Thursday

April 17, 2014

April 17, 2014

Number of results: 296

**Classical Mechanics Physics**

error, so we'll have: F-ff=m.a -> ff=F-m.a (1) I.alpha=F.r+ff.R (2) alpha=a/R then: a=F.(R+r)/(I/R+m.R) a=positive ff=negative, and for a got green mark, but not for f, should I write ff with "-" or just the value? I got a=16.3 f=-4.6
*Friday, December 6, 2013 at 3:41pm by elli*

**physical science**

a. Fap-Ff = m*a = m*0 Fap - Ff = 0 The velocity is constant. Therefore, the acceleration is zero. Fap = Force applied. Ff = Force of friction. b. Fap - Ff = 0 100 - Ff = 0 Ff = 100 N.
*Sunday, February 9, 2014 at 5:21pm by Henry *

**Genetics 7th**

Eye color Fur Color black white black 24 24 red 25 27 The table above describes offspring of the same two parents. What are the most likely genotypes of the parents? Ff Ee and ff ee Ff Ee and Ff Ee FF EE and ff ee
*Thursday, December 13, 2012 at 2:08am by unknown.gonegirl*

**Physics**

What minimum coefficient of friction between the tires and the road will allow a 3700 kg truck to navigate an unbanked curve of radius 25 m at a speed of 45 km/h? My answer: Fnet=ma Ff=m(v2)/r) Ff=(3700)(452/25) Ff=(3700)(2025/25) Ff=(3700)(81) Ff=299700 Frictional force Ff=Fn...
*Friday, November 19, 2010 at 11:32pm by Ed*

**Biology DUE TOMORROW**

Since R and F are dominant, so the mom and the wife must have both recessive genes, namely rr and ff. Similarly, since the mom has rr and ff, the man must have at least one r and one f. Since he rolls his tongue and has free hanging earlobes, his genes must be heterozygous, ...
*Sunday, January 23, 2011 at 8:12pm by MathMate*

**Physics**

F - Ff = M a Ff is the friction force Ff = F - Ma = 15.00 - 1.64 N = 13.36 N
*Wednesday, October 19, 2011 at 4:58pm by drwls*

**Physics**

Fap - Ff =ma. 1640 - Ff = 5900*0.215 = 1269 N. -Ff = 1269 - 1640 = -371 Ff = 371 N. = Force of friction.
*Wednesday, October 10, 2012 at 2:11pm by Henry*

**Figured out question**

Can't find my other post and I know the person who helped me wanted to know if it was right or no. And it wasn't. So Ff=UsFN Ff=49N -Ff=ma a=-6.86 on the penguin Fnet=ma Fa-Ff=ma Fa-49=6.12*6.86 Fa=90.98N There it is. Thank you.
*Thursday, December 7, 2006 at 7:12pm by Armando*

**physics**

Let Ff = Force of Friction Let C = Coefficient of Friction Let N = Normal force Ff = CN The required force to just get a block moving along at a constant velocity will also be equal to the force of friction (Ff) if there is nothing else acting on the horizontal plane. N can be...
*Monday, January 17, 2011 at 1:44pm by EssKay*

**Physics**

Fap-Ff = m*a. 43-Ff = 3.3*0 = 0 Ff = 43 N. = Force of friction.
*Sunday, February 10, 2013 at 1:25pm by Henry*

**hydromechanic**

Fap-Ff = m*a 124-Ff = m*0 = 0 Ff = 124 N. = Force of friction. u = Ff/Fl = 124/1026 = 0.121 = Coefficient of friction. Fap = Force applied.
*Tuesday, August 13, 2013 at 12:20am by Henry *

**Physics**

a. Fn = ma = 100 * 2 = 200N = Net force. Fn=Fap - Ff=200, Fap = Force applied. 250 - Ff = 200, -Ff = 200 - 250 = -50, Ff = 50N = Force of friction. b. Fc = mg = 100kg * 9.8N/kg = 980N = Force of the crate = Normal force. c. u = Ff/Fc 50/980 = 0.05 = coefficient of friction. d...
*Monday, June 20, 2011 at 9:01pm by Henry*

**Physics**

Use the equation Fnet,x = M*a, to solve for M. Fnet,x is the net force on the crate in the +x direction. That is equal to 752 cos22 - Ff The friction force Ff is Ff = (M*g-752sin22)*(0.347) Put it all together and turn the algebra crank to get M.
*Sunday, March 20, 2011 at 11:48pm by drwls*

**Physics**

a. Wb = m*g = 8kg * 9.8N/kg = 78.4 N. = Wt of block. Epmax = mg*h-Fk*d Epmax = 78.4*1.6*sin30 - Fk*1.6 Epmax = 62.72 - Fk*1.6 Ek = 62.72 - Ff*1.6 0.5m*V^2 = 62.72 - Ff*1.6 4*5^2 = 62.72 - Fk*1.6 100 = 62.72 - Ff*1.6 Ff*1.6 = 62.72-100 = -37.28 Ff = -23.3 N. = Force of friction...
*Monday, March 25, 2013 at 1:51am by Henry*

