# Posts by Scott

Total # Posts: 2,433

**algebra**

thanks for the catch, Reiny too late to be working on this stuff

**algebra**

(90 / c) + [130 / (c + 20)] = 4 90 c + 180 + 130 c = 4 c^2 + 80 c c^2 - 35 c - 45 = 0 use quadratic formula to find c

**6 grade math area perimeter**

area looks good perimeter might be 16 ... check it

**6 grade math area perimeter**

looks good

**Physics**

it accelerates for 4 s ... up to 8 rad/s then spins for 11 s ... 88 rad the ave vel during acceleration is ... ) + 8) / 2 = 4 rad/s ... spins 16 rad 104 rad total ... divide by 2? to find revolutions

**calculus**

the friend's distance north is 30 t your distance east is ... 40 - 20 t it's a right triangle use Pythagoras to find the hypotenuse ... which is the distance find the minimum - with respect to t

**Physics**

the centripetal force (m v^2 / r) is equal to the gravitational force ... G M m / r^2 v^2 = G M / r ... = 6.67E-11 * 5.98E24 / 7.12E6 the period is the speed divided into the orbital circumference

**Physics , static fluids**

the ice is 92% as dense as the water so 92% of the block is submerged when it floats that leaves 8% exposed

**Algebra**

s + L = 40 s + 6 = L subtracting the equations ... eliminating s L - 6 = 40 - L

**Physics**

the supports each exert 98 N to support the plank one support exerts [490 N * m / 3] to support the person the other support exerts ... 490 N - [490 N * m / 3]

**Physics , static fluids**

the buoyancy force is the difference in the air weight and immersed weight

**Physics , static fluids**

hello (again) 8.6 meters is 860 cm if you picture a stack of 860 cm cubes of water, what is the force at the bottom? ... 860 gm * g ... this is the force on one cm^2 .86 kg * 9.8 m/s^2 = f a Pascal is one Newton per m^2 ... a kPa is .001 N per cm^2

**Physics**

you can use the density of water and the depth to find the pressure the "rule of thumb" for diving is one atmosphere for every ten meters of depth

**Physics**

e = 1/2 k x^2 ... (kind of like KE) 2.0 = 1/2 * 100 * x^2

**Physics**

look up Doppler shift formula

**Maths**

2 = (1 + .06)^t log(2) = t log(1.06) t = [log(2)] / [log(1.06)]

**Algebra 2 (Please help Ms. Sue!)**

use synthetic division

**Algebra**

(d / 55) + (d / 35) = 50 - 10

**Physics**

momentum is conserved find the momenta in x and y for the two decay products the two y momenta add to zero, and the two x momenta equal the original add the vectors to find the 234Th velocity

**pre calc**

100% - 30% = (1/2)^(2 / h) log(.7) = (2 / h) log(1/2) h = 2 log(1/2) / log(.7) (after a week) = (1/2)^(7 / h)

**Algebra**

it stops at the top, so the average velocity is ... (103 + 0) / 2 gravitational acceleration is 32 ft/s^2 , downward so the time to reach the peak is ... 103 ft/s / 32 ft/s^2 r * t = d

**Physics**

momentum is conserved, so the truck's loss is the car's gain 7500 kg-m/s / 750 kg = ?

**physics**

radio waves travel at the speed of light (c) ... 3E8 m/s the distance from Venus to Earth depends on where each planet is in its orbit around the Sun

**technical drawing**

85 m is 85000 mm 85000 mm / 10000 = ?

**science**

x + y = 10 3.1 x + 2.5 y = 25 solves easily by inspection (observation)

**physics**

6000 Nm/s = 6 kw

**Physics**

the object displaces 1/5 of its weight in water, so it is 5 times the density of water the weight change in the other liquid is 2/3 of the change in water, so the other liquid is 2/3 the density of water

**Chemistry**

the Ar will effectively be "squeezed" into 80% of its previous volume this will increase the tank pressure to 125% of the original the pp of Ar is 80% of the total the pp of nitrogen is 20%

**Physics**

F accelerates the three boxes equally since C is half of the total system mass, F equals twice the force on C

**physics**

the initial vertical velocity is ... 50 m/s * sin(37º) the (vertical) freefall equation is . h = 1/2 g t^2 + [50 sin(37º)] t + 55 plug in zero for h (ball on ground) and solve for t (use quadratic formula) the horizontal velocity is ... 50 m/s * cos(37º) ... ...

