is
x=2*tan(b) *u^2/g =2*tan(20) *(200/9)^2 /9.8 =36.68m;
the distance traveled in vertical direction is
y=-tan(b)*x = -13.35m;
the distance traveled along the ramp is
d= √(x^2 +y^2) = 39.04m;
; ------------------------------------------------------
2) At the scene of a car accident, police find skid marks 40 m long. Speed limit is 70 km/hr. was the vehicle speeding before the accident?
static coefficient of friction=0.9; kinetic coefficient of friction=0.8;
♠ if traveling with speed
v=70km/h= 70*1000/3600 = 175/9 m/s,
the kinetic energy of the car should be
E=0.5*m*v^2, where m is mass of the car;
♦since the break pedal pushed until coming to stop the car was being exerted a force of friction
F=u*m*g, where
u=0.8 is kinetic (not static!) coefficient of friction,
m*g is weight of the car, g=9.8m/s^2;
♣this F produced work F*x necessary to reduce the kinetic energy to zero, that is
E=F*x; or; 0.5*m*v^2 = u*mg*x, hence the skid length should be
x= 0.5*v^2 /(u*g) = 0.5*(175/9)^2 /(0.8*9.8) ≈24m < 40m;
the driver shall be prosecuted!
;-------------------------------------------
3) A girl pushes a merry go round with diameter 5 m from its edge. she pushes for 2s with a constant tangential acceleration of 2m/s^2 before jumping on. What is the magnitude of her acceleration after 1s?
What angle has the merry go round gone through when she hops on (in radians)?
♣ while she’s still on the ground her tangential speed is
v(t) = a*t, where a=2m/s^2;
as she’s running along a circular arc her centripetal acceleration is
b=v^2/R, where R=5m;
the magnitude of her acceleration is c=√(a^2 +b^2) =
= √(a^2 +(v^2/R)^2) = √(a^2 +((a*t)^2/R)^2) =
= {at t=1} = √(2^2 +((2*1)^2/5)^2) =0.4*√29 m/s^2;
♥ the angular acceleration of merry-go-round is h=a/R;
the angle is u=0.5*h*t^2 = 0.5*(a/R)*t^2; thus
u(t=2s) = 0.5*(2/5)*2^2 = 0.8 rad;
;-----------------------
4) a ride at the fair puts participants inside a cylinder with radius 10m, tapered at 10°. Each rider leans back so that he may move up and down without friction. At what angular velocity do the riders lift off of the floor?
♠ suppose a rider being just a particle with mass m, and draw a picture;
the half-angle of the conic he’s inside is b=10/2 =5° to vertical;
♣Centrifugal force on him F=m*w^2*R is parallel to the platform and is directed to outside, the component of F parallel to conic surface being F*sin(b);
♣gravity on him P=mg is normal to the platform and is directed downward, the component of P parallel to conic surface being P*cos(b);
♦he should start sliding when both components are equal, that is
P*cos(b)=F*sin(b); or;
mg*cos(b) = m*w^2*R*sin(b), hence
w=√(g*cot(b)/R) =√(9.8*cot(5°)/10) = 3.35 rad/s;





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