Sunday

April 20, 2014

April 20, 2014

**Recent Homework Questions About Physics**

Post a New Question | Current Questions

**physics?**

.
*Sunday, March 9, 2014 at 5:30pm*

**physics**

If a wire has a resistance of 132ohms a length of 110cm and an area of cross- sectional of 0.00415cm square find the resistivity of the material of which it is made
*Sunday, March 9, 2014 at 4:33pm*

**Physics**

The differential gear of a car axle allows the wheel on the left side of a car to rotate at a different angular speed than the wheel on the right side. A car is driving at a constant speed around a circular track on level ground, completing each lap in 20.2 s. The distance ...
*Sunday, March 9, 2014 at 4:17pm*

**physics**

A construction worker riding in the tray of a truck traveling at 10 m/s, sees one of his friends passing and decides to jump off the truck. The magnitude of his initial speed of jumping is: (a) zero (b) Less than the truck's (c) Equal to that of the truck (d) Greater than ...
*Sunday, March 9, 2014 at 3:27pm*

**Physics**

See previous post: Wed,3-5-14.9:29 PM.
*Sunday, March 9, 2014 at 3:23pm*

**Physics**

An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y axis and the total distance to the plane from you. Choose the x-axis east, y axis north, and z axis up.
*Sunday, March 9, 2014 at 3:16pm*

**Physics-Kinetic Energy**

Can someone help me solve this kinetic energy problem? A 2050 kg truck is traveling east through an intersection at 2.3 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 4.5 m/s. The ...
*Sunday, March 9, 2014 at 2:48pm*

**physics**

A 4.0-uF capacitor is charged to 600 volts. how much energy is stored in the electric field of the capacitor?
*Sunday, March 9, 2014 at 1:16pm*

**Physics**

conservation of momentum says that the result is 2050(2.3i+0j)+1200(0i+4.5j)+1600(10i+j) = (2050+1200+1600)(ai+bj) 4715i+5400j+16000i = 4850(ai+bj) 20715i+5400j = 4850(ai+bj) 4.27i+1.11j = ai+bj speed is 4.42 m/s direction E14.5°N
*Sunday, March 9, 2014 at 12:46pm*

**Physics**

A 2050 kg truck is traveling east through an intersection at 2.3 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 4.5 m/s. The other is a 1600 kg midsize traveling east at 10 m/s. The...
*Sunday, March 9, 2014 at 12:01pm*

**Physics**

TY! TY!
*Sunday, March 9, 2014 at 8:09am*

**Physics**

work = integral f dx = change in kinetic energy if this is the only force v =dx/dt = 4.1 -7.2 t + 13.5 t^2 at t = 0, v = 0 at t = 8.6 v = 4.1 - 7.2(8.6) + 13.5 (8.6)^2 = 941 m/s^2 Ke = .5 m v^2 = .5(1.1)(942)^2 = 486,642 Joules
*Sunday, March 9, 2014 at 8:09am*

**physics**

120cm
*Sunday, March 9, 2014 at 3:38am*

**Physics**

hey what is the formula for cw,Cc, and Calum plss answer it plss
*Sunday, March 9, 2014 at 3:21am*

**Physics**

question: do you multiply the coefficient for steel in the equation and the coefficient for brass in a separate equation and add them together to get the answer?
*Saturday, March 8, 2014 at 11:34pm*

**Physics**

A single force acts on a 1.1 kg particle-like object in such a way that the position of the object as a function of time is given by x = 4.1t - 3.6t2 + 4.5t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.6 s.
*Saturday, March 8, 2014 at 11:33pm*

**Physics**

The force on a particle is directed along an x axis and given by F = 1.7(x/2.8 - 0.95) where x is in meters and F is in Newtons. Find the work done by the force in moving the particle from x = 0 to x = 6.9 m.
*Saturday, March 8, 2014 at 11:32pm*

**physics-lenses**

we are different
*Saturday, March 8, 2014 at 10:01pm*

**Physics**

you will first of all conevert km/h to m/s which is 105 multiply 1000 then divide by 3600= 29.7. use the formula f= ma. 1250 multiply by 9.8=12250 then use formula for work w=f multiply by distance w=12250 multiply by 29.7=363825J.
*Saturday, March 8, 2014 at 9:58pm*

