General
19 Nov 07
19 Nov 07
Two automobiles of weight 7.12 kN and 14.24 kN are traveling along horizontally at 96 km/h when they both run out of gas. Luckily, there is a town in a valley not far off, but it's just beyond a 33.5 m high hill. Assuming that friction can be neglected, which of the cars will make it to town?
19 Nov 07
3 edits
Originally posted by Dutch DefenseThey both can make it over.
Two automobiles of weight 7.12 kN and 14.24 kN are traveling along horizontally at 96 km/h when they both run out of gas. Luckily, there is a town in a valley not far off, but it's just beyond a 33.5 m high hill. Assuming that friction can be neglected, which of the cars will make it to town?
The first one has a mass of 7.12x10^3 N / 9.81 N kg^-1 = 725.79 kg
The second one has a mass of 14.24x10^3 N / 9.81 N kg^-1 = 1451.58 kg
Kinetic energy of first one = 1/2 x 725.79 x (26.67 ms^-2) ^2 = 258.06 kJ
KE of second one = 1/2 x 1451.58 x (26.67 ms^-2) ^2 = 516.25 kJ
Gravitational potential energy of first one at top of hill = 725.79 x 9.81 x 33.5 = 238.52 kJ
PE of second one at top of hill = 1451.58 x 9.81 x 33.5 = 477.04 kJ
So both have enough energy to get to top and still have momentum.
19 Nov 07
Originally posted by Dutch Defensetake it to posers and puzzles spanky
Two automobiles of weight 7.12 kN and 14.24 kN are traveling along horizontally at 96 km/h when they both run out of gas. Luckily, there is a town in a valley not far off, but it's just beyond a 33.5 m high hill. Assuming that friction can be neglected, which of the cars will make it to town?
19 Nov 07
Originally posted by Dutch Defensefriction is negligible so if this is your homework they are just trying to throw you off.
A ball having a mass of 0.50 kg is thrown straight up at a speed of 25.0 m/s.
a) How high will it go if there is no friction?
b) If it rises 22m, what was the average force due to air friction?
19 Nov 07
Originally posted by Dutch Defensea)
A ball having a mass of 0.50 kg is thrown straight up at a speed of 25.0 m/s.
a) How high will it go if there is no friction?
b) If it rises 22m, what was the average force due to air friction?
v0 = 25ms^-1
a = -9.81ms^-2
0 = 25^2 - 2 * 9.81h
h = 31.85m
b)
Can't be bothered, as it is late here and I am off to bed. Although if it rises only 22m, then friction isn't negligible.
19 Nov 07
Originally posted by Dutch Defensea) It depends where you're throwing it from. The escape velocity from the earth's surface is only 7 mps so it would go into orbit about the sun. Sorry, but I don't know the sun's escape velocity.
A ball having a mass of 0.50 kg is thrown straight up at a speed of 25.0 m/s.
a) How high will it go if there is no friction?
b) If it rises 22m, what was the average force due to air friction?
19 Nov 07
Originally posted by Lukerikm/s stands for meters per second not miles per second.
a) It depends where you're throwing it from. The escape velocity from the earth's surface is only 7 mps so it would go into orbit about the sun. Sorry, but I don't know the sun's escape velocity.
19 Nov 07
Originally posted by lauseyb)
a)
v0 = 25ms^-1
a = -9.81ms^-2
0 = 25^2 - 2 * 9.81h
h = 31.85m
b)
Can't be bothered, as it is late here and I am off to bed. Although if it rises only 22m, then friction isn't negligible.
To get 'a' in this situation will be:
0 = 25^2 + 44a
a = -625 / 44 = -14.20ms^-2
Difference of -4.39ms^-2
F = ma
F = 0.5 x 4.39 = 2.19N