Wednesday

November 25, 2015
Total # Posts: 22,909

**Social Studies**

I doubt it but north and west of what? If Europe then 1.
*November 2, 2015*

**Social Studies**

Yes, that is correct.
*November 2, 2015*

**Math**

2 x = 34 x = 34/2 x = 17
*November 2, 2015*

**Physics**

gravity force down slope = m g sin 25 so m a = 24 - 3(9.81) sin 25
*November 2, 2015*

**physics 6**

It holds 90=70 = 20 grams of water. That is 20 cm^3 or cc of water. The liquid has mass of 96 - 70 = 26 grams so density = 26 g/20 cc = 1.3 g/cc
*November 1, 2015*

**school..maikona physics question**

u = speed * cos theta = speed horizontal which never changes u/speed = cos theta = 1/5 = o.2 theta = coa^-1 ( 0.2) = 78.5 degrees
*November 1, 2015*

**Physics**

90,000 m / 3600 s = initial speed = Ui distance = .75 Ui + (Ui/2) t to find t: u = Ui - 10 t when u = 0 we are stopped so t = Ui/10 so in the end distance = .75 Ui + (1/2) Ui^2/10
*October 31, 2015*

**A mistake**

then rise time = fall time = H/.5Vi so total time in air = 4 H/Vi
*October 31, 2015*

**AP PHSYICS DUE LATE HOMEWORK HELP!!!**

5. A projectile is launched a t 35 m/s at an angle of 33 degrees. Determine (a) the maximum height of the projectile. (B) the maximum distance of the projectile goes assuming it lands on the same horizontal plane. FREE RESPONSE ============================= u = horizontal ...
*October 31, 2015*

**Physics**

Ke = (1/2) m v^2 = (1/2) m (21^2) gain of potential energy during rise = m g h = m (9.81)(15.8) so Ke at 15.8 = (1/2) m v^2 = (1/2)m(21^2) - m (9.81)(15.8) m cancels, so compute v at 15.8 v= Vi - g t so 0.81 t = 21 - v
*October 31, 2015*

**AP PHYSICS HELP**

Your problems are all the same. range = u t t is time to fall one meter h = (1/2) g t^2 so t = sqrt (2 g h) remember that so t = sqrt (2*9.81*1) then 5 meters = u t and we know t so solve for u
*October 31, 2015*

**AP PHYSICS HELP**

These two questions are really the same. in the first the horizontal velocity remains 1.5 the whole time then vertical problem 20 = 4.9 t^2 solve for t Thern use that t to get v, speed down at ground v = g t v = 9.81 t then speed = sqrt (2.25 + v^2) The second problem is ...
*October 31, 2015*

**Algebra**

slope = m = [ -5 - 4 ] / [ 3 - -6] = -9/9 = -1 so y = -x + b 4 = 6 + b so b = -2 so equation is y = -x -2 x axis intercept when y = 0 so (-2,0) y axis intercept when x = 0 so (0, -2)
*October 31, 2015*

**Pre calculus**

s = speed south = 190 - 30 cos 40 w = speed west = 30 sin 40 speed magnitude = sqrt (s^2 + w^2)
*October 30, 2015*

**Algebra 2**

yes
*October 30, 2015*

**Algebra 2**

9 t^2 + 6 t + 1 </= 0 agree now y = 9 t^2 + 6 t + 1 Is a parabola with the vertex down (holds water) It satisfies the condition (y </=0) between the two intercepts with the x axis. BUT there is only ONE at x = -1/3 thus ONLY at that point (t = -1/3)
*October 30, 2015*

**Conway high**

http://davidmlane.com/normal.html
*October 30, 2015*

**Physics**

Assuming no air resistance the forward speed of the helicopter is constant and the forward speed of the package is the same as that of the helicopter. Therefore the bomb hits directly below the helicopter. (5) ALWAYS turn and climb after releasing your bomb!
*October 30, 2015*

