Saturday

April 19, 2014

April 19, 2014

Total # Posts: 16,161

**PHYICS**

average angular velocity during stop = 45/2 = 22.5 rad/sec so total angle = 22.5 rad/s * 5 s = 112.5 radians 112.5 radians / 2 pi radians/revolution = 17.9 revolutions

**Physics**

m g h = increase of potential energy 80 * 9.81 * 10^3 = 785*10^3 Joules 785*10^3 = (1/4) total fuel energy burned so total fuel burned = 3.14 * 10^6 Joules 3.14 * 10^6 J / 37*10^6 J/kg = .085 kg

**Calculus 3**

I got sqrt 5 on the bottom

**Algebra**

10 log (2.4*10^-5/10^-16) 10 [ log 2.4 + log 10^11] 10 [11+.38] = 114

**Chemistry**

1 atm = 10.1 Newtons/cm^2 to find that type 1 atm = Newtons/cm^2 into a Google search box 10.1 Newtons/cm^2 * 17.5*10^3 cm^2 = 177,000 Newtons use the Google search box again: 177000 newtons = pounds and you will get back 39,791 pounds

**Simplify 2**

You are welcome :)

**Simplify 2**

19^.5 * 19^.5 = 19^(.5+.5) = 10^1 = 19

**Intermediate Algebra**

here is the method. Put your numbers in: http://www.jiskha.com/display.cgi?id=1294619397

**Physics**

u = 16.1 cos 44 = horizontal velocity the whole time Vi = 16.1 sin 44 = initial vertical velocity h = Hi + Vi t - 4.9 t^2 3 = 0 + Vi t - 4.9 t^2 4.9 t^2 - (16.1 sin 44) t + 3 = 0 solve quadratic for t, use the big one v = Vi - 9.8 t u = 16.1 cos 44 still speed^2 = u^2 + v^2

**Geometric Constructions HELP!**

well, wings are symmetrical port and starboard but not identical replacement cartridges for your printer. DVDs most replacement machinery parts for example new nut for bolt.

**Math**

f-g = -3 x +irrelevant - 9 x - irrelevant = -12 x + b slope = m = -12 Now you knew these last two perfectly well. Give your answer and let us just check.

**precal**

r = 16 , I assume grazes ground h = 16 - 16 cos T where T is angle up from straight down h = 16(1 - cos T) let's find out what angle hits 24 meters 24 = 16 (1 - cos T) 1.5 - 1 = - cos T cos T = -.5 T = 120 degrees so between 120 and (360 - 120) = 240 which is 120 or 1/3 of...

**Math**

constant slope is a straight line

**Math**

f*g = (x+4)(x-4) = x^2 -16 b) y = x^2 - 16 x^2 is always >/= 0 so smallest y is -16

**Ethics**

By the way you did not say if you were to pick only one. I am not sure I would count happiness as a moral value, although some would.

**Ethics**

1. yes, values 2. yes, physical strength

**Social Studies--Check My Answers Please--**

secede

**Check my work plz?**

#11 As far as I know they all are. #14 a square is a quadrilateral 4*90 = 360 I do not understand the earlier questions anyway.

**Check my work plz?**

#11 As far as I know they all are. #14 a square is a quadrilateral 4*90 = 360

**Math**

y = 5 x or any other linear function with finite slope y = x^3 or any other odd power of x

**Physics**

G (m )(4-m) /.25^2 = 2.5*10^-10 (4 m-m^2)=2.5*10^-10*.0625*10^11 /6.67 4 m - m^2 = .0234 * 10^1 = .234 m^2 - 4 m + .234 = 0 m = [ 4 +/- sqrt(16 - .0937) ]/2 = 2 +/- 1.997 so I suspect one of them is tiny and the other is about 4 .00293 and 3.997

**Math**

a) they are odd but 1/x is undefined at x = 0 b) linear but x is odd and |x| is even c) d/dx (2^x) = 2^x ln 2 so both derivatives keep repeating but sin x is periodic

**Calculus**

Note - I made a mistake in earlier problem, note correction, scroll down

**Calculus**

yes, that is where f(x) hits the x axis

**Math**

5 (25/12) = 125/12

**Math**

ok f'(x) = v(x) = .0015 x - .5825 at x = 100 v(x) = .15 -.5825 = -.4325 = -.43 dh/dt is negative, going downhill :)

**Math ???????**

hey - do you mean height OVER (meaning divided by) time or height as a function of time? If the latter, we need to take the derivative.