**physics**

Component of force along the plane (downward), F = mgsin(θ) Let frictional force be ff, then ff=μmgcos(θ) Net acceleration (F-ff)/m = 1.38 m/s² Solve for μ
*Sunday, December 5, 2010 at 5:19pm by MathMate*

**physics**

Fb=mg = 20kg * 9.8N/kg = 196N @ 30deg. Fp = 196*sin30 = 98N = Force parallel to plane. a=(Vf-Vo) / t = (21-0) / 6 = 3.5m/s^2. Fn = Fp-Ff = ma, 98 - Ff = 20 * 3.5, Ff = 28N = Force of friction.
*Monday, April 25, 2011 at 5:55pm by Henry*

**Physics**

a) Relevant equation: Emec = DU + DK If there are no nonconservative forces, this equation equals zero. Or in this case, some K is transformed into Eint (heat). So the equation becomes: Emec = Ff.d. (F in Newton, distance in meters and N.m = J). -> DU + DK = Ff.d Then, 0,5 ...
*Thursday, March 8, 2012 at 12:19pm by Saycks*

**physics**

Wc = mg = 20kg * 9.8N/kg=196N.= Weight of crate. a. Fc = [196N,0deg). Fp = 196sin(0) = 0 = Force parallel to floor. Fv = 196cos(0) = 196N. = Force perpendicular to floor. Fn = Fap*cosA - Fp - Ff = 0, Fn = 50cos25 - 0 - Ff = 0, 50cos25 - Ff = 0, Ff = 50cos25 = 45.32N. = Force ...
*Monday, October 24, 2011 at 8:43am by Henry*

**physics**

Wb = mg = 1.7kg * 9.8N/kg = 16.66N. Fb = (16.66N,0deg) = Force of book.. Fp = 16.66sin(0) = 0 = Force parallel to table. Fv = 16.66cos(0) = 16.66N. = Force perpendicular to table. 1. Fap - Ff = 0, Fap = Force applied. 2.60 - Ff = 0, Ff = Force of friction. Ff = 2.60, u*Fv = 2....
*Saturday, October 15, 2011 at 12:55pm by Henry*

**Physics**

I agree with your answers. Fn = Fap - Ff = ma = 0. a = 0. Since there is no change in velocity or speed, there is no acceleration(a=0). Fn = 25,000 - Ff = 0, Ff = 25,000N. Fn = Net force. Fap = Force applied by tractor. Ff = Force of friction. Remember the force of the tractor...
*Sunday, November 6, 2011 at 1:30pm by Henry*

**phy**

Sum forces in the vertical Fc= force closest Ff=force furtherest Fc+Ff-W-mg=0 where mg is the weight of the beam, W is the weight of the piano Then sum moments about any point. I will sum about the closest end. W*L/3+mgL/2-Ff*L=0 solve for Ff, then use the first equation for Fc
*Friday, October 22, 2010 at 3:37pm by bobpursley*

**Physics **

Fb = 10kg * 9.8N/kg = 98N. Fb =98N @ 0 deg. Fp = 98sin(0) = 0 = Force parallel to plane(hor). Fv=98cos(0) = 98N.=Force perpendicular to the plane. a. Fn = Fap - Fp - Ff = 0, 25 - 0 - Ff = 0, Ff = 25N = Force of friction. b. Ff = uFv = 25, u*98 = 25, u = 0.26 = Coefficient of ...
*Monday, July 18, 2011 at 12:51am by Henry*

**physics**

a) I can't do that b) Ff = mu (mass) (force of gravity) = (0.85)(0.5)(9.8) =4.17N c) W=Ff(distance) =4.17(2) =8.34 J
*Saturday, January 12, 2013 at 3:15pm by Matt M*

**Physics Classical Mechanics**

@Fima Hello fima... it seems to be the right equation, you see... The force only starts to push the block when it "wins" the Friction Force (Ff). So, its only after 2 seconds has passed. Using your numbers: Ff = m*g*mu = 2*10*0.2 = 4N... F=bt^2 = 1 * (2)^2=4N... So right after...
*Saturday, December 7, 2013 at 6:51am by Anoninho*

**Physics**

Wb = mg = 0.55kg * 9.8N/kg = 5.39N. Fb = (5.39N.,0 deg.). Fp = Fh = 5.39sin(0) = 0 = Force parallel to plane = Hor. comp. Ff = Force due to friction. a = (Vf^2 - V9^2) / 2d, a = (0 - (3^2)) / 1.06 = -8.49m/s^2. Fn = ma = 0.55 * (-8.49) = -4.67N. = Net force. Fn = Fp - Ff = -4....
*Thursday, September 29, 2011 at 12:53pm by Henry*

**Physics**

I am not sure that I understand what you mean. If Ff is the friction force and Fg is the weight (on a level surface), then Ff = Uk* Fg if the object is moving. Ff = Us* Fg is the maximum static friction force that keeps it from moving. Us and Uk are the static and kinetic ...
*Thursday, April 12, 2012 at 2:11pm by drwls*

