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Lorico [155]
4 years ago
6

Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as

the initial height of the block. Will A make it completely around the loop without losing contact with the track?

Engineering
1 answer:
Zolol [24]4 years ago
4 0

Answer:No

Explanation:

Given

Block A is at height h and released from rest

Initial Energy possessed by block A is equal to Potential energy of Block which is given by

E_i=mgh

where m=mass of block

After releasing the block, block first reaches to the bottom of the circle and then uses this energy to reach a top point of the loop.

But as soon as block reaches the top point of the loop it acquires the energy which is equal to Initial energy i.e. all the energy is stored in the form of potential energy and there is no kinetic energy so the block will not able to move further and fall from the top point.

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13- Convert the following numbers to the indicated bases. List all intermediate steps.
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Answer:

Following are the conversion to this question:

Explanation:

In point (a):

\to \frac{36459080}{8} = 4557385   + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673   + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \  \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\

\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}

In point (b):

\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \  \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}

In point (c):

\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\

\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}

Symbols of Base 25 are as follows:

0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O

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