**physics-lenses**

You must be having an identity crisis, Marian/Melissa/Thalia/Sofia. Please use the same name for your posts.
*Saturday, March 8, 2014 at 9:57pm*

**physics-lenses**

trace and describe the image formed by a convex lens (object between 2f and f)
*Saturday, March 8, 2014 at 9:52pm*

**physics**

I = V/R = 1.5/3 = 0.5A. The voltage across each resistor is 1.5 volts.
*Saturday, March 8, 2014 at 9:46pm*

**physics**

R = V/I = 2/3.2*10^-3
*Saturday, March 8, 2014 at 9:42pm*

**physics**

tension=mg+ma=mg+mv^2/r=you do it. at hightest point, gain of PE= KE at bottom, gain of PE=1/2 .380*2.03^2/.831 mgh=gain of PE h=gainofPE/(.380*9.8) cos Theta= (r-h) /r tension at highest point. mg*cosTheta
*Saturday, March 8, 2014 at 9:28pm*

**physics-lenses**

A refractor's magnification is calculated by dividing the focal length of the optical tube with the focal length of the eyepiece.[1] M=fo/fe fo=length of tube=2*60cm
*Saturday, March 8, 2014 at 9:20pm*

**physics**

Vs = 3m/s[135o] + 2m/s[270o] X = 3*cos135 = -2.12 m/s. Y = 3*sin135 + 2*sin270o = 0.1213 m/s. Tan Ar = Y/X = 0.1213/-2.12 = -0.05723 Ar = -3.275o = Reference angle. A = -3.275 + 180 = 176.7o Vs = X/cosA = -2.12/cos176.7=2.123 m/s. [176.7o]
*Saturday, March 8, 2014 at 9:13pm*

**physics-lenses**

a certain Galilean telescope has a magnification of 2 times. if the focal length of the objective lens is 60 cm, how long is the telescope? thanks :))
*Saturday, March 8, 2014 at 9:11pm*

**physics**

A converging lens is 50cm away from the object. if the distance of the image from the object is 120cm and greater than four times the focal length of the lens, what is the focal length of the said lens?
*Saturday, March 8, 2014 at 9:09pm*

**Physics**

a. V^2 = Vo^2 + 2g*h = 0 @ max ht. Vo^2 - 19.6*1.45 = 0 Vo^2 = 19.6*1.45 = 28.42 Vo = 5.33 m/s = Initial velocity after striking the floor. Impulse = m*V = 0.20*5.33 = 1.067 kg-m/s b. V^2 = Vo^2 + 2g*h = 0 + 19.6*1.95 = 38.22 V = 6.18 m/s. = Velocity just before striking the ...
*Saturday, March 8, 2014 at 8:45pm*

**physics**

A 0.380 kg pendulum bob passes through the lowest part of its path at a speed of 2.03 m/s. What is the tension in the pendulum cable at this point if the pendulum is 83.1 cm long? When the pendulum reaches its highest point, what angle does the cable make with the vertical? ...
*Saturday, March 8, 2014 at 8:16pm*

**physics**

A 75.5 kg skier encounters a dip in the snow's surface that has a circular cross section with radius of curvature of $r$ = 10.1 m. If the skier's speed at point A in the figure below is 7.82 m/s, what is the normal force exerted by the snow on the skier at point B? h...
*Saturday, March 8, 2014 at 8:14pm*

**Physics**

a. D = d1 + d2 + d3 = 70*(45/60) + 95*(10/60) + 55*(30/60) = 95.83 km. T = 45 + 10 + 30 + 20=105 min=1.75 h. Avg. Speed=D/T=95.83km/1.75h=54.8 km/h. NOTE: My procedure is the same as Bob's, but he forgot to include the 20 min for eating and buying gas. Your answer in Part...
*Saturday, March 8, 2014 at 7:06pm*

**Physics**

A 2050 kg truck is traveling east through an intersection at 2.3 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 4.5 m/s. The other is a 1600 kg midsize traveling east at 10 m/s. The...
*Saturday, March 8, 2014 at 6:59pm*

