Monday

December 22, 2014

December 22, 2014

Total # Posts: 563

**physics**

65 = .5 g t^2 130 = 9.8 t^2 the distance is 7.8*t
*November 11, 2012*

**physics**

.4 = .5 g t^2 .8 = 9.8 t^2 the x-velocity is 76/t
*November 11, 2012*

**physics**

A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 173 N at an angle of 27.5 above the horizontal. The box has a mass of 27 kg, and μk between the box and the floor is 0.2. The acceleration of ...
*November 6, 2012*

**physics**

Consider the 57 N weight held by two cables shown below. The left-hand cable is horizon- tal. 57 N 52 a) What is the tension in the cable slanted at an angle of 52? Answer in units of N
*November 6, 2012*

**physics**

A 2.67 kg block starts from rest at the top of a 27.6 incline and accelerates uniformly down the incline, moving 2.89 m in 2.45 s. The acceleration of gravity is 9.81 m/s2 . a) What is the magnitude of the accelera- tion of the block? Answer in units of m/s2
*November 5, 2012*

**physics**

A 31 kg box slides down a 33.0 ramp with an acceleration of 1.30 m/s2. The acceleration of gravity is 9.81 m/s2 . Find the coefficient of kinetic friction between the box and the ramp.
*November 5, 2012*

**math(please help)**

Al + As + B = 124 Al + B = 102 subtracting equations ___ As = 22 Al + As = 86 subtracting equations ___ Al = 64 subtracting equations ___ B = 38
*November 4, 2012*

**Physics**

the vertical component of the brick's velocity is 13.9 * sin(27.6º) 0 = (-.5 g 3^2) + [13.9 * sin(27.6) * 3] + (building height)
*November 4, 2012*

**Math**

d/dx [cos(x)]^2 = 2 cos(x) * [-sin(x)]
*November 4, 2012*

**Pre- AP Physics**

the gravitational force between two objects is the same for both objects
*November 4, 2012*

**Math**

a. w = f * d = 150 * 4 = 600 Nm b. 600 / .5 = 1200 Nm
*November 3, 2012*

**physics**

A 1.6 kg otter starts from rest at the top of a muddy incline 95.9 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline? Answer in units of N
*November 3, 2012*

**physics**

A shopper in a supermarket pushes a loaded 32 kg cart with a horizontal force of 10 N. The acceleration of gravity is 9.81 m/s2 . a) Disregarding friction, how far will the cart move in 3.3 s, starting from rest?
*November 3, 2012*

**physics**

A freight train has a mass of 3.1 × 107 kg. If the locomotive can exert a constant pull of 4.9 × 105 N, how long would it take to increase the speed of the train from rest to 71.9 km/h? Disregard friction.
*November 3, 2012*

**physics**

f = m * a ___ a = f / m a = .052 N / 474 kg v = a * t ___ t = v / a t = 880 m/s / (.052 N / 474 kg) t will be in sec__convert to days
*October 30, 2012*

**physics**

Salmon often jump waterfalls to reach their breeding grounds. Starting 1.77 m from a waterfall 0.253 m in height, at what minimum speed must a salmon jumping at an angle of 44.8 leave the water to continue upstream? The acceleration due to gravity is 9.81 m/s2 . Answer in...
*October 16, 2012*

**physics**

A basketball player 1.9 m tall wants to make a basket from a distance of 7.8 m. The hoop is at a height of 3.05 m.If he shoots the ball (from a height of 1.9 m) at a 23.5 angle, at what initial speed must he shoot the basketball so that it goes through the hoop without ...
*October 16, 2012*

**math**

if an airplane is traveling at 245 meters per second, how many kilometers per second does it travel?
*October 10, 2012*

**Geometry**

interior + exterior = 180º
*October 3, 2012*

**7th grade Science help**

1. a 2. b 3. d
*September 26, 2012*

**math**

divide 1st (multiply by inverse) (1/3 * 7/5) + (5/14 * 7/5) 7/15 + 1/2 = 14/30 + 15/30
*September 26, 2012*

