SAT Math Part 17 - Solving Formulas For a Specified Variable - Free Educational videos for Students in K-12

SAT Math Part 17 - Solving Formulas For a Specified Variable - By The Organic Chemistry Tutor

Transcript


00:0-1	right 58 . The gravitational potential energy of an object
00:05	is directly proportional to the mass of an object ,
00:09	as well as the gravitational acceleration of the planet and
00:14	the height of the object above the ground . The
00:17	formula is shown below which of the following expressions correctly
00:22	represents the height of the object above ground level .
00:28	Now something that you need to be able to do
00:30	is you need to know how to solve for a
00:33	certain variable in an equation . So in this case
00:38	we want to solve for the height H . How
00:42	can we isolate age in this equation ? So what
00:46	we need to do is we need to divide both
00:49	sides by M . G . So that diesel cancer
00:54	. So age is going to equal to what we
00:56	see on the left . It's the gravitational potential energy
01:00	you divided by the mass and divided by the gravitational
01:05	acceleration of the planet . So this represents the height
01:09	of the object , which means that answer choice A
01:12	is the correct answer . 59 . The final velocity
01:18	of a moving vehicle is equal to the sum of
01:22	the initial velocity and the product of its acceleration and
01:26	time , which is the following expressions represent the acceleration
01:32	of the moving vehicle . So we need to isolate
01:37	the variable a acceleration . How can we do some
01:42	? Yes . We need to get a by itself
01:45	somehow . Now what we need to do in order
01:49	to do this is subtract both sides by the initial
01:53	, so we're going to have the final minus the
01:57	initial Is equal to 18 . Now we need to
02:01	separate A from T . Since they're multiplied to each
02:04	other , we need to divide so let's divide both
02:08	sides by T . So the acceleration is the final
02:13	velocity minus the initial velocity , which is basically the
02:18	change in velocity divided by the time . So this
02:22	right here is our answer , which corresponds to answer
02:25	choice B . Number 60 . The combined gas law
02:30	equation shown below describes the relationship between the pressure volume
02:36	and Calvin temperature of an ideal gas , Which of
02:40	the following expressions can be used to calculate T .
02:43	two . So here we have two fractions separated by
02:50	an equal sign . Something that we can do is
02:53	cross multiply . So this will give us P two
02:57	V two times T one , And that's going to
03:00	be equal to P one if he wanted T to
03:04	. Now , in this form , it's going to
03:06	be very easy to isolate teach you all we need
03:09	to do is divide both sides By P one .
03:12	V . 1 . Yeah . And so T two
03:16	is equal to what we see on the left side
03:19	. It's P two V 2 , T one Over
03:22	P one V 1 , which means answer choice D
03:25	. Is the answer 61 . The electric force between
03:31	two charged particles is proportional to the magnitude of each
03:36	electric charge and inversely proportional to the square of the
03:42	distance between the center of the charges . Which equation
03:48	can be used To calculate the magnitude of the first
03:51	charge Q one . So how can we isolate Q
03:57	one in this problem ? What's the first thing that
04:02	we need to do ? What I would recommend is
04:06	multiplying both sides by R squared . So on the
04:10	right side these will cancel . Next we can divide
04:16	both sides by K and Q tune . Yeah .
04:23	Yeah . So cable cancel and Q two will cancel
04:28	. So all we have left over on the right
04:30	side is Cuban . So Q one is equal to
04:33	F . Times are squared divided by K and divided
04:38	by Q two , which means A . Is the
04:41	right answer . Choice 62 . The gravitational force between
04:47	two planets is directly proportional to the mass of the
04:51	planet and inversely related to the square of the distance
04:56	between their centers . Which equation can be used to
05:01	calculate art . So how can we isolate are in
05:08	this equation ? Well , first we don't want it
05:14	to be on the bottom of a fraction . So
05:16	let's multiply both sides by our square . Yeah .
05:22	Now the next thing we want to do is divide
05:24	both sides by F . So right now we have
