Physics, asked by pepitocena9, 8 months ago

An object 8.9 N in weight is released from a height H = 30.1 m so that it can rotate in the loop of radius r = 4.8 m and continue on its way to a rough horizontal section of Length L = 36.2 m, to stop at D. Only the CD section is rough. h = 1 m a) the speed of the object at the highest part of the loop (point B). b) the kinetic friction coefficient μC of the section CD. pls help Or my teacher will think that I'm a human trash for the rest of my days

Attachments:

Answers

Answered by HariesRam

The formula to calculate the coefficient of friction is μ = f÷N. The friction force, f, always acts in the opposite direction of the intended or actual motion, but only parallel to the surface.

❤️

$<!DOCTYPE html><html lang="en"><head><title>Koala</title></head><body><div class="character"><div class="head"><div class="ear ear-left"> </div><div class="ear ear-rigth"> </div><div class="eyes"></div><div class="nose"></div><div class="lip"></div></div><div class="body"><div class="hand hand-right"></div><div class="hand hand-left"></div></div></div><style>body {display: flex;justify-content: center;padding-top: 150px;}.character {position: relative;}.character .head {height: 125px;width: 155px;background: #94e1fd;border-radius: 80px 89px 55px 55px;position: relative;border: 4px solid black;top: 6px;}.character .head:after {content: "";position: absolute;bottom: -43%;width: 40px;height: 50px;background: white;border-bottom-left-radius: 50%;border-bottom-right-radius: 50%;left: 42%;}.ear {width: 78px;height: 78px;background: white;border-radius: 38% 62% 57% 43% / 45% 48% 52% 55%;border: 4px solid black;position: absolute;}.ear-left {top: -24%;left: -24%;border-bottom-color: #94e1fd;transform: rotate(-49deg);box-shadow: inset 6px 20px 0px 12px #94e1fd, inset -20px 18px 0 0px #94e1fd, inset 3px -9px 0 0px #94e1fd;animation: ear-left 2s infinite linear;}.ear-left:before {content: "";width: 4px;height: 6px;border-radius: 50%;background: black;position: absolute;top: 31px;left: 110%;transform: rotate(-76deg);box-shadow: 15px 12px, 30px 0px;opacity: 0;animation: ear-left-after 2s infinite linear;}.ear-rigth {left: 65%;top: -24%;transform: rotate(-227deg);z-index: -1;box-shadow: inset 20px -4px 0px 13px #94e1fd, inset 20px -11px 0 0px #94e1fd, inset 20px 19px 0 0px #94e1fd;animation: ear-right 2s infinite linear;}.eyes {width: 14px;height: 10px;background: black;position: absolute;border-radius: 50%;top: 46%;left: 28%;box-shadow: 65px 0px;animation: eyes 2s infinite linear;}.eyes:after {content: "";width: 14px;height: 3px;background: black;position: absolute;border-radius: 50%;bottom: -57%;box-shadow: 65px 0px;}.nose {width: 31px;height: 35px;background: #6f69ef;border-radius: 53% 47% 59% 41% / 73% 76% 24% 27%;border: 4px solid;position: absolute;top: 43%;left: 43%;animation: nose 2s infinite linear;}.lip {width: 34px;height: 4px;background: black;position: absolute;border-radius: 25%;top: 83%;left: 45%;animation: lip 2s infinite linear;}.body {position: relative;z-index: -1;}.body:after,.body:before {content: "";border-radius: 50%;width: 44px;height: 105px;border: 4px solid;position: absolute;background: #94e1fd;border-right-color: transparent;left: 24%;top: -8px;}.body:after {left: 46%;border-left-color: transparent;border-right-color: black;}.hand {border-right-color: transparent;width: 34px;height: 55px;border-radius: 71% 29% 22% 78% / 17% 78% 22% 83%;background: #94e1fd;border: 4px solid;position: absolute;left: 8%;top: -2px;border-top-color: #94e1fd;transform: rotate(39deg);}.hand-left {left: 63%;transform: rotate(-27deg);border-radius: 59% 41% 32% 68% / 20% 62% 38% 80%;border-left-color: transparent;}@keyframes eyes {100% {height: 30px;top: 22%;width: 10px;}}@keyframes nose {100% {top: 22%;}}@keyframes lip {100% {background: #f77676;height: 25px;width: 25px;top: 72%;border-top-left-radius: 50%;border-top-right-radius: 50%;transform: translateY(-10px);}}@keyframes ear-left {100% {border-bottom-color: black;top: 165%;transform: rotate(-175deg);height: 24px;}}@keyframes ear-right {100% {top: 165%;transform: rotate(-175deg);height: 24px;}}@keyframes ear-left-after {95% {opacity: 0;}100% {opacity: 1;}}</style></body></html>$

Answered by SwaggerGabru

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

An object 8.9 N in weight is released from a height H = 30.1 m so that it can rotate in the loop of radius r = 4.8 m and continue on its way to a rough horizontal section of Length L = 36.2 m, to stop at D. Only the CD section is rough. h = 1 m a) the speed of the object at the highest part of the loop (point B). b) the kinetic friction coefficient μC of the section CD.

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

We know that according to polygon law of vector addition, the resultant of these six vectors is zero.

Here A = B = C =D=E=F (magnitude) So, Rx = A cosi + A cos /3 + A cos 2r/3 + A cos 3r/3 + A cos 4/4 + A cos 5x/5=0 As resultant is zero. X component of resultant R = 0] = cos 8 + cos /3 + cos 2n/3 + cos 33 + cos 4x/3 + cos 5/3 = 0 Note: Similarly it can be proved that sin 0+ sin /3 + sin 2/3 + sin 3/3 + sin 4/3 + sin 5a/3 = 0

__________________

@HarshPratapSingh

Previous Question

Next Question