**algebra**

slope = (15-7)/(5-1) = 8/4 = 2 f(x) = 2 x + b now find b 7 = 2(1) + b so b = 5 so f(x) = 2x + 5
*October 30, 2015*

**Algebra**

14 = 4 x + 6 8 = 4 x 2 = x
*October 30, 2015*

**Math**

every time sin 4x = 0 and every time cos x = 0 sin 4 x = 0 when x = 0, pi/4 , pi/2 , 3 pi/4, pi , 5 pi/4.... 8 pi/4 which is 2 pi cos x = 0 when x = pi/2 , 3 pi/2 but we already have those two points
*October 30, 2015*

**Physics**

side force port = 135 sin 37 side force starboard = F sin 45 so F = 135 (sin 37 / sin 45) for no net force to either side
*October 30, 2015*

**math**

7/8 - 6/8 = 1/8
*October 30, 2015*

**Math**

if x = 4, x-2 = 2 if x = 3, x-2 = 1 if x = 2, x-2 = 0 if x = 1, x-2 = -1 if x = 0, x-2 = -2 if x = -1, x-2 = -3 so + or - depending on if x greater than or less than 2
*October 29, 2015*

**No help**

I would have taken the same approach. I wonder though if something was left out of the problem statement.
*October 29, 2015*

**precalc**

1)sin pi/2 = 1 csc pi/2 = 1 and 1+4 = 5 sure enough 2) 1/sin = -5 sin = -1/5 I get -11.537 deg
*October 29, 2015*

**precalc**

#3 trick is to get sin^2 + cos^2 = 1 in there sin^2/cos^2 + 2 /cos^2 = 3 sin^2 + 2 = 3 cos^2 sin^2 + cos^2 + 2 = 4 cos^2 1 + 2 = 4 cos^2 cos^2 = .75 cos = .866 then theta = 30 degrees
*October 29, 2015*

**Math typo**

26 - 6 x = 8 (x-1.5)
*October 29, 2015*

**Math**

Sarah runs 6 x sarah is 26 - 6 x from school Ryan ran x -1.5 hours ryan is 8 (x-1.5) from school when they meet 26 - 6 x = 8 (x-11.5) can you take it from there?
*October 29, 2015*

**Math**

Sarah runs 6 x sarah is 26 - 6 x from school Ryan ran x -1.5 hours ryan is 8 (x-1.5) from school when they meet
*October 29, 2015*

**Physics**

s = that speed u = s cos 35 = .819 s Vi = s sin 35 = .574 s 20 = u t so t = 20/u = 20/.819 s = 24.4/s h = 6.5 = Vi t -4.9 t^2 6.5 = .574 s (24.4/s) - 4.9 (24.4/s)^2 6.5 = 14 - 2917 /s^2 2917 = 7.5 s^2 s = 19.7 m/s
*October 29, 2015*

**math**

sqrt 3 = 1.73
*October 29, 2015*

**math**

hypotenuse sqrt (12^2 + 9^2) sqrt ( 144 + 81) sqrt (225) = 15
*October 29, 2015*

**That is not an equation**

perhaps it is a expression you want evaluated - 4 8/9
*October 29, 2015*

**MATH help**

well, sure looks that way
*October 29, 2015*

**MATH help**

try first 5 (2) - (-9/3) =10 + 3 =13 sure enough try second 5(3) - (-6/3) = 15 + 2 = 17 no way
*October 29, 2015*

**Pre-Calculus**

18 = 23 - 5 so when is 20 cos (π/4 (t-3)) = -5 ? cos (pi/4(t-3)) = -1/4 = -.25 use inverse cos in radians mode pi/4(t-3) = 1.823 radians and also 2 pi - 1.823 = 2.889 radians (that is the second time, on the way down) now I think you can get t from (pi/4)(t-3) = 1.823 etc...
*October 29, 2015*