**Math**

-(5 f/6) > 6/3 we want to multiply both sides by -6/5 If we multiply by a negative, then arrow reverses f < (6/3)(-6/5) f < -36/15

**maths**

distance = rate * time 300 = 60 t t = 5 hours

**Pre-Cal**

I already did this today with a 7 minute period anyway They gave you PERIOD = 9 minutes Amplitude = wheel radius = 25 meters midline at 5 + 25 = 30 meters h = 30 - 25 cos (2 pi t/T )

**Math**

well wherever the derivative is zero, the INSTANTANEOUS rate of change is zero. That means whenever the cos x = 0 which is at pi/2 , 3pi/2 etc so those do work but..... HOWEVER the average change is zero between any two points a period apart in other words between x = 0 and x...

**whoops error**

a^2 + 4 a = -a -3 a^2 + 4 a + 3 = 0 a = [ -4 +/- sqrt (16-12)]/2 a = -2 +/- 1 a = -1 or -3

**Math**

You are welcome :)

**Math**

h(a) = a^2 + 3 a + 2 h(-3)= 9 - 9 + 2 h(a) -h(-3) = a^2 + 3 a so (a^2 + 3 a) /(a+3) = -1 a^2 + 3 a = -a - 3 a^2 + 4 a + 4 = 0 (a+2)(a+2) = 0 a = -2

**Math**

just for fun, let's see when v = 0 9.8 t = 23 t = 2.3 seconds at top so our problem is all on the way up. so h(2) = -4.9(4) + 23(2) + 2 h(1.5)=-4.9(2.25)+23(1.5) + 2 h(2)-h(1.5) = -4.9(4-2.25) + 23(.5) = 4.5 so v average = 4.5/.5 = 9 m/s part b well I already differentiate...

**Pre-Cal**

r = 20 period = 7 min = T when t = 0 h = 5 at t = 0 and wheel does rotation between h = 5 and h = 45 center at h = 25 so of form h = 25 - 20 cos (2 pi t/T )

**Algebra**

L = 2 x + 4 A = L w = (2x+4)(x+2) = 2 x^2 + 8 x + 8 using foil or use distributive property 2x(x+2) = 2 x^2 + 4 x 4 (x+2) = 4 x + 8 -------------------------add 2 x^2 + 8 x + 8 remarkable :)

**Physical Science Need Help Quick PLEASE:)**

Al;so go back and find I added to answers to your earlier questions

**Physical Science Need Help Quick PLEASE:)**

Newton - THIRD LAW use Google !!!!!

**physics**

High school texts --- BOO ! HEADING is the direction you point a boat or plane BEARING is the angle you measure to a lighthouse FROM your boat or plane. That said: V across = 2.11 V down stream = 1.84 |V| = sqrt (2.11^2 + 1.84^2) |V| = 2.8 m/s

**physics**

momentum at start = 0 momentum at end = .048*69.4 F = change in momentum/change in time (That is the original form Newton gave for the second law. If m is constant that is F=mA) = .048*69.4 / .00101 Newtons

**physics**

Ac = v^2/r F = m Ac = 1200 * 144/45

**B-Tech mechanical**

Google "Mohr's circle". for example: http://www.youtube.com/watch?annotation_id=annotation_851576&feature=iv&src_vid=1plaFAWN8rk&v=qqgfKHBpj1k

**Physical Science**

a = change in velocity/change in time = (11 - 0) / 10 = 1.1 m/s^2

**Physics**

However in general find x or east component Cx =A cos(angle to x axis of A) + B cos (angle to x axis of B) find y or north component Cy =A cos (angle to y axis of A) + B cos (angle to y axis of B) (note cos of angle to y axis is sin of the angle to x axis) then |C|^2 = Cx^2+Cy...

**Physics**

I do not know what those angles are measured from and in what direction from the x axis or north or whatever

**Science**

divide the total distance by the total time. That gives the average SPEED. If you divide the total DISPLACEMENT (which includes the effect of the directions of the VELOCITY vectors) by the total time you get the average VELOCITY for example walk one mile north at 1 mile/hr the...