**Math**

given f(x)=3x-5 and g(x)=2x^2+5x and h(x)=1/(x+3) find: ff(-2) =(3(-2)-5)(3(-2)-5) =(-6-5)(-6-5) =121 is this correct or is it wrong? I don't really know how to write ff(-2) x if f(x)=-7 =??? I don't know how to start this please help and thank you
*Tuesday, March 13, 2012 at 1:55pm by Anonymous*

**physics**

Wb = mg = 4kg * 9.8N/kg = 39.2N = Weight 0f block. Fb = 39.2N@0deg. Fp = 39.2sin(0) = 0 = Force parallel to surface. Fv = 39.2cos(0) = 39.2N. = Force perpendicular to surface. Fap - Fp - Ff = 0, 20 - 0 - Ff = 0, Ff = 20N. = Force of friction. u = Ff / Fv = 20 / 39.2 = 0.510...
*Wednesday, October 26, 2011 at 2:43pm by Henry*

**physics**

Wb = mg = 3kg * 9.8N/kg = 29.4N. Fb = (29.4N,33deg). Fp = 29.4sin33 = 16.o1N. = Force parallel to incline. Fv = 29.4cos33 = 24.66N. = Force perpendicular to incline. d = Vo*t + 0.5at^2 = 2.0m, 0*1.75 + 4.9a*(1.75)^2 = 2, 15a = 2, a = 0.133m/s^2. Fn = Fp - Ff = ma, 16.01 - Ff...
*Saturday, October 22, 2011 at 6:40pm by Henry*

**physics**

Calculate the friction force using the kinetic coefficient of friction, Uk = .7. Call the friction force Ff Ff = M*g*cos15*Uk Let the distance it travels be X When Ff*X equals the initial kinetic energy plus any gravitational potential energy decrease, all of the available ...
*Saturday, October 9, 2010 at 12:04am by drwls*

**college physics**

To solve this you have to know something about the additional friction where the string goes over the edge. Perhaps they want you to assume it is zero. That is why I asked if there is a pulley wheel at the edge. If there is no string friction, 1.67 g - T = 1.67 a T - Ff = 3.34...
*Wednesday, February 16, 2011 at 12:04am by drwls*

**physics**

Wb = mg = 2.54kg * 9.8N/kg = 24.89N. = Weight of block. Fb = 4.89N @ 0deg. Fp = 24.89sin(0) = 0 = Force parallel to floor. Fv = 24.89cos(0) = 24.89N. = Force perpendicular to floor. a. W = Fap*cos29.8 * d, W = 5.87cos29.8 * 4.48 = 5.1J. c. Fn = Fap*cos29.8 - Fp - Ff = 0, 5....
*Wednesday, November 2, 2011 at 12:20am by Henry*

**physics**

Calculate the friction force Ff = m*a The coefficient of kinetic friction is Uk = Ff/(m*g) = a/g = ___ You don't need to know the mass.
*Thursday, October 6, 2011 at 12:26am by drwls*

**Science**

Wb = mg = 200kg * 9.8N/kg = 1960N. Fb = 1960N @ 0deg. = Force of bobsled. Fp = 1960sin(0) = 0 = Force parallel to surface. Fv = 1960cos(0) = 1960N. = Force perpendicular to surface. Fn = Fap - Fp - Ff = ma = 0, a = 0. 1200 - 0 - Ff = 0, Ff = 1200 - 0 = 1200N. = Force of ...
*Tuesday, November 8, 2011 at 9:19pm by Henry*

**calculus**

ff' = fg gg' = gf ff' = gg' 2f f' - 2g g' = 0 f^2 - g^2 = c
*Sunday, October 21, 2012 at 2:13pm by Steve*

**physics**

Wc = mg = 1670kg * 9.8N/kg = 16,366N. = Weight of car. Fc = (16,366N.,10.3 deg.), Fp = 16,366sin10.3 = 2826.3N. = Force parallel to plane. a. Fv = 16,366cos10.3 = 16,102.3N. = Force perpendicular to plane = Normal force. b. Ff = Force due to friction. Fn = ma = 0 = Net force, ...
*Thursday, September 29, 2011 at 3:06pm by Henry*

**Physics**

a. Fc = mg = 1500kg * 9.8N/kg = 14700N. Wc = 1500kg / 0.454kg/lb = 3304lbs. = weight of car in lbs. b. Fv = mgcos20 = 14700cos20 = 13,814N. = Force perpendicular to the hill. c. Ff = Fp = Force parallel to the hill. Ff = Fp = mg*sin20 = 14,700sin20 = 5028N. d. u = Ff / Fv = ...
*Monday, June 20, 2011 at 9:03pm by Henry*

**physics**

Wb = mg = 125kg * 9.8N/kg = 1225N. = Weight of box. Fb = (1225N,0deg.). Fp = Fh = 1225sin(0) = 0 = Force parallel to plane = hor. force. Fv = 1225cos(0) = 1225N = Force perpendicular to plane = The normal. Fn = ma = 125 * 1.2 = 150N = Net force. Fn = Fap - Fp - Ff = 150N, 350...
*Sunday, October 9, 2011 at 12:41pm by Henry*