**Physics**

a person's metabolic processes can usually operate at a power of 6 W/kg of body mass for several hours at a time. If a 60 kg woman carrying a 12 kg pack is walking uphill with an energy-conversion efficiency of 20 percent, at what rate, in meters/hour, does her altitude ...
*Saturday, March 8, 2014 at 6:17pm*

**Physics**

I didn't understand what you did for part 2?
*Saturday, March 8, 2014 at 5:07pm*

**Physics**

a
*Saturday, March 8, 2014 at 4:50pm*

**Physics**

A 2050 kg truck is traveling east through an intersection at 2.3 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 4.5 m/s. The other is a 1600 kg midsize traveling east at 10 m/s. The...
*Saturday, March 8, 2014 at 3:29pm*

**physics**

P2/P1 = ((2V)^2/R)/(V^2/R)= (4V^2/R)*R/V^2=4V^2/V^2 = 4 Alternate Method: P1 = V1^2/R = 10^2/100 = 1. Watt. P2 = V2^2/R = 20^2/100 = 4 Watts. P2/P1 = 4W/1W = 4 So the power is increased by a factor of 4.
*Saturday, March 8, 2014 at 3:16pm*

**Physics**

A 5.36 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for the following. (a) first meter of fall initial kinetic energy 0 J final kinetic energy ___...
*Saturday, March 8, 2014 at 3:16pm*

**Physics**

A car of mass m moving at a speed v1 collides and couples with the back of a truck of mass 3m moving initially in the same direction as the car at a lower speed v2. (Use any variable or symbol stated above as necessary.) (b) What is the change in kinetic energy of the car&#...
*Saturday, March 8, 2014 at 3:15pm*

**physics**

A stationary rocket is launched vertically upward. After 4s, the rocket's fuel is used up and it is 225.6 m above the ground. At this instant the velocity of the rocket is 112,8 m/s. Then the rocket undergoes free fall. Ignore the effect of the air friction.Assume that g ...
*Saturday, March 8, 2014 at 3:08pm*

**Physics**

A penny rests on top of a bowling ball. You give the bowling ball a slight bump and the penny slides off the frictionless, spherical surface as shown below. At what angle θ does the penny leave the surface of the bowling ball? (Hint: When the penny leaves the surface, the...
*Saturday, March 8, 2014 at 2:35pm*

**physics**

A ) m g h = 2 * 10 * 17 = 340 Joules B ) zero C ) at top Ke = 17 and Pe = 340 so at bottom where Pe = 0 then Ke = 357 I do not understand why you are asking because the problem leads you step by step to the solution.
*Saturday, March 8, 2014 at 1:33pm*

**Physics**

first how long before it hits ground? 83 = 4.9 t^2 t = 4.12 s so it fell for 4.12 - 1.60 = 2.52 seconds so how far did it fall in 2.52 seconds? d = 4.9*(2.52)^2 = 31 meters so it is 83 - 31 = 52 meters up
*Saturday, March 8, 2014 at 1:21pm*

**Physics**

A stone is dropped from the top of a 83.0 m tall building. How far above the ground is the stone 1.60 seconds before it reaches the ground?
*Saturday, March 8, 2014 at 1:06pm*

**physics**

You reach out a second-story window that is 17 m above the sidewalk and throw a 2-kg ball straing upward with 17 J of kinetic energy. Use g = 10m/s^2 Part A) What is the ball's gravitational potential energy when it is released? Use a height measured from the sidewalk. ...
*Saturday, March 8, 2014 at 11:50am*

**Physics**

Unfortunately, no figure here
*Saturday, March 8, 2014 at 11:25am*

**Physics**

Not much of a cannon :) time to fall to ten meters = time to rise ten meters 10 = (1/2) g t^2 = 4.9 t^2 t = 1.43 seconds up or down v = Vi - g t 0 = Vi -9.81 (1.43) Vi = 14 m/s muzzle velocity now at 33 deg u = 14 cos 33 = 11.8 m/s forever Vi = 14 sin 33 = 7.62 initial speed ...
*Saturday, March 8, 2014 at 11:24am*

**Physics**

Block A in the figure weighs 48.0N . The coefficient of static friction between the block and the surface on which it rests is 0.250. The weight w is 6.00N and the system remains at rest. -Find the friction force exerted on block A. -Find the maximum weight w for which the ...
*Saturday, March 8, 2014 at 11:10am*

