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Gharib
Questions (5)
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.
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Skydiver Joseph Kittinger, record for the longest jump, reaches a terminal speed of 60m/s with his arms and body fully extended.
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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
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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
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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
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Answers (11)
Thanks Bobpurley, I didn't know it was that simple.
@risha, equate (1/2)mv^2=8.5F m and v are known, solve for F, therefore F=604N
Is there any angle with the horizon? if I am not mistaken, then T= ma
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
If you draw the FBD for both object separately. Then, Positive being up and right, T=m1a Eq 1 T-m2g=-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
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,
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,
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,
M=0.393kg m=0.181kg Draw the FBD, you will see that T1+T2-W(of M)=0 Eq 1 T2-W(of m)=0 Eq 2 (T2 is what we need) now, Eq1-Eq2 => T1+Wm-WM=0 T1= -g(m-M) T1= x Newtons now, plug in T1 in Eq 1, T2=WM-T1 *Reminder W=mass*9.8 Ans: T2=1.77N
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
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