**physics**

Use F = m a, along the direction of motion. The net force pusing the box downhill is M g sin39 - Ff = M a Solve for Ff, the friction force. Please do not put your grade level as the subject.
*Saturday, November 27, 2010 at 9:21pm by drwls*

**Physics**

Fg = m*g = 20kg * 9.8N/kg = 196 N. Fp = 196*sin40 = 126 N. = Force parallel to incline. Fn = 196*cos40 = 150 N. = Normal. h = L*sin40 = 6.5*sin40 = 4.18 m. Wg = Fg * h = 196 * 4.18 = 819.3 Joules. Wn = = Fn * h = 150 * 4.18 = 628 N. Fp-Ff = m*a = m*0 = 0 126-Ff = 0 Ff = 126 N...
*Saturday, November 2, 2013 at 11:23pm by Henry *

**PHYSICS**

Whats the difference between the formulas for kinetic and static friction. Kinetic: Ff= ukFn Static: Ff= usFn Are the 'f' and the 'n' the mass of the object? Are the 'k' and the 's' just the
*Thursday, April 16, 2009 at 11:26pm by Lena*

**Physics**

Inward force needed =m a = m v^2/R = 600 (900)/120 = 4500 Newtons inward down slope component = 4500 cos 25 Force up slope = Ff friction weight down slope = m g sin 25 = 2488 Net force down slope = 2488 - Ff so 2488-Ff = 4500 cos 25 Ff = 2488 - 4078 Ff = -1590 N so 1590 DOWN ...
*Thursday, February 20, 2014 at 6:22am by Damon*

**foundation of mechanics**

there's a mistake in substitution: The friction force opposing motion is Ff = M g cosA*(1/3) = (1/5)*M*g Kinetic energy at Q = (Potential Energy Loss) - (Frictional work) (1/2)MV^2 = M g L sinA - Ff*L = M*g*L (3/5) - (1/5)M*g*L = (2/5)*M*g*L its supposed to be M*g*L (4/5) when...
*Tuesday, December 27, 2011 at 12:10pm by smart guy*

**College Physics**

Wb = mg = 16.36kg * 9.8N/kg = 160.2N = Wt. of block. Fp = 160sin(0) = 0 Newtons. = Force parallel to plane. Fv = 160cos(0) = 160N. = Force perpen- dicular to plane. Fn = Fap - Fp - Ff = ma = 0(a = 0). 9.36 - 0 - Ff = 0, Ff = uFv = 9.36, u*160 = 9.36, u = 0.0585 = Force of ...
*Monday, September 12, 2011 at 12:02am by Henry*

**physics**

If it is sliding to a stop along a horizontal surface, the initial kinetic energy equals the work done against friction. I will assume the weight is in Newtons (M/2)Vi^2 = Ff*S M = (19600 N)/g = 2000 kg Ff = friction force = 1000*900/100 = 9000 N
*Monday, January 7, 2013 at 6:41am by drwls*

**physics repost**

A car of mass 1330 kg is traveling at 28.0 m/s. The driver applies the brakes to bring the car to rest over a distance of 79.0 m. Calculate the retarding force acting on the car. same as before just posting again so you wont have to scroll. heres my work KE=1/2*m*v^2 KE=521360...
*Wednesday, April 16, 2008 at 8:46pm by Anonymous*

**Physics**

Subtract the final kinetic energy from the work done pushing the book, 25*0.88 joules The difference will be work done against friction, which is Ff*X Ff is the friction force, M*g*Muk Solve for Muk
*Sunday, February 20, 2011 at 12:03am by drwls *

**Classical Mechanics Physics**

It might be better to indicate the equation rather than answers, because the data changes. So, we have: F-ff=ma (1) I.alpha=F.R-ff.r (2) and we have two equation ant variables. but alpha is a/R or a/r
*Friday, December 6, 2013 at 3:41pm by elli*

**physics**

(a) You claim you already have the correct answer for ěk. (b) The friction force is Ff = (ěk)*(M g -380 sin20) Net horizontal pulling force = 380 cos20 -Ff = M*a Solve for a. M = 1440/9.8 = 146.9 kg
*Friday, October 12, 2012 at 7:18pm by drwls*

**Physics**

During the Olympic ice competition, Boris (m = 75 kg) glides at 1.8 m/s to a stationary Juliette (52 kg) and hangs on. How far will the pair slide after the “collision” if the coefficient of kinetic friction between the ice and their skates is .042? My answer: Conservation of ...
*Tuesday, November 23, 2010 at 8:13pm by Mike*

**Physics**

Since the kinetic energy does not change, Friction work = (Friction force)*(sliding distance) = Potential energy loss M g*3.0 m = Ff*7.0 m Ff = (3/7) M g = 105 J is the friction force that resists motion.
*Tuesday, November 22, 2011 at 11:51pm by drwls*

