I did my Physics HW and it’s worth a test grade so can someone check and make sure that I did them correctly?

Question

I did my Physics HW and it’s worth a test grade so can someone check and make sure that I did them correctly?
PLEASE, I WILL APPRECIATE ALL THE HELP THAT I CAN GET –IT’S NOT REALLY THAT LONG. I NEED THESE ANSWERS TO BE RIGHT AND I HOPE YOU GET THE EQUATIONS, IT’S HARD TO TYPE PHYSICS PROBLEMS ON HERE …. GOD BLESS!


#1) A child on a merry-go-round is moving with a speed of 1.35 m/s when 1.20m from the center of the merry-go-round. Calculate (a) the centripetal acceleration of the child, and (b) the net horizontal force exerted on the child (Mass=25kg)

For (a) I did Centripetal Acc= V^2/ (r) and got 1.5 m/s^2
And for (b) I got Fnet=mv^2(r) and I got 38N

#2) A girl sits in a tire that is attached to an overhanging tree limb by a rope 2.10m in length. The girl’s father pushes her with a tangential speed of 2.50 m/s. If the magnitude of the force that maintains her circular motion is 88N, what is the girl’s mass?

I did Fc=m v^2/(r) then m=Fc r/(v)^2 (I got this equation from our packet) and my answer came out to be 29.5kg.

#3) A bicyclist is riding at a tangential speed of 13.2m/s around a circular track w/ a radius of 40m. If the magnitude of the force that maintains the bike’s circular motion is 377N, what is the combined mass of the bicycle and rider?

I did m= Fc (times) r/(v)^2 (sorry, I can’t type out the exact formula) and I got that the mass is 86.5kg but I don’t get what it means by COMBINED MASS DO I HAVE TO ADD SOMETHING TO THAT? HELP?

#4) A 20.0kg child wishes to balance on a seesaw with a child of 32kg. If the smaller child sits 3.2m from the pivot, where must the larger child sit?

I did (m1)(g)(d1) = (m2)(g)(d2) and so
(20)(9.8)(3.2) =(32)(9.8)(d2) and I got d2=2m …. I hope this makes sense, most importantly, I HOPE IT’S RIGHT!

#5) Tarzan plans to cross a gorge by swinging in an arc from a hanging vine. If his arms are capable of exerting a farce of 1400N on the rope, what is the maximum speed he can tolerate at the lowest point of his swing? His mass is 80kg and the vine is 4.8m long.

I used the same equation …. Since Fc=mv^2/(r) then v^2= Fc r/m and then I got v^2 =9.16 m/s ….
#6)Tarzan tries to cross a river by swinging from one bank to the other on a vine that is 10m long. His speed at the bottom of the swing, just as he clears the surface of the river, is 8m/s. Tarzan does not know that the vine has a breaking strength for 1.0x10^3 N. What is the largest mass Tarzan can have and make it safely across the river?

I did m=Fc r/vt^2 (as in Tang. Velocity) … and I got m=156.25kg. Is it right?

Answer 1

Problems 2, 5, and 6 involve objects moving in vertical circles. (All the rest of your answers look good, by the way.) I will comment on #5, and you can apply what I say to 2 and 6.

When Tarzan is at the bottom of his swing, the forces acting on him are his weight = mg (pointing down) and the vine pulling up on him. According to Newton's second law, the force pulling him toward the center of the circle minus the force pulling him down = m v^2/r

So, 1400 - 80(9.8) = 80 v^2 / (4.8)

Solve this equation and you will get the maximum speed he can have and still hang on at the bottom of his swing. (Think about this, if he was not moving at all, he would have to exert a force equal to his weight to hang on to the rope, your approach neglected his weight and gave too high a value for his maximum speed.)

Applying the same reasoning to #6, you get

1000 - m(9.8) = m 8^2/10 and you can solve for m.

I think your value is too high.

In #2, I think the same reasoning applies,

88 - m(9.8) = m 2.5^2 / 2.10 (And then solve for m)

(According to your approach, the girl's mass would be infinite, if the velocity was zero, can that be true?)

Answer 2

There are 2 hassle-free approaches to do it. you are able to multiply eighty 4.38 cases .seventy 5, then multiply ninety one.18 cases .25 and upload the outcomes collectively. which will make your universal 86.08. THEN yet otherwise: you are able to upload the eighty 4.38 thrice, upload in the ninety one.18 as quickly as and divide by utilising 4. you will nonetheless arise with 86.08 as your universal universal. desire this helps!