**typo**

your DISTANCE walked is 2 miles in two hours so your average speed was 1 mile/hour

**Science 8th Grade**

divide the total distance by the total time. That gives the average SPEED. If you divide the total DISPLACEMENT (which includes the effect of the directions of the VELOCITY vectors) by the total time you get the average VELOCITY for example walk one mile north at 1 mile/hr the...

**Science 8th Grade**

You are welcome.

**Science 8th Grade**

The hiker spent one hour going 6 km west.

**Algebra-help**

You are welcome.

**typing ?**

could you mean y = 2/(7x-4) or do you mean what you typed y = [2/(7x)] - 4 or possibly could you mean y = (2/7) x - 4 ???????????????? if the last, which I am guessing from the 28 in all the answers, then: 7 y = 2 x - 28 2 x - 7 y = 28

**Math HELP**

sin 30 = 1/2 so 1/2 = short side/hypotenuse = short side/18 so short side = 9 which is none of the above.

**Physics**

u = .9 cos 23 = .828 until it hits ground Vi = - .9 sin 23 = -.352 so h = Hi + Vi t -4.9 t^2 0 = 5.4 - .352 t - 4.9 t^2 t^2 + .0718 t - 1.1 = 0 t = [ -.0718 +/- sqrt(.00516+4.4) ]/2 t = [ -.0718 +/- 2.1 ] /2 = 1.01 second d = u t = .828*1.01 = .839 meters

**Physics**

How long does it take him to reach the landing spot? d = 0 + 0 + (1/2) a t^2 8 = .5 * 2 * t^2 t^2 = 8 t = 2 sqrt 2 seconds he has to start 2 sqrt 2 seconds before the ball reaches the catching point.

**Physics**

u = 30 cos 38.3 Vi = 30 sin 38.3 Hey, this is very much like the pizza on the roof problem I just did. Follow my solution to that one, then do this one.

**Physics**

All three of these problems are really the same Joy. Solve one and you solve them all.

**Physics**

Vi = 20 h = Hi + Vi t - 4.9 t^2 -5 = 0 + 20 t - 4.9 t^2 t^2 - (20/4.9) t - 5/4.9 = 0 solve quadratic for two values of t, use the positive one.

**Physics**

I showed you how to get the magnitude. Scroll down. tan theta = sumy/sumx

**Physics**

x components of each 57 cos 130 = -36.6 38 cos -130= -24.4 25 cos 29= 67 cos -64= sum of x components = sumx = y components of each 57 sin 130= 38 sin -130= 25 sin 29= 67 sin -64= sum of y components = sumy = |F|^2 = sumx^2 + sumy^2

**Physics**

x = r cos T y = r sin T 4 = r cos 46 r = 4/cos 46 = 5.76 y = 5.76 sin 46 = 4.14

**Science**

You are welcome :)

**Science**

Yes

**Science**

Google "photosynthesis".

**Edu 656-Education My take on it**

Well, let me see. I read "The Sea and Civilization" which reminded me to go south from Crete, then east along the North African coast, drop my wares at Alexandria, sail east to the Levant, North along the Levant to Asia Minor, west along the coast of Asia Minor to ab...

**English help explaining**

It is gibberish but I think the author is trying to say that the immigrant is isolated from the new community and remembers the old life as he/she is learning to live in the new place and its differences from the former life.

**Math**

x^2 + 8 x + 1 = -3y x^2 + 8 x = -3y - 1 x^2 + 8 x + 16 = -3 y + 15 (x+4)^2 = -3 (y-5) vertex at (-4,5) symmetric about x = -4 y = 0 when (x+4)^2 = 15 x+4 = +/- sqrt 15 x = -4 +/- sqrt 15 are the zeros

**Math**

As it is stated it is impossible to solve because you can not start at both 20 and at 1.25 feet :)

**Math**

Where did you find this problem? at t = 0, the senior has to be at 20 feet I think. however your function shows her at 1.25 feet at t = 0 That is not right. Moreover in real life h = Hi + Vi t -(1/2) g t^2 in the feet/seconds system -(1/2) g t^2 is about -16 t^2 NOT -t^2

**math**

r = 2 is positive root from http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php so w = 4

**math**

width = 2 r length of rectangle = 2 r + 2.25 area of semicircle = (1/2) pi r^2 area of rectangle = 2r(2r+2.25) total area of free throw section = (pi/2) r^2 + 4 r^2 + 4.5 r Pi/2 + 4 = 5.57 5.57 r^2 + 4.5 r = 31.28 5.57 r^2 + 4.5 r - 31.28 = 0 use quadratic equation to solve or...