**Pre-Calculus**

f(x) + g(x) = (x^2) + (3x + 9) = x^2 + 3x + 9 = h(x) you can break it into whatever pieces you like h(x) = (x + 5)(x + 3) = x^2 + 8x + 15 f(x) = x^2 + 9 ; g(x) = 8x + 6 h(x) = f(x) + g(x) a. f(x) = 3x - 1 ; g(x) = 3 - 2x __ (3x - 1) + (3 - 2x) = x + 2 = h(x) b. f(x) = 7x - 3...
*September 26, 2012*

**physics**

it takes 5 s to stop the car __ 55 / 11 the average speed (Va) is __ (55 + 0) / 2 the stopping distance is __ Va * 5 twice as fast means that the stopping time is doubled, along with the average speed __ this means that the car travels 4 times as far
*September 26, 2012*

**math**

(3 1/2) / (2 1/3) (7/2) / (7/3) = 3/2 = 1 1/2
*September 26, 2012*

**science**

1. a 2. d 3. a 4. c 5. b
*September 26, 2012*

**science**

LARGER
*September 26, 2012*

**Physical Science**

you need 500 N to overcome the static friction as is frequently the case; once something starts, it is easier to keep it moving
*September 26, 2012*

**physics**

the frictional force is __ .29 g the time to stop is __ 5.6 / (.29 g) the average velocity is __ (5.6 + 0) / 2 the distance traveled is the average velocity divided by the time
*September 26, 2012*

**Physics**

t * 2.3 * 120 = (59/60) * 16 * 240
*September 26, 2012*

**math grade 10**

cuberoot(1280) cuberoot(64 * 20) 4 * cuberoot(20)
*September 26, 2012*

**Geometry**

if you double the length, you double the area to have an area of 55 and a perimeter of 32, the original garden was 5 X 11 so the increase in perimeter is 22 __ the percentage is __ (22/32) * 100
*September 26, 2012*

**Physics**

average speed (Va) is __ (23 + 0) / 2 time (t) is distance/speed __ 210 / Va acceleration is speed change/time __ (0 - 23) / t
*September 26, 2012*

**Stats**

74.2 * 85
*September 26, 2012*

**Chemistry**

if you over titrate, you should be able to correct with a titration of a standard acid if you think you may have added too much, the amount of the unknown acid will appear to be more than it actually is __ this would cause you to think you have more moles, so the equivalent ...
*September 26, 2012*

**chemistry**

1000 nm in a µm
*September 26, 2012*

**chemistry**

50C is halfway between the freezing and boiling points of water (0C and 100C) 50F is is slightly above the freezing point (32F) and well below the boiling point (212F)
*September 26, 2012*

**physics**

momentum is conserved 41 * 7 = v * (41 + 30)
*September 26, 2012*

**Physics**

the throwing angle has no effect the KE at impact is the initial KE plus the KE gained from the gravitational potential (m g h) KE = (.5 * .058 * 11.0^2) + (.058 * 9.8 * 12.3) v^2 = (2 * KE) / .058
*September 26, 2012*

**Chem**

the molecular weight of CS2 is 76.13 so the C in 67.1 g is: 67.1 * (12.01 / 76.13) the molecular weight of CCl4 is 153.81 so the C in 45.2 g is: 45.2 * (12.01 / 153.81) divide the CCl4 carbon by the CS2 carbon to find the percent yield
*September 26, 2012*

**Physics**

the frictional force (which must be overcome) is __ m * g * 0.200
*September 26, 2012*

**Physic**

(m v^2) / r = m g sinΘ v^2 = 6.00 m * g * sin(9.0º)
*September 26, 2012*

**Physics**

[2.00 * sin(37.0º)] + [1.785 * sin(153.5º)]
*September 26, 2012*

**Math**

look at 1000 items M1 produces 300 with 18 defects M2 produces 300 with 15 defects M3 produces 400 with 16 defects the probability that a defect is from M1: 18 / (18 + 15 + 16)
*September 26, 2012*

**statistics**

80 ≤ [(88 + 83 + 86 + 72 + 3x) / 7] ≤ 89
*September 26, 2012*

**Physics**

the centripetal force (gravity) is now 4 times as large f = m v^2 / r multiplying f by 4 (and keeping m and r constant) means that v is doubled doubling the velocity would halve the period
*September 26, 2012*

