Saturday

August 30, 2014

August 30, 2014

Total # Posts: 22

**Math**

c^2 = a^2 + b^2 22.1^2= 16.8^2 + b^2 solve for 'b' using pythag. theorem
*August 31, 2012*

**physics**

time to travel horizontal distance (8.7m) = time to fall vertical distance (3.3m) (a) solve for t in the vertical direction (i) initial speed in vertical direction is 0 m/s (ii) d = (initial speed in vertical diection)t + 1/2 g t^2 3.3 m = 0 + 1/2(9.81 m/s) t^2 (iii) solve for...
*August 31, 2012*

**math**

distance= (5280/1 mi)(2 mi)
*August 31, 2012*

**Science**

force = mass x acceleration mass = 5 kg acceleration = 10 m/s2 plug in the values and you have your answer.
*August 30, 2012*

**pre-algebra**

Add all the Ps 4p-13p-p = -10p divide both sides by -10 to isolate P -10P = -150 -10p/-10p = -150/-10 p = 15 check by inserting calculated value into original equation
*August 29, 2012*

**Physics**

Ft = m(final velocity) - m(initial velocity) F = ma so equation changes to mat = m(final velocity) - m(initial velocity) plug in values and solve for 'a'
*August 29, 2012*

**physics**

power = force x velociy. Plug the values in and you have your answer.
*August 29, 2012*

**physics**

initial velocity = 78.3 m/s final velocity = 0 m/s distance traveled - 959 m a = ? use (final velocity)^2 = (initial velocity)^2 + 2ad solve equation for 'a'.
*August 29, 2012*

**physics**

use the work-energy theorem determine mass = 16.0N/9.8 m/s^2 initial velocity = 0; so initial KE = 0 final velocity = 15.0 m/s final height = 0; so final PE = 0 initial height = 150 sin 28.0 degrees substitute values in the work-energy theorem expression
*August 29, 2012*

**physics**

initial velocity = 0 m/s final velocity = 4.92 m/s constant acceleration so, (a) average velocity = (initial velocity + final velocity)/2 (b) distance = average velocity x time substitute and calculate
*August 29, 2012*

**physics**

initial velocity = 42.4 m/s final velocity = 25.0 m/s time = 4.00s a = (final velocity - initial velocity)/time (25.0m/s - 42.4m/s)/ 4.00 s 'sign' of answer determines direction of acceleration
*August 29, 2012*

**Science**

I agree. forces to the right = 20n + 40n = 60n force to the left = 30n net force = 60n - 30n or 30n
*August 27, 2012*

**Science**

I agree. forces to the right = 20n + 40n = 60n force to the left = 30n
*August 27, 2012*

**Science**

just remember that weight is a force that acts down. In this case you would add both of the weights. 120n + 40n = 160 n
*August 27, 2012*

**math**

Let x = number in ones place Let x + 2 = number in tens place Let x - 4 = number in hundreds place add all and set equal to 16 x + x+2 + x-4 = 16 x + x + x = 16 -2 + 4 3x = 18 divide both sides by 3 3x/3 = 18/3 x = 6 Answer = three digit number is 286
*August 27, 2012*

**Algebra 1**

(1) rearrange terms, putting 'x' values on the left side. 4x + 3x = -5 + 9 7x = 4 divide both sides by '7' 7x/7 = 4/7 x = 4/7
*August 27, 2012*

**math**

(1)distance: Change miles to kilometers divide 15.7 mi/0.6214mi/km (2)time: Change minutes to hours 49.0 min /60 min/hr rate = distance/time
*August 27, 2012*

**Science**

Net force = force acting up - force acting down = 30N - 50N Net force = -20N Net force is in the direction of the largest force.
*August 27, 2012*

**Science**

final velocity = initial velocity + (acceleration due to gravity x time) final velocity= 0 m/s +(-9.8 m/s^2)(2 s) final velocity = -19.6 m/s or 20 m/s answer (D)
*August 27, 2012*

**math**

Let Claire = x Julia = 4x Damon = 4x(4x) Damon = (4x)(4x) = 16x^2 16(4)^2 =(16)(16) Exponential = 5.018 x 10^3 stamps Standard = 5018 stamps
*August 27, 2012*

**Math**

(1) draw vectors for speed direction adding vectors head-to-tail Results: a right angle (2) draw the resultant vector (hypothenuse) completing the right triangle. (30 Solve using pythag. theorem
*August 27, 2012*

**Physics**

(1) draw a right triangle. (2) base = 100.0 m (3) height = 31.4 m (4) determine the angle (theta) between the base and the hypothenuse of the triangle. Use tan theta = opposite side/adjacent side. don't forget to inverse tan.
*August 27, 2012*

Pages: **1**