Mixing
![Proving $frac1sin(A/2)+frac1sin(B/2)+frac1sin(C/2)ge 6$, where $A$, $B$, $C$ are angles of a triangle [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Denominator doubled
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Originally $a = 12 space m/s$, then $r$ is doubled. Find the new value for $a$. $a = v^2/r$. If I am right, $v$ is originally $6$ which then turns to $36$ so if you divide it by $3$ it gives you $12$. Later if you double $r$, it would be $6$ which the final answer would be $6$.
algebra-precalculus physics
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
asked Jul 27 at 17:16
Santiago Lopez
11
add a comment |Â
up vote
0
down vote
favorite
Originally $a = 12 space m/s$, then $r$ is doubled. Find the new value for $a$. $a = v^2/r$. If I am right, $v$ is originally $6$ which then turns to $36$ so if you divide it by $3$ it gives you $12$. Later if you double $r$, it would be $6$ which the final answer would be $6$.
algebra-precalculus physics
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
asked Jul 27 at 17:16
Santiago Lopez
11
What part is task, what part is yours?
– mvw
Jul 27 at 17:18
1
This is not clear. If by $a$ you are referring to acceleration then your units are wrong. And what is $r$? Are you talking about acceleration around a circle? Are you keeping speed constant?
– lulu
Jul 27 at 17:19
Please use MathJax to format your answer as your formula for $a$ is currently ambiguous. But if $a$ is inversely proportional to $r$ and then $r$ is doubled, then $a$ will halve to give $6$.
– packetpacket
Jul 27 at 17:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Originally $a = 12 space m/s$, then $r$ is doubled. Find the new value for $a$. $a = v^2/r$. If I am right, $v$ is originally $6$ which then turns to $36$ so if you divide it by $3$ it gives you $12$. Later if you double $r$, it would be $6$ which the final answer would be $6$.
algebra-precalculus physics
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
asked Jul 27 at 17:16
Santiago Lopez
11
Originally $a = 12 space m/s$, then $r$ is doubled. Find the new value for $a$. $a = v^2/r$. If I am right, $v$ is originally $6$ which then turns to $36$ so if you divide it by $3$ it gives you $12$. Later if you double $r$, it would be $6$ which the final answer would be $6$.
algebra-precalculus physics
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
asked Jul 27 at 17:16
Santiago Lopez
11
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
edited Jul 27 at 18:25
VinÃcius Lopes Simões
409211
409211
asked Jul 27 at 17:16
Santiago Lopez
11
asked Jul 27 at 17:16
Santiago Lopez
11
asked Jul 27 at 17:16
Santiago Lopez
11
11
What part is task, what part is yours?
– mvw
Jul 27 at 17:18
1
This is not clear. If by $a$ you are referring to acceleration then your units are wrong. And what is $r$? Are you talking about acceleration around a circle? Are you keeping speed constant?
– lulu
Jul 27 at 17:19
Please use MathJax to format your answer as your formula for $a$ is currently ambiguous. But if $a$ is inversely proportional to $r$ and then $r$ is doubled, then $a$ will halve to give $6$.
– packetpacket
Jul 27 at 17:20
add a comment |Â
What part is task, what part is yours?
– mvw
Jul 27 at 17:18
1
This is not clear. If by $a$ you are referring to acceleration then your units are wrong. And what is $r$? Are you talking about acceleration around a circle? Are you keeping speed constant?
– lulu
Jul 27 at 17:19
Please use MathJax to format your answer as your formula for $a$ is currently ambiguous. But if $a$ is inversely proportional to $r$ and then $r$ is doubled, then $a$ will halve to give $6$.
– packetpacket
Jul 27 at 17:20
What part is task, what part is yours?
– mvw
Jul 27 at 17:18
What part is task, what part is yours?
– mvw
Jul 27 at 17:18
1
1
This is not clear. If by $a$ you are referring to acceleration then your units are wrong. And what is $r$? Are you talking about acceleration around a circle? Are you keeping speed constant?
– lulu
Jul 27 at 17:19
This is not clear. If by $a$ you are referring to acceleration then your units are wrong. And what is $r$? Are you talking about acceleration around a circle? Are you keeping speed constant?
– lulu
Jul 27 at 17:19
Please use MathJax to format your answer as your formula for $a$ is currently ambiguous. But if $a$ is inversely proportional to $r$ and then $r$ is doubled, then $a$ will halve to give $6$.