**Calculus-math**

using slope-intercept, the given slope is -1/2 the slope of the tangent to the function is the 1st derivative f(x) = 1 / sqrt(x) = x^(-1/2) f'(x) = (-1/2) x^(-3/2) substitute to find x and then y at the point of tangency use point-slope to write the equation of the new ...
*September 26, 2012*

**math**

16! / (7! * 1! * 8!)
*September 26, 2012*

**Physics**

the average velocity during deceleration is __ (76 km/h + 0) / 2 the time is the distance divided by the average velocity you will have to convert km/h to m/s to find the answer in s
*September 25, 2012*

**Physics**

the average velocity (Va) during acceleration is __ 3000 m / (3.8 * 60) sec the final velocity (Vf) is related to the initial velocity (Vi) by: Vf = (2 * Va) - Vi
*September 25, 2012*

**statistics**

there are 10C3 ways of drawing 3 chips there are 7C3 ways of drawing 3 red chips there is one way to draw 3 blue chips p = 1 - [(7C3 + 1) / 10C3]
*September 25, 2012*

**math**

(8 * 6) + (½ * 5 * 4)
*September 25, 2012*

**math**

-8bc^4
*September 25, 2012*

**Precalc**

you seem to be finding the axis of symmetry okay when you plugged in to find the vertex in #1, there is an error (should be 2,-4) __ this also affects the range completing the square should be: f(x) = (x - 2)^2 - 4 (x - 2)^2 = x^2 - 4x + 4 so you need to subtract out the extra...
*September 25, 2012*

**physics**

gravity also opposes the motion with a force of __ g * 297 * sin(25.9º) sum the forces; then find the acceleration with __ f = m a
*September 25, 2012*

**physics**

h = -.5 g t^2 + 2.6 t + 1.3 the max height occurs at the value of t that lies on the axis of symmetry (t = -b / 2a) __ t = -2.6 / -9.8 find t and substitute back to find the max height
*September 25, 2012*

**math**

Py = Pz = (2 * 5) + ( 2 * 3) write the equation for the perimeter of y using the length and width (2 * Ly) + (2 * Wy) = Py = Pz write the equation for the area of y using the length and width Ly * Wy = Ay = Py = Pz you now have two equations with two unknowns that you can ...
*September 25, 2012*

**algebra**

subtract the two equations to ELIMINATE the x-term -5y = 9 __ y = -9/5 substitute back to find x 4x + 9/5 = 5
*September 25, 2012*

**science**

gas equation temperatures are in Kelvin (absolute) v = (100 cm^3)[(1 atm) / (2 atm)](263º / 293º)
*September 25, 2012*

**physics**

it appears that the mass of the block is not involved in your final calculation (it cancels out in the second line) your work seems to be correct
*September 25, 2012*

**Physics**

you seem to have forgotten gravity add g to your acceleration number and you should be okay
*September 25, 2012*

**physics**

the acceleration is the change in velocity over time in this case __ (12.0 - 7) / 2.0
*September 25, 2012*

**Math**

Yoko's speed is 9/10 of Yolanda's in the time it takes Yolanda to run 110 yd, Yoko runs 99 yd so Yolanda wins the second race by a yard for a tie, the 100 yd that Yoko runs must be 9/10 of the distance that Yolanda runs .9 x = 100 __ x = 1000/9 Yoko runs 100 yd (900/9...
*September 24, 2012*

**math 176**

(x - 3y + 2z)(x - 3y - 2z)
*September 24, 2012*

**MaTh**

the area of the base is a factor of 42 with a perimeter of 18, the possible block arrangements are __ 8 X 1 __ area 8 7 X 2 __ area 14 6 X 3 __ area 18 5 X 4 __ area 20 14 is the only factor of 42, so the height must be 3
*September 24, 2012*

**Pre-Ap Physics**

the initial vertical velocity is __ (2650 m/s) * cos(41.9º) the time of flight is the initial vertical velocity divided by the gravitational acceleration __ 2 * [(2650 m/s) * sin(41.9º)] / (9.8 m/s^2) the horizontal velocity is __ (2.65 km/s) * cos(41.9º) the ...
*September 24, 2012*