**math**

V = pi r^2 h = pi (x+5)^2 x = pi (x^2 +10 x +25)x = pi (x^3 + 10 x^2 + 25 x = 1920 ???? if so then x^3 + 10 x^2 + 25 x - 611.15 = 0 find roots: http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php Only real root is at x = 5.52 then r = 10.5...

**Math**

if k+1 = 0 , then the product is 0 so k+1 = 0 k = -1 first answer if 4k+5 = 0 then product = 0 so 4k = -5 k = -5/4 second answer

**Physics**

speed up = 1.2 - .5 = .7 speed down = 1.2 + .5 = 1.7 time up = 1000/.7 seconds time down = 1000/1.7 seconds add those times in still water 2000/1.2 seconds

**Math**

I assume D x a means D a I know nothing about medicine but a) at t = 0, e^0 = 1 so C(0) = something(1-1) = 0 It has not had time to be absorbed after a long time, both e^-bt and e^-at approach zero so the concentration approaches zero. At first assuming a is much bigger than b...

**Physics**

ultraviolet is not visible for humans.

**Physics**

http://www.jiskha.com/display.cgi?id=1392230927

**Science ( Need help really bad Please!)**

F = m g or Weight = mass * acceleration of gravity m is mass, sometimes in kilograms. It is the same everywhere and depends only on which proton and neurons are there and how many. g is the acceleration of gravity. On earth it is about 9.81 meters/second^2 On the moon for exam...

**Astronomy**

See below about precession of the equinoxes. http://www.astro.com/astrology/in_praezession_e.htm

**Astronomy**

Leo

**science**

The water exerts a force up on your hand equal to the weight of water displaced. Therefore an equal and opposite force is down on the scale. Ultimately that force comes from your arm pushing the finger down into the water. In the air there will be no effect because you are not...

**physics**

a = F/m = 890/265 d = 0 + 0 + (1/2) a t^2 3 = (1/2)(890/265) t^2 solve for t v = 0 + a t v = (890/265) t

**physics**

C = Q/V 30 * 10^-12 = Q/3 Q = 90 * 10^-12 coulombs

**maths**

lcd = 3*3*5 = 45 8/45

**Algebra 2**

I do not know a better way than to say the terms are 2n/(n+2)

**Math**

(12in/36 hr)* (24*7 hr/wk) = 56 in/wk

**math**

use denominator of 36 5/6(6/6) + 3/4(9/9) = 1/18(2/2)- 6c - 2/3(12/12) + 7c + 1/9(4/4) 30/36 +27/36 =2/36 -6c -24/36 +7c +4/36 75/36 = 1 c

**Math**

15 - .75 n = 13 - .5 n .25 n = 2 n = 8

**Physics**

weight = mass * g = 9.81 m

**Physics**

gravity on earth g = 9.81 m/s^2

**Physics**

You are welcome :)

**Physics**

weight = m g = 1*9.81 = 9.81 N max friction force = .3*9.81 = 2.94 N 2.94 = k x = 29 x x = .101 meter = 10.1 cm

**physics**

P = V*i = 4.04*10^6 * 23.4 * 10^-3 94.5 * 10^3 = 94,500 watts

**PHYSICS**

By the way, I answered your roller question problem long ago.

**PHYSICS**

energy stored in LH spring at max = (1/2)kx^2 = (1/2)130(.16)^2 same amount will be stored in the RH spring when it is compressed all the way (1/2)130(.16)^2 = (1/2)280 x^2 x^2 = (130/280).16^2 x = .11 meter = 11 cm part b (1/2)k x^2 = (1/2)m v^2 130 * .16^2 = .2 v^2 v = 4..08...

**Intermediate Algebra**

F(t) = 600 - 20 t for t </= 30

**PHYSICS**

http://www.jiskha.com/display.cgi?id=1392238771

**physics**

it starts at h = 5 it ends at h = 1.5 it fell 3.5 meters

**MATH**

You are welcome :) be sure you want mm^2, usually that would be converted to cm^2 or m^2

Pages: <<Prev | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | **19** | 20 | 21 | 22 | 23 | 24 | 25 | Next>>