**physics**

Ws = 46N. = Weight of sled. Fs = 46N @ 0deg. = Force of sled. Fp = 46sin(0) = 0=Force parallel to the snow. Fv = 46cos(0) = 46N. = Force perpendicular to the snow. Ff = u(Fv+625), Ff = 0.11(46 + 625) = 671N. = Force of friction. Fn = Fap - Fp - Ff = 0, Fap - 0 - 671 = 0, Fap...
*Thursday, October 27, 2011 at 5:11pm by Henry*

**physics**

a) 325 sin 20 b) it is not accelerating so the sum of forces in any direction is zero. c)Call x positive up the plane. The thing is moving up the plane, in the plus x direction the friction force resists the motion so it acts down the plane, in the -x direction Sum all forces ...
*Sunday, January 13, 2008 at 7:52pm by Damon*

**physics**

Remember the equation: deltaF = ma In this case delta F is the force applied(800N)-Force of friction. Now we need to find out the force of friction(Ff). We can do this through the equation Ff=uN. u is given(.2) N(the normal force)opposes the downward force of the desk(mg)250*9...
*Monday, December 13, 2010 at 9:07pm by Omer*

**Physics**

That is a very high coefficient of friction. Are you sure it is 3 and not 0.3? If it is 3, it won't move. The downward force on the floor is Fn = M g + 60 sin 30 = 128 N The friction force is Ff = µ * Fn The acceleration is (60 cos 60 - Ff)/M
*Monday, April 7, 2008 at 6:41pm by drwls*

**Physics**

Downward force due to gravity is F=1368.1 Friction force resisting movement is FF=1127.6 Therefore net downward force =force required to prevent load from sliding down =F-FF =1368.1-1127.6 =240.45 N
*Friday, July 5, 2013 at 11:47am by MathMate*

**Physics**

You pushed a 1 kg box on floor where= 0.3. If the force you applied was 5 N, will the box move? m = 1kg u= 0.3 Fapp= 5 N Ff= (0.3)(1) (9.81) Ff = 2.943 N So therefore the box would move because the frictional force is less than the applied force. Did I answer this correctly?
*Thursday, April 16, 2009 at 11:40pm by Lena*

**Physics**

okay so this is what i did to find the work done by overcoming friction: Ff = .9 N d = .91 m (Ff x d = Wf --> .9 x .91 = .82 J) hmm okay so how would you find the work done by friction using that calculation that you just said (change in total energy = work done by friction...
*Sunday, December 16, 2007 at 2:26pm by Hayley*

**physics**

Fb=15.6kg * 9.8N/kg = 152.88N @ 0 deg.= Force of the block. Fp = Fh = 155.88sin0 = 0N = Hor comp of block. Fv = 155.88cos0 = 155.88N = Force perpendicular to plane. Applied Force: Fap = 72.2N @ 34.8deg. Fh = 72.2cos34.8 = 59.29N. = Hor comp of applied force. Fv = 72.2sin34.8...
*Friday, February 25, 2011 at 5:02pm by Henry*

**physic**

1. Fb = 3kg * 9.8N/kg = 29.4N @ 30deg = Force of block. Fp = 29.4sin30 = 14.7N = Force parallel to plane acting downward. Fv = 29.4cos30 = 8.82N = Force perpendicular to plane acting downward. Ff = u*Fv = 0.3 * 8.82 = 2.65N. = Force of friction. Fap - Fp - Ff = 0, Fap = Fp + ...
*Friday, February 25, 2011 at 7:51pm by Henry*

**physic**

1. Fb = 3kg * 9.8N/kg = 29.4N @ 30deg = Force of block. Fp = 29.4sin30 = 14.7N = Force parallel to plane acting downward. Fv = 29.4cos30 = 8.82N = Force perpendicular to plane acting downward. Ff = u*Fv = 0.3 * 8.82 = 2.65N. = Force of friction. Fap - Fp - Ff = 0, Fap = Fp + ...
*Friday, February 25, 2011 at 7:51pm by Henry*

**physics**

Wb = mg = 0.45kg * 9.8N/kg = 4.41N. = Weight of block. Fb = (4.41N,0deg.). Fp=Fh = 4.41sin(0) = 0 = Force parallel to hor. surface. Fv = 4.41cos(0) = 4.41N. = Force perpendicular to hor. surface. Ff = u*Fv = 0.67 * 4.41 = 2.95N. = Force of static friction. Ff = u*Fv = 0.360 * ...
*Monday, October 17, 2011 at 10:24am by Henry*

**Physics**

One way that I found was knowing the force overcoming friction (Ff). so the equation to fin dthe work would be Wf = Ff * d.. as you said above (force*distance). But what other calculation would you use to find the wasted work without knowing the friction force?
*Sunday, December 16, 2007 at 2:26pm by Hayley*

**physics**

(a) Without air friction, she would achieve a velocity of V = sqrt(2gH) = 17.15 m/s when entering the water. Instead, it is 14.0 m/s. (Average force)*(distance) = K.E. loss = (57/2)[17.15^2 - 14^2] = 2793 J Average force = 2793/15 = 186 N (b) [(Friction force)+(buoyancy force...
*Saturday, April 13, 2013 at 11:47pm by drwls*