– packetpacket
Jul 27 at 17:20
Please use MathJax to format your answer as your formula for $a$ is currently ambiguous. But if $a$ is inversely proportional to $r$ and then $r$ is doubled, then $a$ will halve to give $6$.
– packetpacket
Jul 27 at 17:20
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
You're right.
Since $a=v^2/r$, $a$ is inversely proportional to $r$.
Doubling $r$ halves $a$, and $6text m/s^2$ is half of $12text m/s^2.$
answered Jul 27 at 17:33
John
21.9k32346
add a comment |Â
up vote
0
down vote
Assuming $v$ stays constant,
$$ar = v^2 = k$$
where $k$ is a constant. So $a$ and $r$ are inversely proportional. If $r$ is doubled, $a$ is halved, and the new acceleration is $$frac12(12fracms^2)=6 fracms^2$$
answered Jul 27 at 17:33
RayDansh
884214
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You're right.
Since $a=v^2/r$, $a$ is inversely proportional to $r$.
Doubling $r$ halves $a$, and $6text m/s^2$ is half of $12text m/s^2.$
answered Jul 27 at 17:33
John
21.9k32346
add a comment |Â
up vote
0
down vote
You're right.
Since $a=v^2/r$, $a$ is inversely proportional to $r$.
Doubling $r$ halves $a$, and $6text m/s^2$ is half of $12text m/s^2.$
answered Jul 27 at 17:33
John
21.9k32346
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You're right.
Since $a=v^2/r$, $a$ is inversely proportional to $r$.
Doubling $r$ halves $a$, and $6text m/s^2$ is half of $12text m/s^2.$
answered Jul 27 at 17:33
John
21.9k32346
You're right.
Since $a=v^2/r$, $a$ is inversely proportional to $r$.
Doubling $r$ halves $a$, and $6text m/s^2$ is half of $12text m/s^2.$
answered Jul 27 at 17:33
John
21.9k32346
answered Jul 27 at 17:33
John
21.9k32346
answered Jul 27 at 17:33
John
21.9k32346
answered Jul 27 at 17:33
John
21.9k32346
21.9k32346
add a comment |Â
add a comment |Â
up vote
0
down vote
Assuming $v$ stays constant,
$$ar = v^2 = k$$
where $k$ is a constant. So $a$ and $r$ are inversely proportional. If $r$ is doubled, $a$ is halved, and the new acceleration is $$frac12(12fracms^2)=6 fracms^2$$
answered Jul 27 at 17:33
RayDansh
884214
add a comment |Â
up vote
0
down vote
Assuming $v$ stays constant,
$$ar = v^2 = k$$
where $k$ is a constant. So $a$ and $r$ are inversely proportional. If $r$ is doubled, $a$ is halved, and the new acceleration is $$frac12(12fracms^2)=6 fracms^2$$
answered Jul 27 at 17:33
RayDansh
884214
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assuming $v$ stays constant,
$$ar = v^2 = k$$
where $k$ is a constant. So $a$ and $r$ are inversely proportional. If $r$ is doubled, $a$ is halved, and the new acceleration is $$frac12(12fracms^2)=6 fracms^2$$
answered Jul 27 at 17:33
RayDansh
884214
Assuming $v$ stays constant,
$$ar = v^2 = k$$
where $k$ is a constant. So $a$ and $r$ are inversely proportional. If $r$ is doubled, $a$ is halved, and the new acceleration is $$frac12(12fracms^2)=6 fracms^2$$
answered Jul 27 at 17:33
RayDansh
884214
answered Jul 27 at 17:33
RayDansh
884214
answered Jul 27 at 17:33
RayDansh
884214
answered Jul 27 at 17:33
RayDansh
884214
884214
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864604%2fdenominator-doubled%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What part is task, what part is yours?
– mvw
Jul 27 at 17:18
1
This is not clear. If by $a$ you are referring to acceleration then your units are wrong. And what is $r$? Are you talking about acceleration around a circle? Are you keeping speed constant?
– lulu
Jul 27 at 17:19
Please use MathJax to format your answer as your formula for $a$ is currently ambiguous. But if $a$ is inversely proportional to $r$ and then $r$ is doubled, then $a$ will halve to give $6$.
– packetpacket
Jul 27 at 17:20