**PHYSICS**

F2^2 + 12^2 = F1^2 right triangle, F2 smaller F1 = (18 - F2) F2^2 + 144 = 324 -36 F2 + F2^2 36 F2 = 180 F2 = 5 then F1 = 18-5 = 13
*Saturday, March 8, 2014 at 11:09am*

**Physics**

There is no net force on the object if it is moving at constant speed Therefore if it is moving in the direction of F2 then the sum of the other two forces must be equal and opposite to F2
*Saturday, March 8, 2014 at 10:33am*

**Physics**

A bronze cannon ball with a mass of 4.5 kg is fired vertically from a small cannon and reaches a height of 10.0 m. If the canon is now aimed at an angle 33 degrees above the horizontal, calculate the horizontal distance the cannon ball would travel before it reaches the ground.
*Saturday, March 8, 2014 at 10:28am*

**physics**

i dont know about the answer, i need a solution for this question
*Saturday, March 8, 2014 at 10:26am*

**Physics**

I don't Understand your answer
*Saturday, March 8, 2014 at 10:24am*

**PHYSICS**

the sum of the magnitude of two forces acting at a point is 18 and the magnitude of their resultant is 12.if the resultant is at 90 degree with the force of the smaller magnitude, then their magnitudes are:
*Saturday, March 8, 2014 at 10:16am*

**Physics**

|F1 + F3| = |F2| since the resultant force on the object is zero (constant speed, no acceleration, no force)
*Saturday, March 8, 2014 at 10:06am*

**Physics**

Three forces, F1 = 9.2 N, F2 = 7.8 N and F3 = 8.3 N are acting on a small box of mass 47.0 kg. If the box is moving in the direction of F2 at constant speed, what is the magnitude of the combination of F2 and F3
*Saturday, March 8, 2014 at 10:00am*

**physics**

Not be that person or anything... but your final calculation is incorrect... it is 4342.
*Saturday, March 8, 2014 at 9:32am*

**physics**

120 mi/h (1609 m/mi)(1 h/3600s) = 53.6 m/s F = change in momentum/time = 60 (53.6)/56 = 57.5 Newtons
*Saturday, March 8, 2014 at 6:22am*

**Physics-motion**

distance accelerating = (1/2)at^2 same distance de-accelerating so L = a t^2 t = (L/a)^(1/2) I do not agree with the 4
*Saturday, March 8, 2014 at 6:18am*

**Physics-motion**

A car accelerates uniformly from rest at a. After a certain time, its acceleration becomes -a. It continues to travel and stops finally. If the total displacement of the car is L, what is the total time of travel? The answer is (4L/a)^(1/2). But why??
*Saturday, March 8, 2014 at 5:28am*

**physics**

A) K(hat) B) -K(hat)
*Saturday, March 8, 2014 at 12:39am*

**physics**

A jet plane takes about 56 s to go from rest to the takeoff speed of 120 mph (miles per hour). What is the average horizontal force that the seat exerts on the back of a 60-kg passenger during takeoff? First make sure that each of the values are in SI units. Use 1 mile = 1609...
*Saturday, March 8, 2014 at 12:39am*

**physics**

A small block of mass 42 kg slides along a frictionless loop-the-loop track. The radius of the loop is 5 meters. At what height above the bottom of the track should the block be released from so that it just makes it through the loop-the-loop without losing contact with the ...
*Friday, March 7, 2014 at 11:45pm*

**PHYSICS**

A frictionless horizontal air track has a spring at either end. The spring on the left has a spring constant of 1600 N/m; the one on the right has a spring constant of 1060 N/m. A glider with a mass of 2.9 kg is pressed against the left-hand spring, compressing it by 3.7 cm. ...
*Friday, March 7, 2014 at 11:40pm*

**physics**

Ricky, the 5.8 kg snowboarding raccoon, is on his snowboard atop a frictionless ice-covered quarter-pipe with radius 9 meters. If he starts from rest at the top of the quarter-pipe, what is his speed at the bottom of the quarter-pipe? Use SI units in your answer. v_{Ricky} =
*Friday, March 7, 2014 at 11:19pm*