**math**

the intercepts are points with x and y coordinates for x-intercepts, the y coordinates are zero...but they still exist

**Physics**

avg vel = (14 + 0) / 2 = 7 rev/s stop time = 1/7 s distance = 1/2 a t^2 ... 1 = 1/2 * a * (1/7)^2

**Maths**

6 pts total per event 24 pts total awarded...so 4 events Betty has three 1st's and a 2nd Cathy has a 2nd and three 3rd's

**physics**

momentum (as usual) is conserved .05 * v = (5 + .05) * 10

**applied physics**

g = -980 cm/s^2 the object reaches its peak at 2 s average velocity is 980 cm/s ... so peak height is 1960 cm time up equals time down

**Physics**

KE = 1/2 m vB^2 6 KE = 1/2 * m/2 * vA^2 12 = (vA / vB)^2

**Physics**

twice the force, twice the stretch

**Math**

what design?

**Algebra**

the general equation is ... x = a(y - k)^2 + h the vertex is at the origin, so h and k are both zero a = 1 / 4p ... where p is the distance from the vertex to the focus b) the center depth is the value of x at the edges of the dish (where y equals ±7)

**Physics**

the acceleration (j) of the jet and gravitational acceleration (g) are two sides of the triangle j / g = tan(25º) ... j = 9.8 m/s^2 * tan(25º) multiplying j by 11 s will give the takeoff speed

**Psychics**

d = 1/2 a t^2 88.3 = 1/2 * 23.6 * t^2

**Math**

your "weird" answer is correct

**Math**

An = [10 * 3^(n-1)] + 1

**Technology**

the moments about the fulcrum must be equal for balance moment = mass * distance from fulcrum

**university physics**

momentum is conserved with the pellet/block find the velocity of the pellet/block the KE of the p/b is the work that compresses the spring 1/2 m v^2 = 1/2 k x^2

**Differential calculus**

the general vertex form is ... y = a(x - h)^2 + k ... (h,k) is the vertex a = 1 / 4p ... p is the distance from the directrix to the vertex

**Trig**

use Pythagoras (a^2 + b^2 = c^2) to find the angle functions then use the trig identity to find the tangent

**Advanced physics**

the period of the swing (pendulum) is related to its length the speed is not a factor

**Algebra**

a = 2 b 3(a + b) = 90

**Chemistry**

an interesting question will probably make for a lively class discussion a couple of pertinent facts ... normal exhaled air is about 15% O2 ... humans can "survive" in air down to about 6% O2

**Physics**

the work (energy) is equal to the KE of the object f d = 1/2 m v^2 123 * 45.6 = 1/2 * 10.0 * v^2

**Maths**

12 ... 4/10

**Math (Derivative)**

a) y = 6 x^-1 ... dy/dt = -6 x^-2 dx/dt ... dy/dt = (-6 / 64) * 12 b) missing y a typo?

**Math**

p(p + 1) = 3192 p^2 + p - 3192 = 0

**Science**

yes certainly if 180º apart

**Physics**

work (energy) to stop the wagon ... w = 42 * g * .18 * 16 w equals the initial KE of the wagon ... 1/2 * 42 * v^2 = w the speed of the boy is the same as the initial speed of the wagon

**Chemistry**

you are correct

**Maths**

eight "3" cards gives 24 ... 7 shy of 31 replace four "3" cards with three "5" cards and a "4" card

**Physics**

d = 1/2 a t^2 t = ?(2 * d * a)

**Physics**

the period (of the pendulum) is dependent on the length the speed of the "bob" is not a factor

**math- Trigonometry**

looks good...TALL lighthouse

**Chemistry**

.325 L * .05 M/L * 55.8 g/M

**Math**

time is inversely proportional to the number of machines and directly proportional to the number of raviolis t = 15 min * 15/75 * 6000/500

**science**

would call this a "trick" question circular path means radial (centripetal) acceleration there is no tangential component