**physics**

Wb = mg = 2kg * 9.8N/kg = 19.6N. = Weight of block. Fb = 19.6N. @ 0deg. = Force of block. Fp = 19.6sin(0) = 0 = Force parallel to surface. Fv = 19.6cos(0) = 19.6N. = Force perpendicular o surface. Ff = u*Fv = 0.4 * 19.6N. = 7.84N. = Force of friction. a. W = Fap * d = 20 * 2...
*Thursday, November 3, 2011 at 10:52am by Henry*

**Physics**

First calculate the friction force, Ff. Ff = (0.32)*M*g = 146.14 Newtons Then calculate the net horizontal force on the box Fnet = 211.67 - 146.14 = 65.53 N Then calculate the acceleration a = Fnet/M = 1.41 m/s^2 Then use Distance moved = (1/2) a t^2 (for a starting velocity ...
*Tuesday, January 22, 2013 at 8:03pm by drwls*

**Physics**

fundamental frequency (ff) = 440 Hz frequency at C (fc)= 523.3 Hz L = 32.7 cm L(C) = ? x(C) = ? L (C) = L * ff/fc L (C) = 27.50 cm x (C) = L - L(C) = 5.20 cm Finger has to be placed at 5.20 cm from the end.
*Wednesday, December 8, 2010 at 5:22pm by Anonymous*

**physics**

Fs = (75N,24deg). Fp=75sin24 = 30.51N. = Force parallel to plane acting down the plane. Fv = 75cos24 = 68.52N. = Force perpen- dicular to plane. Ff = u*Fv = 0.25 * 68.52 = 17.13N. = Force of static friction. Fk = 0.13 * 68.52 = 8.91 = Force of kinetic friction. a. Fap - Fp - ...
*Sunday, October 23, 2011 at 6:55pm by Henry*

**PHYSICS PLEASE HELP !!!!!**

Wo = mg = 13kg * 9.8N/kg = 127.4N = Weight of object. Fo = 127.4N@odeg. = Force of object. Fp = 127.4sin(0) = 0 = Force parallel to surface. Fv = 127.4cos(0) = 127.4N. = Force perpendicular to surface. W = Fn*d, Fn = W / d = 499 / 5 = 100N. Fn = F2cos28 - Fp - Ff = 100. ...
*Thursday, October 27, 2011 at 2:46pm by Henry*

**physics**

Sin¤= 1/3 ¤=19.47' Fnet= Fg// - Ff Fnet= mg.sin¤ -Ff =2(9.8)sin19.47-0.35 Fnet= 6.183N a= Fnet/m a= 6.183/2 a=3.091m/s^2 (Vf^2)=(Vi^2) +2as Vf^2= (0)+2(3.091)(3) Vf^2= 18.546 Vf= 4.31m/s downward. ***GOOD LUCK*** For info call me on 0723624826
*Monday, August 1, 2011 at 2:16am by Thakgalo sakwana*

**physics**

Given: m=11.3kg. A = 29.2 deg., Ff = 44.6N. Wc = mg = 11.3kg * 9.8N/kg = 110.74N. = Weight of cart. Fc = (110.74N,0deg.). Fp = 110.74sin(0) = 0 = Force parallel to surface. Fv = 110.74cos(0) = 110.74N. = Force perpendicular to surface. Ff = 44.6N = Force of fiction. Fn=Fap*...
*Wednesday, October 19, 2011 at 5:22pm by Henry*

**physics**

Wc = mg = 11.9 kg * 9.8 N/kg = 116.6 N= Weight of crate. Fc = 116.6 N @ (0) Deg.=Force of crate. Fp = 116.6*sin(0) = 0 = Force parallel to floor. Fv = 116.6*cos(0) = 116.6 N. = Force perpendicular to floor. a. Ff = u*(Fv+31.9*sin23.6), Ff = 0.0975*(116.6+12.77) = 12.61 N. Fn...
*Wednesday, January 18, 2012 at 6:29pm by Henry*

**physics**

I don't see how "the angle" can be 30 degrees if the surface is level. Are you talking about the angle that his pulling force vector makes with the horizontal? In any case, the work done on a level surface is the friction force times the distance that the sled is pulled (10 m...
*Saturday, January 26, 2008 at 7:32pm by drwls*

**Physics**

Wc = mg = 25kg * 9.8N/kg = 245N. = Weight of crate. Fc = 245N @ 0deg. Fp = 245sin(0) = 0 = Force parallel to floor. Fv = 245cos(0) = 245N. = Force perpendicular to floor = Normal. Ff = u*Fv = 0.23 * 245 = 56.35N. = Force of friction. a. Fn = Fap - Fp - Ff = ma = 0, a = 0. Fap...
*Tuesday, November 1, 2011 at 12:52pm by Henry*

**physics**

The normal force is Fn = 12 lb * cos 30 The friction force is Ff = Fn sin 30 if the object is not moving. If the object starts moving at that tilt angle, the coefficient of static friction is Mus = Ff/Fn tan 30
*Saturday, January 26, 2008 at 10:31pm by drwls*