05:29	R squared is equal to G . M one M
05:33	2 Over F . Now to get rid of the
05:36	square , we need to take the square root of
05:39	both sides , so the square root of R squared
05:43	is simply are so R is equal to what we
05:48	see here , which means that De is the right
05:51	answer . 63 . The difference in pressure between two
05:58	points and a moving fluid depends on the density of
06:02	the fluid . The height difference between the two points
06:06	and The fluid velocities V two and V one of
06:10	those two points , according to Bernoulli's equation , which
06:15	of the following expressions can be used to calculate the
06:18	density of the fluid . So the density of the
06:22	fluid is represented by the greek symbol role , which
06:26	looks like a lower case P . Now we need
06:29	to isolate the role , but there's two of them
06:32	. So what should we do whenever you want to
06:36	isolate a variable ? And if you have more than
06:39	one of that variable , what you need to do
06:41	is factor . So we're going to factor out the
06:45	density . So if we take out roll from rogue
06:49	, ehh it's gonna be just negative G . H
06:53	. And if we take out road from this expression
06:55	, it's gonna be everything acceptable . So this is
07:03	what we now have . The last thing that we
07:06	need to do is divide both sides by what we
07:10	see here . That is by negative G . H
07:15	minus one half V . Two minus V . One
07:17	squared . And let's do the same on the left
07:22	side . So on the right side all of this
07:31	will disappear . So all we have left over on
07:34	the right side is the density rho on the left
07:38	is the answer . So this corresponds to answer choice
07:43	A . 64 . So here we have the simplified
07:48	version of the Doppler effect formula . And it describes
07:53	the relationship between the frequency of sound measured by a
07:57	stationary observer . That is fo emphases basically the source
08:03	frequency , that's the object generating the source . I
08:08	mean the object generating the sound and V . Represents
08:14	the speed of sound in the air . V .
08:17	S . Is the speed of the object generating the
08:21	sound . So in this problem we want to determine
08:25	the speed of sound and air . We need to
08:27	isolate the variable V . So how can we do
08:31	that ? Especially since we have two of them .
08:35	So at some point we need to factor out V
08:38	. But we can't do that in its current form
08:42	. So we're going to have to rearrange the equation
08:46	. So let's begin . The first thing we need
08:48	to do is multiply both sides by v minus vitas
08:55	. So on the right side , these will cancel
08:59	. So right now , what we have left over
09:01	is V minus V . S . Times fo and
09:06	that's equal to the product of these two V times
09:09	FFS mm . Our next step is to distribute fo
09:20	so we have fo times V . And then fo
09:24	times of the S . Yeah , now I'm going
09:33	to take this term , move it to the other
09:35	side . It's negative on the left side , but
09:38	it's going to be positive on the right side .
09:41	And this term I'm going to move it to the
09:43	left side , it's positive on the right , but
09:45	it will be negative on the left . So if
09:48	you want to show your work , we're going to
09:51	add V . S . F . O . To
09:55	both sides , and at the same time we're going
09:57	to subtract VF F s from both sides . Yeah
10:02	. Mhm . So we're gonna have the F .
10:05	O minus V . F . S . And that's
10:09	equal to V . S . F . O .
10:12	Our next step is to factor out with the ,
10:18	so if we take out V we're going to have
10:21	F . O minus F . S and then that's
10:23	equal to V . S . Fo Yeah , so
10:28	that's how we can take out or convert to v
10:31	variables into one . It's by factor in that .
10:34	Our next step is to the viable size by f
10:36	o , f s or fo minus f s .
10:41	Yeah , so now we have our final answer .
10:44	V is equal to what we see on the right
10:46	side , which means as a choice , C is
10:49	the right answer .

Summarizer

DESCRIPTION:

OVERVIEW:

SAT Math Part 17 - Solving Formulas For a Specified Variable is a free educational video by The Organic Chemistry Tutor.

This page not only allows students and teachers view SAT Math Part 17 - Solving Formulas For a Specified Variable videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.