**Science chemistry**

if n is the number of half-lives (1/2)^n * 1.000 = 0.121 (1/2)^n = 0.121/1.000 n [log(1/2)] = log(0.121) n = [log(0.121)] / [log(1/2)] half life = [1h17m] / n
*September 24, 2012*

**chem**

both compounds have one Cu, so the moles should be the same to find grams, multiply by the molar mass of CuO
*September 24, 2012*

**chemistry**

density is measured in g/cm^3 d = 19.07 / (3.1 * 1.9 * 4.6)
*September 24, 2012*

**Physics(Please help, thank you)**

the force that was needed to balance out F1 and F2 is equal in magnitude to the resultant, but opposite in direction so the resultant would be 280g at 77º since F1 and F2 have equal magnitudes and are 90º apart; the expected resultant would be 283g at 75º __ so ...
*September 24, 2012*

**physics**

momentum is conserved the mass of the clay, times its velocity; is equal to the mass of the clay+block, times its velocity 0.5 * v = (6 + 0.5) * 3
*September 23, 2012*

**Physics**

at 4s, the speed is 24 + 4g at 5s, the speed is 24 + 5g so the average speed over the interval (for one second) is 24 + 4.5g the distance is the speed times the time
*September 23, 2012*

**Physics**

h = .5 * g * t^2 34 = .5 * 9.8 * t^2
*September 23, 2012*

**physics**

constant velocity means no acceleration, which means zero net force so the third force is equal in magnitude to the resultant of the given forces; but opposite in direction the magnitude of the resultant of the given forces is __ sqrt(61.9^2 + 52.0^2) the tangent of the angle ...
*September 23, 2012*

**physics**

the work done on the car is reflected by the change in kinetic energy w = [.5 * 2000 * 20^2] - [.5 * 2000 * 10^2] the units are joules
*September 21, 2012*

**Science**

the kinetic energy at the bottom is the kinetic energy at the top plus the gravitational potential across the circle .5 m (Vb)^2 = [.5 m (Vt)^2] + [m g h] dividing by .5 m (Vb)^2 = (Vt)^2 + 2 g h v^2 = 3^2 + [2 * 9.8 * (2 * .8)]
*September 21, 2012*

**precal help**

n = d - 4 [(d - 4) / d] + [d / (d - 4)] = 5/2 multiplying by [2 * d * (d - 4)] [2 (d - 4)^2] + (2 d^2) = 5 d (d - 4) 2d^2 - 16d + 32 + 2d^2 = 5d^2 - 20d 0 = d^2 - 4d - 32 factoring __ 0 = (d - 8)(d + 4) 0 = d - 8 __ 8 = d
*September 21, 2012*

**Physics**

this is like a projectile problem turned sideways 0 = -(.5 * 3.7 * t^2) + (30 * t) solve the quadratic for t (you want the non-zero solution)
*September 21, 2012*

**physics**

the crate is sliding, so μk (the kinetic coefficient) is the relevant one the direction of the frictional force is OPPOSITE to the direction of motion the magnitude of the force is: the normal force of the crate on the ramp [m * g * cos(30º)]; multiplied by the ...
*September 20, 2012*

**phsycis**

the force diagram reflects the physical setup the walker is 15.0 m from either end; and is displacing the wire 1.0 m this means that the vertical component of the tension is one-fifteenth the tension in the wire (on BOTH sides) so the force exerted by the walker is 3675/15 * 2...
*September 20, 2012*

**physics**

because it starts from rest and accelerates uniformly, the average velocity is 1/2 the final velocity convert the km/h into m/s [multiply by 1000/3600] the distance is the average velocity, multiplied by the time (8.5 s)
*September 20, 2012*

**Math**

looks good
*September 13, 2012*

**Physics**

horizontal velocity (Vh) is the velocity (V) times the cosine of 20.0º vertical velocity (Vv) is V times the sine of 20.0º or __ Vv = Vh * tan(20.0º) = (104/4.60) * tan(20.0º) = 8.23 when the stone hits the ground, H = 0 0 = (-.5 * g * 4.60^2) + (8.23 * 4....
*September 13, 2012*