**Math**

d = 6 233 - 53 = 180 ... so, 31 terms s = (233 + 53) * 31 / 2

**Physics**

1 kcal = 4184 joules m g h = 300 * 4184 h = (300 * 4184) / (1000 * g)

**Physics**

the gravitational potential energy becomes kinetic energy m g h = 1/2 m v^2 v = ?(2 g h)

**Physics**

20º is 1/18 of a circle (20/360) the length of the arc is ... 1/18 * 2 * ? * 1 m

**Physics**

momentum is conserved the smaller mass is 1/3 of the larger ... so it must have 3 times the speed

**Math**

check the related questions

**Physics**

f = m a ... a = f / m a = (33 - 12) / (6 + 15)

**physics**

time up equals time down h = 1/2 g t^2 = 1/2 * 32 ft/s^2v * (2 s)^2

**math**

the x-coordinate doesn't change the sign of the y-coordinate is reversed drawing a picture can help

**Math**

complex roots (zeros with i) occur in conjugate pairs f(x) = (x-6+3i)(x-6-3i)(x^5)

**math**

i = 7400 * [1 - (1 + .06/4)^(4*4)]

**algebra**

try again on the 1st one the slope formula does work

**physics**

the frictional force is ... 5 * 9.8 * .4

**Physics: Centre of Mass**

(4 * 4) + (8 * 5) > 7 * 5 ... so the CM is > 0 [4*(4-x)]+[5*(8-x)]-[7*(5+x)] = 0

**Math**

Tom has a 40 km head start ... 80 km/h * .5 h Cindy is catching up at 20 km/h ... 100 km/h - 80 km/h

**Physics**

the velocity is the vector sum of the horizontal and vertical components horizontal ... 88 m / 3.8 s vertical ... time up equals time down ... (3.8 s / 2) * 9.8 m/s^2 the max height is half the initial vertical component, multiplied by half the flight time

**Math**

distributing the 10 ... 40x + 30 - 9x

**Math**

undefined slope means the line is vertical the equation is of the form ... x = ?

**Physics**

a = v^2 / r r = v^2 / a = (54 m/s)^2 / 90 m/s^2

**Physics**

the object is moving at 20 m/s when the force is removed, the object will be decelerated to a stop (over some distance) by friction find the frictional force (m * g * .2) find the deceleration ... a = f / m find the stopping time ... t = v / a = 20 / a find the stopping ...

**pre-calc.**

100 = 10^2 10 = 10^1 logarithms are exponents -- powers of some base regardless of the base, the logarithm of 100 will always be twice the logarithm of 10

**Pre-Calc**

the values for sine and cosine are limited to the range ±1

**physics**

h = 1/2 g t^2 40 = 1/2 * 9.8 * t^2

**Physics**

gravity follows an inverse square relation two radii above the surface is three times the distance from the center (of mass) to the surface so the force would be ... g / (3^2)

**physics**

Pi * Vi / Ti = Pf * Vf / Tf the T is absolute (Kelvin) temperature

**physics**

you're adding two vectors 2^2 + 4^2 = s^2

**Physics**

need the drawing for a solution or, at least, more data

**Physics**

(b) f = m a ... a = f / m = 900 / 1800 a = .5 m/s^2 so it takes 40 s to stop ... 20 / .5 ave velocity is ... (20 + 0) / 2 distance = v * t

**Algebra**

.12 x + .10(7500 - x) = 890 .02x + 750 = 890

**Science Psysics**

the gravitational potential energy at the top of the ramp becomes kinetic energy at the bottom of the ramp (minus the work done by friction) m g h - f d = 1/2 m v^2 [6 * 9.8 * 3.5 sin(31º)] - (17 * 35.) = 1/2 * 6 * v^2

**Algebra 2**

x^2 - 8x + 14 = 0 use the quadratic formula to find x x = {8±?[64-(4 * 1 * -14)]} / (2 * 1)

**Algebra 2**

use the quadratic formula to find the value(s) of x

**physics**

momentum is conserved (1200 * 14) + (2000 * 25) = ... (1200 + 2000) * v

**Chemistry**

8 half-lives 625 * (1/2)^8

**Chemistry**

3 half-lives m * (1/2)^3 = 6

**Chemistry**

2 days (48 hr) is 8 half-lives 10.0 * (1/2)^8