**Physics**

your AVERAGE velocity = 38/8 = 4.75 That means at the start your velocity is twice that Vi = 2*4.75 = 9.5 m/s and at finsh v = 0 SO a = -9.5/8 = -1.1875 m/s^2 so F = m a = - 21.375 Newtons m g = 18*9.81 = 176.58 Newtons mu = 21.375 / 176.58 = 0.121
*October 29, 2015*

**MATH HELP**

f(3) = 6 f(4) = 10 [f(4)-f(3) ] / (4-3) = [f(x) -f(4]/(x-4) [ 10 - 6 ]/(1) = [f(x) - 10] /(x-4) 4 = (f(x)-10)/(x-4) 4 x - 16 = f(x) - 10 f(x) = 4 x - 6
*October 28, 2015*

**phsyics**

force down slope = m g sin 21 .5 = m a
*October 28, 2015*

**Physics**

the ends are 6 feet from the fulcrum at the center about fulcrum it is 50 * 6 foot pounds 100 d = 50*6 d = 6/2 = 3 ft
*October 28, 2015*

**I am supposed to do magic?**

I have no idea what you did in the experiment or what "x" is in this problem.
*October 28, 2015*

**math**

However you did not have to ask .2*31 = 6.2 31 - 6.2 = 24.80
*October 28, 2015*

**math**

yes, 24.8/.8 = 31
*October 28, 2015*

**math**

0.8 x = 24.80 so x = 24.8/.8
*October 28, 2015*

**math logarithm**

1.632*10^1 * 6.35*10^-3 / 4.82*10^-2 = (1.632*6.35/4.82) * 10^(1-3+2) = (1.632*6.35/4.82) * 10^0 = (1.632*6.35/4.82) * 1 now do your base 10 log thing remember log 10^x = x so log 1 = 0 log (answer)= log 1.632 + log 6.35 - log 4.82 + 0
*October 28, 2015*

**Statistics - PLEASE HELP!**

p = p win = .324 1-p = p lose = .676 binary distribution P(x = k) = C(n,k) p^k (1-p)^(n-k) C(2,3) = 3!/2!(3-2)! = 3 so = 3(.324)^2 (.676) = .213
*October 28, 2015*

**Trigonometry application**

tan 60 = height /distance = h/d so h = 1.73 d tan 25 = h/(1.7 + d) so .466 (d+1.7) = 1.73 d
*October 28, 2015*

**physics**

forces down slope = Fd Fd = m g sin 19.4 + 541 cos 19.4 forces normal to slope = m g cos 19.4 - 541 sin 19.4 so force up slope = Ff = mu ( m g cos 19.4 - 541 sin 19.4) no acceleration so Ff=Fd mu (m g cos 19.4 - 541 sin 19.4)=m g sin 19.4 + 541 cos 19.4
*October 28, 2015*

**Uiiecest bama**

the same as the rectangle you can draw using that base and the height. The triangle left off on one side is added on the other.
*October 27, 2015*

**Uiiecest bama**

base times height
*October 27, 2015*

**science**

Ke at top = Ke at bottom - energy lost
*October 27, 2015*

**science**

I am sure that you could have done this problem three times in the time you spent typing it. v = 27 km/hr * 1000m/km * 1 hr/3600 s Ke = (1/2) m v^2 loss of energy = m g h Ke at top = Ke at bottom i energy lost (1/2) m u^2 = Ke at top
*October 27, 2015*

**physics**

T = tension in string in air: T = m g in water: T = m g - rho g V where rho is density of water (about 1000kg/m^2) g is about 9.81 m/s^2 V is volume in meters^3 google Archimedes Principle
*October 27, 2015*

**physics**

work = change in energy = integral of P dt = t^3 - t^2 + t at 4 - at 1
*October 27, 2015*

**Math**

If you only had five packages you would have 25 pencils which is not enough. You need at least six packages. If you have six packages, that is 30 pencils and 3 will remain after you give out 27.
*October 27, 2015*