**physics**

A bartender slides a glass mug with a mass of 520 g along the top of a level bar. The coefficient of kinetic friction between the glass and the bar is 0.09. When the glass leaves the bartender's hand, it is moving, it is moving at a speed of 3.9 m/s. How far does the ...
*Friday, March 7, 2014 at 11:03pm*

**Physics**

F = 10N[45o] + 8N[330o] -7.5N X = 10*cos45 + 8*cos330 - 7.5 = 6.50 N. Y = 10*sin45 + 8*sin330 = 3.07 N. Tan A = Y/X = 3.07/6.50 = 0.47247 A = 25.29o Fr = X/cosA = 6.50/cos25.29=7.19N[25.29] = Resultant force. a = Fr/m = 7.19[25.29o]/49=0.147[25.29]
*Friday, March 7, 2014 at 10:19pm*

**physics**

Two forces, 1 = (3.85 − 2.85) N and 2 = (2.95 − 3.65) N, act on a particle of mass 2.10 kg that is initially at rest at coordinates (−2.30 m, −3.60 m). (a) What are the components of the particle's velocity at t = 11.8 s? = m/s (b) In what direction...
*Friday, March 7, 2014 at 9:26pm*

**Physics**

L = V/F = 0.333 m. L = V/15000 = 0.333 V = 0.333 * 15,000 = 4995 m/s. 4995 - 343 = 4652 m/s faster.
*Friday, March 7, 2014 at 9:23pm*

**physics**

X = 130 km/h Y = 275 km/h V^2 = 130^2 + 275^2 = 92,525 V = 304.2 km/h.
*Friday, March 7, 2014 at 9:12pm*

**Physics**

F = 4.8 - 0.6 = 4.2 N.
*Friday, March 7, 2014 at 8:57pm*

**physics**

the cheetah is is one of the fastest accelerating animals, for it can go from rest to 34.5 m/s in 5.00 seconds. if its mass is 106 kg, determine the average powee developes by the cheetah during the acceleration phase of its motion
*Friday, March 7, 2014 at 8:44pm*

**Physics**

Know matter how i work it out, i keep getting the wrong answers. please help!!! i only have two more submissions before it closes on me. A sample of blood is placed in a centrifuge of radius 14.0 cm. The mass of a red blood cell is 3.0 x 10^-16 kg, and the magnitude of the ...
*Friday, March 7, 2014 at 8:04pm*

**Physics**

Know matter how i work it out, i keep getting the wrong answers. please help!!! i only have two more submissions before it closes on me. A sample of blood is placed in a centrifuge of radius 14.0 cm. The mass of a red blood cell is 3.0 ✕ 10−16 kg, and the ...
*Friday, March 7, 2014 at 8:03pm*

**Physics**

Floor pushes up total weight = (39.7 + 65.3)(9.81) crate pushes up on person = (65.3)(9.81)
*Friday, March 7, 2014 at 5:31pm*

**physics**

Two can play this passive agressive thing coward. Aint the info age grand.
*Friday, March 7, 2014 at 4:57pm*

**Physics**

A 39.7-kg crate rests on a horizontal floor, and a 65.3-kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person.
*Friday, March 7, 2014 at 4:52pm*

**physics**

Using conservation of energy. Energy transfers from potential spring to kinetic. 1.85N/cm=185N/m 25cm=.25m PE=.5*185*(.25^2)=5.78125 PE=KE(final) KE(final)=1/2(toatl mass)*velocity^2 KE=.5(2*1.6)(v^2) 5.78125=.5*2*1.6*v^2 algebra....v=1.901m/s
*Friday, March 7, 2014 at 3:59pm*

**Physics**

4
*Friday, March 7, 2014 at 2:55pm*

**Physics**

#2
*Friday, March 7, 2014 at 12:59pm*

**physics**

You must show your work for full credit.) Calculate the circular speed of an object that orbits the Sun at a distance of 0.2 AU (1 AU = 1.5e+11 meters, the distance between the Earth and the Sun)
*Friday, March 7, 2014 at 8:54am*

**Physics**

opposite charges, they attract. Use coulombs law to find the force. c. excells= chargeonCork/chargeononeElectron
*Friday, March 7, 2014 at 6:05am*