**Physics**

Fap = 131N = Applied force. Fc=mg=14kg * 9.8N/kg=137.2N @12.3deg. Fp = 137.2sin12.3 = 29.2N = Force parallel to the plane. Fv = 137.2cos12.3 = 134.1N = Force perpendicular to the plane. Ff = u*Fv = 0.26 * 134.1 = 34.9N = Force of friction. Fn = Fap - Fp - Ff, Fn = 131 - 29.2...
*Wednesday, March 2, 2011 at 11:25am by Henry*

**Physics**

A force sensor is used to measure a force applied to a 7.5 kg box on a hard, horizontal surface. It records that the box starts to move when the force is equal to 16.5 N. This force makes the box start to move with an acceleration of 0.5 m/s2. Find s and k respectively. I keep...
*Wednesday, September 30, 2009 at 5:14pm by Bella*

**foundation of mechanics**

Let X = 10 m be the distance from P to Q. There will be energy loss due to friction as the particle slides from P to Q. The angle from horizontal is A = arctan(4/3) = 53.13 deg sin A = 4/5 cosA = 3/5 i) The friction force opposing motion is Ff = M g cosA*(1/3) = (1/5)*M*g ...
*Tuesday, December 27, 2011 at 12:10pm by drwls*

**Physics**

If it was dragged at constant velocity, then you are right with Ff=3N (note, not negative). Net force is the accelerating force, above the friction force. If it were at constant velocity, you have no idea what net force is. With constant veloicty, net force is zero (it is not ...
*Wednesday, March 18, 2009 at 7:58pm by bobpursley*

**Physics**

Since the box is moving through sea water at a constant speed, the net force on it must be zero. The forces are: i) W= mg its weight downwards ii) Fb - buoyant force upwards iii) Ff - frictional force offered by water acting upwards. now W = mg = V*Db*g (Db-density of box) and...
*Friday, June 29, 2012 at 2:11pm by ajayb*

**physics**

(a) Solve X = (1/2)a t^2 to get the acceleration, a = 1.34 m/s^2 (b) If there were no friction, the acceleration would be g sin30 = 4.9 m/s^2. There must be an opposing friction force Ff = M(4.9 - 1.34)= 10.7 N (That answers part (c)) The coefficient of kinetic friction is muk...
*Thursday, March 10, 2011 at 12:28am by drwls*

**physics**

Here's a hint: (Work done by Juana) - (Work done against friction) = (Increase in kinetic energy) Use the final kinetic energy to get the speed. The friction force is Ff = M*g*(0.12) = 66.8 N The work done against friction is Ff*1.3 = 209.6 J The work done pulling is 124N*1.3m...
*Wednesday, May 11, 2011 at 9:47pm by drwls*

**dt -food**

Psomi is one. Thank you for using the Jiskha Homework Help Forum. Here are some recipes for Greek Bread: http://www.cooks.com/rec/search/0,1-0,greek_bread,FF.html Sra
*Tuesday, September 30, 2008 at 11:48am by SraJMcGin*

**phy**

At a factory, a noon whistle is sounding with a frequency of 484 Hz. As a car traveling at 77 km/h approaches the factory, the driver hears the whistle at frequency fi. After driving past the factory, the driver hears frequency ff. What is the change in frequency ff − fi...
*Friday, April 12, 2013 at 9:59pm by kira*

**finance**

FF has an annual credit sales of $250,000 units with an average collection period of 70 days. The company has a per unit variable cost of $20 and a per unit sale price of $30. Bad debts currently are 5% of sales. The firm estimates that a proposed relaxation of credit would ...
*Sunday, January 1, 2012 at 5:38pm by Vanessa*

**PHYSICS**

QUIZ 1 AND 2, PROBLEM 3 A block of mass M is pulled with a rope that exerts a force of a magnitude F, causing it to slide over horizontal ground with kinetic coefficient of friction μ (use "mu" when you input your answer). The force is applied at angle θ (use "theta...
*Saturday, June 29, 2013 at 3:48am by XYZ*

**physics**

Fc=mg=1460kg * 9.8N/kg=14308N @ 10deg. Fp = 14308*sin10 = 2484.56N = Force parallel to the plane downward. Fv = 14308*cos10 = Ff = .63N = Force perpendicular to the plane and acting downward. Ff = 497N = Force of friction acting opposite to the motion. W = Fn*d = 149000J. Fn...
*Monday, February 28, 2011 at 11:44am by Henry*

**Physics**

Correction: mass of the car = 2000kg: NOT 2kg. Fc = 2000kg * 9.8N/kg = 19,600N. Fp = 19,600sin(0) = 0. Fv = 19,600cos(0) = 19,600N. Ff = u*Fv = 0.5 * 19,600 = 9800N. Fn = Fp - Ff = o - 9800 = -9800N. a = Fn/m = -9800 / 2000 = -4.9m/s^2. The remaining calculations are correct ...
*Sunday, July 17, 2011 at 4:09pm by Henry*

**physics!**

Ok, so after struggling through the above explanation I decided to repost a (imho) better worded, generalized explanation: Firstly let me define my variables: Fb=force of BOTH back wheels on the ground=2F(1backwheel) Ff=force of BOTH front wheels on the ground=2F(1frontwheel) ...
*Saturday, January 5, 2008 at 7:27pm by Arno*