**Calculus**

p(t) = 40 + 280t + 4t^2 + 3t^3 I think it is p(t) = 40 + 280t + 4t^2 + (1/2)t^3
*October 25, 2015*

**physics**

initial momentum east =.4*5.38 the final momentum east is the same so .4 * 5.38 = .4 u + .95 v where u and v are the two resulting speeds also kinetic energy is the same (1/2)(.4)(5.38)^2 = (1/2)(.4)(u)^2 + (1/2)(.95)(v)^2 use those two equations to solve for u and v
*October 25, 2015*

**Geography**

http://www.weather.com/tv/shows/amhq/video/economic-impact-of-hurricane-patricia
*October 25, 2015*

**Geography**

Well, what sort of impact would you guess? What wold happen here if such a storm hit a beach resort area?
*October 25, 2015*

**math 1350**

10 x + 5 x + x = 1216 cents 16 x = 1216 x = 76
*October 25, 2015*

**Physics**

u = constant horizontal velocity = 36.3 so 36.3 = 38 cos theta so theta = 17.2 degrees above horizontal Vi = initial vertical velocity = 38 sin theta = 38 sin 17.2 = 11.2 m/s v = Vi - 9.81 t at top v = 0 so t = 11.2/9.81 = 1.15 seconds to top h = Vi t - 4.9 t^2 = 11.2(1.15) - ...
*October 25, 2015*

**physics**

u = horizontal velocity = 17.4 cos 33 Vi = initial vertical velocity = 17.4 sin 33 v = Vi - 9.8 t at top v = 0 so at top t = Vi/9.8 it spends the samer asmount of time falling from the top so time in air = 2 t = 2 Vi/9.8
*October 25, 2015*

**socail studies**

It is called the "United States" , a union of states.
*October 25, 2015*

**Algebra**

4 (48 - 6 y) + 5 y = 21 192 - 24 y + 5 y = 21 171 = 19 y y = 9 etc
*October 25, 2015*

**physics**

If the initial velocity V has components Vi and u then d = u t or t = d/u and at time t the banana must be at height h so h = Vi t - 4.8 t^2 so h = Vi d/u - 4.9 d^2/u^2
*October 25, 2015*

**physic**

there are two pages per sheet of paper (front and back) so 548 sheets of paper = 3.2 cm so 3.2/548 about 6/1000 = .006 or .01 cm
*October 25, 2015*

**Physics**

(1/2) m Vi^2 = m g h so Vi = sqrt(2 g h) v = Vi - g t 0 = Vi - 9.81 t solve for t, time to rise to 3.2 m then time in air = 2 t
*October 25, 2015*

**Physic**

assume speed of light is infinite for this purpose two seconds * 343 m/s = 686 meters
*October 25, 2015*

**Algebra 1**

B is the only one that includes all the possibilities
*October 25, 2015*

**Physics**

yes, a = F/m the units are a mess of course :( remember force up = m g + m a you need mg up just to hold them
*October 25, 2015*

**Chemistry**

The others are very similar.
*October 25, 2015*

**Chemistry**

1. You have 0.1 mol of HNO3 because you have one liter H = 1 N = 14 O3 = 3*16 = 48 so 63 grams/mol so 6.3 grams
*October 25, 2015*

**Math fix typo**

= 7! /{2*5!} = 7 * 6 /2 = 21 well that checks so
*October 25, 2015*

**Math**

combinations of 2 at a time? for example for 7 students n!/[ 2! (7-2)! ] = 7 /{2*5!} = 7 * 6 /2 = 21 well that checks so 90! /[ 2!(88!) ] = 90 * 89/2 = 45*89 = 4005
*October 25, 2015*

**math**

Ke = (1/2) m v^2 = 20000 volts * electron charge
*October 25, 2015*

**physics**

speed = s = 22.7 m/s Vi = initial vertical component of speed = 22.7 m/s sin 21 v = Vi - g t at top v = 0 so going up t = Vi /9.81 it spends the same time coming down so 2 t = time in air = 2 Vi/9.81 or 2 t = 2 * 22.7 * sin 21 /9.81
*October 24, 2015*

