Posts by Bartholomew
Total # Posts: 11
A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to move off at an angle of 45 degrees. The neutron's initial speed is 5.9×105 m/s. Determine the speeds of the ...
A bullet of mass 1.9×10−3 kg embeds itself in a wooden block with mass 0.987 kg, which then compresses a spring (k = 120 N/m) by a distance 3.5×10−2 m before coming to rest. The coefficient of kinetic friction between the ...
#1. Ten Christmas lights are connected in series and plugged into a 120 V wall outlet. Each light is identical and can be thought of as a resistor. a) What should the value of this resistance be if the total string is to dissipate 50 W? b) How much power is dissipated in each ...
I plugged in my value for x, but both sides of the equation had different values, so I think I have the wrong value for x. How do I solve 200x^2-49x+490?
A 5 kg mass is thrown up with a velocity of 4 m/s from a height of 30 m onto a spring with a relaxed length of 10 m, and a constant of 400 N/m. What will the maximum compression of the spring be? What will the speed of the object be when the spring is compressed 0.3 m? HERE...
There are 3 feet in a yard. Square yards to get square yards.
(V2^2)=(Vi)^2+2ad 0=16-19.6d -16=-19.6d d=0.816 m + 30 m = 30.816 meters above the ground mgh = mgh + 1/2kx^2 = (5 kg)(9.8 m/s^2)(30.816) = 200x^2-49x+490 x=2.75 m mgh=1/2kx^2+mgh+1/2mv^2 (5)(9.8)(30.186)=1/2(400)(0.3)^2+(5)(9.8)(9.7)+1/2(5v^2) v=20.17 m/s
A 5 kg mass is thrown up with a velocity of 4 m/s from a height of 30 m onto a spring with a relaxed length of 10 m, and a constant of 400 N/m. What will the maximum compression of the spring be? What will the speed of the object be when the spring is compressed 0.3 m?
Sorry about the lines, I was expecting it to look like this: img517.imageshack.us/img517/9280/picture1in2.png
ANSWER A = 133.333333 B = 33.3333333 C = 13.3333333 WORK A = 4B C = B-20 A + B + C = 180 | | | | (4B) + B (B-20) = 180 \ | / \ | / \ | / \ | / \ | / \|/ 6B - 20 = 180 B = 33.33333 ^ A is 4 times that C is 20 less than that
A 50 kg block is on a 30 degree incline. The coefficient of friction between the block and the incline 0.05. If it is given an initial velocity of 10 m/s up the incline, how far will it travel until it stops?