**English**

http://www.cooks.com/rec/search/0,1-0,apple_turnover,FF.html
*Sunday, December 7, 2008 at 6:09pm by Ms. Sue*

**physics**

Fr = 283N.[41o] + 283N.[0o] X = 283*cos41 + 283 = 496.6 N. Y = 283*sin41 = 185.7 N. Fr^2 = X^2 + Y^2 Fr^2 = 496.6^2+185.7^2 = 281,096.05 Fr = 530.2 N. Ff = u*mg = 0.283 * 264.6 = 74.9 N. a = (Fr-Ff)/m=(530.2-74.9)/27=16.9 m/s^2
*Sunday, November 10, 2013 at 5:47pm by Henry *

**Physics (dynamics)**

The block on top can support a pull of 0.50*18*g = 88.2 N without slipping. Since the applied force is greater than that, it will slip and there will be a friction force of Ff = 0.4*18*g = 70.56 N between the blocks. That will help pull the lower block forward. The lower block...
*Saturday, March 10, 2012 at 9:41pm by drwls*

**Physics**

Wb = mg = 5kg * 9.8N/kg = 49N. Fb = (49N,30deg.). Fp = 49sin30 = 24.5N. = Force parallel to incline. Fv = 49cos30 = 42.4N. = Force perpendicular to incline = The normal. Ff = u*Fv = 0.436 * 42.4 = 18.50N. = Force of friction. 1. Work=Ff * d = 18.50 * 2.5 = 46.3J. 2. d = h = 2....
*Wednesday, October 12, 2011 at 5:27pm by Henry*

**Physics**

Solve: vbr{lr<5vvx><f(nn)/.bbn> vbr<lr><5.12>(9.8)^3 .bbn=%ff %ff=F(9x) .bbn=%ff9x .bbn=.133%tri .bbn=.tri t.ri=vbr v=31.4 b=59.3 r=28.6 KE___?
*Monday, November 19, 2012 at 9:59pm by Mikey*

**Help- Physics**

Wc = 130kg * 9.8N/kg = 1274N. Fc = 1274N @ 0deg. Fp = 1274sin(0) = 0 = Force parallel to floor. Fv = 1274cos(0) = 1274N. = Force perpendicular to floor. Ff = u*Fv = 0.5 * 1274 = 637N. = Force of friction. Fn = Fap - Fp - Ff = ma = 0. a = 0. Fap - 0 - 637 = 0, Fap = 637N. = ...
*Friday, November 11, 2011 at 11:23am by Henry*

**Physics**

a. Fap*cos25.4 - Ff = m*a. Fap*cos25.4 - 53.1 = 19.5*0 = 0 Fap*cos25.4 = 53.1 Fap = 53.1/cos25.4 = 58.8 N. = Force applied by shopper. b. Work = Fx * d=58.8*cos25.4*35.6=1891 Joules. c. Work = Ff*d = 53.1 * 35.6 = 1890 J.
*Friday, October 5, 2012 at 7:48pm by Henry*

**physics**

Fb = mg = 7.50kg * 9.8N/kg = 73.5N Fb = 73.5N @ 25 deg. = Weight of box. Fp=73.5s8in25 = 31.06N.=Force parallel to the plane(ramp). Fv = 73.5cos25 = 66.6N = Force perpendicular to the plane(ramp). Ff = u*Fv = 0.312 * 66.6 = 20.78N. a. Fn = F/cos25 - Fp - Ff = ma. 1.1F - 31.06...
*Saturday, August 6, 2011 at 9:29am by Henry*

**Physics**

How does torque equal to <frictional force times radius [of solid cylinder]>? " A potter's wheel, a thick stone disk of radius 0.50 m and mass 100 kg, is freely rotating at 50 rev/min. The potter can stop the wheel in 6.0 s by pressing a wet rag against the rim and ...
*Wednesday, October 30, 2013 at 11:36pm by Ánfer*

**Physics **

1. Fb = mg = 15kg * 9.8N/kg = 147N. 147N @ 0 deg. Fp = 147sin(0) = 0 = Force parallel to plane(hor.). Fv = 147cos(0) = 147N = Force perpendicular to plane. Ff = u*Fv = 0.35 * 147 = 51.45N. Fn = Fap - Fp - Ff, Fn = 100 - 0 - 51.45 = 48.6N = Net force parallel to the hor plane.
*Monday, July 18, 2011 at 12:38am by Henry*

**Physics**

sinA = Y/r = 4.2 / 8 = 0.525, A = 31.67deg. Wb=10.3kg * 9.8N/kg = 100.9N @ 31.7deg = Weight of block. Fp = 100.9sin31.7 = 53N = Force parallel to plane. Ff = 0 = Force due to friction. Fn=Fp - Ff = 53 - 0 = 53N = Net force. a = Fn / m = 53 / 10.3 = 5.15m/s^2. Vf^2 = Vo^2 + 2ad...
*Tuesday, February 22, 2011 at 10:17am by Henry*