**Physics**

mass is irrelevant when centripetal acceleration = 1 g, the vehicle leaves the road Va = v^2/r = g so r = v^2/g = 400/9.81 around 40 meters
*October 24, 2015*

**Maths**

graph cos x and sin 2 x first cos x is negative the whole way from pi/2 to 3pi/2 sin 2x goes through an entire period, first negative, then positive so you must figure out where one curve is below the other and where above
*October 24, 2015*

**problem incomprehensible**

An angle can not be parallel to another, only lines - moreover I have no idea what your drawing looks like
*October 24, 2015*

**Physics**

Perhaps it means force and acceleration are related uniformly but acceleration and position are not simply related and in fact position is the second integral of acceleration for example: If force is constant, acceleration is constant. If acceleration is constant, velocity is ...
*October 24, 2015*

**Astronomy**

You are welcome.
*October 24, 2015*

**Astronomy**

yes, here http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0CCQQFjABahUKEwjI1qjtttzIAhVLGz4KHfxjBKY&url=http%3A%2F%2Fwww.atmos.washington.edu%2F~davidc%2Fpapers_mine%2FCatling2009_SciAm_AtmEscape_Preprint.pdf&usg=AFQjCNE-...
*October 24, 2015*

**Astronomy**

Yes, I would guess those light gasses would exceed escape velocity due to heat.
*October 24, 2015*

**Typos, math, physics**

forgot x -.28 * 9.81 * x = (1/2) * (.13^2 - .28^2)
*October 24, 2015*

**physics**

work done = force * distance = change in kinetic energy F = -.28 m g = -.28 * 18 * 9.81 work = F * x final - initial Ke = (1/2) *18 * (.13^2 - .28^2) so -.28 * 9.81 = (1/2) * (.13^2 - .28^2)
*October 24, 2015*

**Physics (Biomechanics)**

weight component down hill = 75*9.81 * sin 30 a = (75*9.81*.5 - 11 -9 )/75
*October 24, 2015*

**Physics**

west wrap right hand fingers around wheel so the top of the tire scrapes your hand from palm to finger tips. Your thumb points left or west. Rotation is clockwise around westward pointing axis.
*October 24, 2015*

**physics**

I do not do magic. I do not have your diagram. Is the monkey falling at the same time you release the banana? If so, then shoot straight at the monkey because the banana and the monkey fall at the same rate. Thus you only need the time for the monkey to fall H h = (1/2) g t^2 ...
*October 24, 2015*

**History**

Sorry - use Ms Sue reply
*October 24, 2015*

**History**

It is an estimate because it is virtually impossible to gather data on every single citizen
*October 24, 2015*

**astronomy**

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/wien3.html
*October 24, 2015*

**astronomy**

well, what is wavelength of visible light? for resolution it is the ratio of wavelength to distance between slots that matters. for your radio it is 1^10^-3 meters / .6*10^ m the lower the wavelength to baseline, the sharper
*October 24, 2015*

**astronomy**

by the way p is pi, 3.14159 etc
*October 24, 2015*

**astronomy**

http://www.schoolphysics.co.uk/age16-19/Mechanics/Gravitation/text/Geostationary_satellite/index.html
*October 24, 2015*

**Physics**

west - right fingers around wheel, thumb points west
*October 24, 2015*

**algebra**

n - 3 = 39 - n 2 n = 42 n = 21
*October 24, 2015*

**math**

well, it does not ask for x but x - 11 = 134 - 2 x - 22 3 x = 123 x = 41
*October 24, 2015*

**math**

C+D = 67 so C = 67 - D (D-11) = 2 (C-11) call D x x-11 = 2(67-x -11)
*October 24, 2015*

**math help azap!!!!!!!!!!!!**

yes, subtract 15 from both sides and you get x = -7
*October 23, 2015*