Gharib
Most popular questions and responses by Gharib
Physics
Kevin stands at the edge of a cliff, holding one ball in each hand. He throws one of the balls straight up with speed v, and at the same time he throws the other ball straight down, also with speed v. Ignoring air resistance, which ball hits the ground
asked on October 26, 2010 
Physics
A large plate falls to the ground and breaks into three pieces. m1 falls with v1=3m/s, 25 degrees to the left from the vertical. m2 falls with v2=1.79m/s, 45 degrees up from the horizon. And m3=1.3kg falls with v3=3.07m/s vertically down. How would I find
asked on October 26, 2010 
Physics
Berry Bonds hits a baseball ball with an initial speed of 35m/s at an angle of 50 degrees with respect to the horizon. The playing field is surrounded by a 4.0m tall wall. Assume that when hit, the ball is initially very close to the ground. If the
asked on October 26, 2010 
Physics
A tennis ball of mass 57. g bounces off a wall. Right before hitting the wall, the ball is moving to the right and up with vx=+40. m/s and vy=+10. m/s. Right after bouncing off the wall, the ball is moving to the left and up with vx=35. m/s and vy=+10.
asked on October 25, 2010 
Physics
Skydiver Joseph Kittinger, record for the longest jump, reaches a terminal speed of 60m/s with his arms and body fully extended. Then, he switches to a head first dive, reducing his cross section area by 80% (that is, his cross section area when headfirst
asked on October 26, 2010

Physics
Thanks Bobpurley, I didn't know it was that simple.
posted on October 25, 2010 
Physics
@risha, equate (1/2)mv^2=8.5F m and v are known, solve for F, therefore F=604N
posted on October 25, 2010 
physics
Is there any angle with the horizon? if I am not mistaken, then T= ma
posted on October 25, 2010 
Physics
You have to make two FBD; 1st for the force being applied to the leg, and 2nd for the mass hanging from cable. Positive being up and right. Y direction for FBD #2: T= mg Eq 1 x direction for FBD #1: F+2Tcos(60)=0 Eq 2 T= F/2cos(60) T=x From Eq 1, m=x/g
posted on October 25, 2010 
Physics
If you draw the FBD for both object separately. Then, Positive being up and right, T=m1a Eq 1 Tm2g=m2a Eq 2 T= m2a+m2g Eq 3 Eq 3 = Eq 1 m1a= m2a+m2g a=m2g/(m1+m2) since, a=x then, T=m1x
posted on October 25, 2010 
Physics
Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method]. In that case,
posted on October 25, 2010 
PHYSICS
Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method]. In that case,
posted on October 25, 2010 
Physics
Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method]. In that case,
posted on October 25, 2010 
physics
M=0.393kg m=0.181kg Draw the FBD, you will see that T1+T2W(of M)=0 Eq 1 T2W(of m)=0 Eq 2 (T2 is what we need) now, Eq1Eq2 => T1+WmWM=0 T1= g(mM) T1= x Newtons now, plug in T1 in Eq 1, T2=WMT1 *Reminder W=mass*9.8 Ans: T2=1.77N
posted on October 25, 2010 
woburn
m=77kg angle=26 degrees Normal force holds half the weight. Positive being up, Y direction: (1/2)w(F1)cos(26)(F2)cos(26) =(1/2)(77kg)(9.8m/s^2)2Fcos(26) [since F1=F2 => F=377.3N/2cos(26) F=210N
posted on October 25, 2010 
woburn
m=77kg angle=26 degrees Normal force holds half the weight. Positive being up, Y direction: (1/2)w(F1)cos(26)(F2)cos(26) =(1/2)(77kg)(9.8m/s^2)2Tcos(26) [since T1=T2 => T=377.3N/2cos(26) T=210N
posted on October 25, 2010