**Physics**

a) vertical __ H = (-.5 * g * 2.00^2) + 20.60 __ g is gravitational acceleration horizontal __ d = 19.0 * 2.00 b) H = 0 when the ball hits the ground 0 = (-.5 * g * t^2) + 20.60 solve for t to find the flight time
*September 13, 2012*

**Physics**

the k.e. is equal to the initial k.e. plus the potential energy from the drop k.e. = (.5 * .100 * 5.00^2) + (.100 * g * 10.0) g is gravitational acceleration
*September 13, 2012*

**Math**

a) 128 = (5 * 25) + (3 * 1) = (5 * 5^2) + (0 * 5^1) + (3 * 5^0) in base 5 __ 503 b)128 = 2^7 = (1 * 2^7) + (0 * 2^6) + (0 * 2^5) + (0 * 2^4) + (0 * 2^3) + (0 * 2^2) + (0 * 2^1) + (0 * 2^0) in base 2 __ 10000000 c) 128 = (10 * 12) + (8 * 1) = (10 * 12^1) + (8 * 12^0) in base 12...
*September 13, 2012*

**Physics**

the vertical component of the velocity (Vv)is __ 20.0 * sin(32.0º) the horizontal component (Vh) is __ 20.0 * cos(32.0º) the time (T) to max height is __ Vv / g a) Hmax = (-.5 g T^2) + (Vv * T) + 1.1 b) at Hmax, the ball has no vertical velocity; only horizontal (Vh)
*September 13, 2012*

**PHYSICS**

the time to accelerate to takeoff speed is __ 35 / 3.8 the average speed during acceleration is __ (0 + 35) / 2 the distance (length of runway) is speed times time
*September 13, 2012*

**physics**

0 = (-.5 * g * 5^2) + (10 * 5) + h g is gravitational acceleration
*September 13, 2012*

**physics**

Ha = -.5 g t^2 + 75 Hm = -.5 g t^2 + 30 t 75 = 30 t __ this is when substitute t into either equation to find where
*September 13, 2012*

**Physics**

a) [(2.25 m/s * 4.98 min) + (1.58 m/s * 2.18 min)] * 60 s/min b) (answer from a) / (4.98 min + 2.18 min) / (60 s/min)
*September 13, 2012*

**physics**

5.20 m/s * cos(51.5º)
*September 12, 2012*

**PHYSICS**

the final velocity of the ball must be zero at the edge of the table (after 1.32 m of travel) the average velocity is initial plus final,divided by two v = (2.1 cm/s + 0 cm/s) / 2 = 1.05 cm/s the time to stop the ball is the distance divided by the velocity t = 132 cm / 1.05 ...
*September 12, 2012*

**Physics**

after 1st stage: final velocity __ Vf1 = 26.4 * 3.92 final height __ Hf1 = .5 * 26.4 * 3.92^2 after 2nd stage: final velocity __ Vf2 = (12.3 * 4.48) + Vf1 final height __ Hf2 = (.5 * 12.3 * 4.48^2) + (Vf1 * 4.48) + Hf1 free-fall time is __ 21.2 s - 3.92 s - 4.48 s , or 12.8 H...
*September 12, 2012*

**science**

the angular momentum of the system is constant, so the velocity is inversely proportional the the radius of the orbit Va = Vp * (Rp / Ra)
*September 12, 2012*

**physics**

f = m a __ the force (weight) is directly proportional to the acceleration (g) weight = 685 * (9.8 / 12)
*September 12, 2012*

**physics**

f = m a __ acceleration is inversely proportional to the mass a = 2.9 m/s^2 * (2.7 / 14.9)
*September 12, 2012*

**physics**

time up equals time down (still) initial vertical velocity is __ 1650 * sin(66.1º) flight time is __ 2 * 1650 * sin(66.1º) / 9.8
*September 12, 2012*

**Physics**

ignoring the height of contact time up equals time down h = .5 g t^2 = .5 * 9.8 * 2.1^2
*September 12, 2012*

**math**

1,000,000 = 10^6 = 2^6 * 5^6
*September 12, 2012*

**physics**

average velocity equals initial plus final, divided by two Va = (12 + 24) / 2 = 18 m/s distance equals velocity times time d = 18 * 6.5
*September 12, 2012*

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