**Physics**

A small cork with an excess charge of +6.0 μC (1 μC = 10^-6 C) is placed 0.12 m from another cork, which carries a charge of -4.3 μC. a. What is the magnitude of the electric force between the corks? b. Is this force attractive or repulsive? c. How many excess ...
*Friday, March 7, 2014 at 2:06am*

**Physics**

10.6
*Friday, March 7, 2014 at 1:27am*

**Physics**

After a 0.200-kg rubber ball is dropped from a height of 1.95 m, it bounces off a concrete floor and rebounds to a height of 1.45 m. (a) Determine the magnitude and direction of the impulse delivered to the ball by the floor. (b) Estimate the time the ball is in contact with ...
*Friday, March 7, 2014 at 12:24am*

**physics **

The lens-to-retina distance of a woman is 1.96 cm, and the relaxed power of her eye is 55.2 D. a) What is her far point? b)What eyeglass power will allow her to see distant objects clearly, if her glasses are 1.80 cm from her eyes?
*Thursday, March 6, 2014 at 11:41pm*

**Physics**

No
*Thursday, March 6, 2014 at 11:29pm*

**physics**

23
*Thursday, March 6, 2014 at 11:17pm*

**physics**

A 6.0 Ù resistor and a 3.0 Ù resistor are connected in parallel to a 1.5 V battery of negligible internal resistance. What is the current in the 3.0 Ù resistor?
*Thursday, March 6, 2014 at 9:22pm*

**physics**

A potential difference of 2.0 V is applied across a wire of cross sectional area 2.5 mm2. The current which passes through the wire is 3.2 × 10-3 A. What is the resistance of the wire?
*Thursday, March 6, 2014 at 9:20pm*

**physics**

At a critical junction in a supercomputer, the current as a function of time is given as i(t) = 3t^2-2t, where i is measured in milliamps and t is measured in seconds. How much charge passes through this critical junction during the interval 0 < t < 5.00 s?
*Thursday, March 6, 2014 at 9:20pm*

**Physics**

Wt. of crate = m * g = 3.2kg * 9.8N/kg = 31.36 N. Fp = 31.36*sin35 = 17.99 N. = Force parallel to the incline. F = Fp*cos35 = 17.99*cos35 = 14.74 N.
*Thursday, March 6, 2014 at 9:19pm*

**physics**

V^2 = Vo^2 + 2g*h V^2 = !0^2 + 19.6*75 = 1570 V = 39.62 m/s KE = 0.5m*V^2 = 1*39.62^2 = 1570 J.
*Thursday, March 6, 2014 at 8:36pm*

**Physics**

A potential difference of 2.0 V is applied across a wire of cross sectional area 2.5 mm^2. The current which passes through the wire is 3.2 × 10-3 A. What is the resistance of the wire?
*Thursday, March 6, 2014 at 8:34pm*

**Physics**

N identical resistors are connected in parallel. The resistance of one of them is equal to R. What is the equivalent resistance of all N resistors?
*Thursday, March 6, 2014 at 8:27pm*

**physics**

a=(V^2-Vo^2)/2d=(0-15^2)/14=-16.07 m/s^2 d = (V^2-Vo^2)/2a d = (0-(32^2))/-32.14 = 31.86 m.
*Thursday, March 6, 2014 at 7:30pm*

**physics**

speed at bottom = sqrt(2 g h) = sqrt (2 * 9.81 * .38) = 2.73 m/s momentum before crash = .6*2.73 = 1.64 kg m/s same momentum after of course new mass = .6 + .94 = 1.54 kg so 1.64 = 1.54 v v = 1.06 m/s
*Thursday, March 6, 2014 at 5:39pm*

**physics**

weight = m g friction force = mu m g so static mus 75 = mus (20)(9.81) and kinetic muk 60 = muk (20)(9.81)
*Thursday, March 6, 2014 at 5:29pm*

**Physics**

well, If he jumped up, all bets are off. If he just fell off then: v = sqrt (2 g h) because m g h = (1/2) m v^2 conservation of energy g h = (1/2) v^2 so v = sqrt (2*9.81*15) = 17.2 meters/second
*Thursday, March 6, 2014 at 5:27pm*

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