A block of wood is compressed 1.2 nm when inward forces of magnitude 103 N are applied to it on two opposite sides.
(a) What is the effective spring constant of the block? (HINT: Please use Hooke's law equation 6-19 from chapter 6 of your text book.)
(b) Assuming Hooke's law still holds, how much would the block be compressed by inward forces of magnitude 490 N?
2006-08-25 17:07:02 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Hooke's law is F=-kx. F is the force, x is the deformation, and k is the spring constant. So you rearrange it to be:
k = -F/x
k = -(103) / (-1.2 E-9)
1.2 E-9 means 1.2 * 10^-9; it's a conversion from nanometers to meters. For part b just change the force to 490N.
2006-08-25 17:48:14 · answer #1 · answered by Anonymous · 0⤊ 0⤋
spring constant= 103/ (1.2x10^-9)=8.6x10^10newtons/metre.
490 newtons would compress it 490/103 x 1.2nm=5.7 nm.
2006-08-25 18:47:15 · answer #2 · answered by zee_prime 6 · 0⤊ 0⤋
not a lot, by average standards.
2006-08-25 17:10:22 · answer #3 · answered by damndirtyape212 5 · 0⤊ 0⤋
DO YOUR OWN HOMEWORK OR I'LL TAKE AWAY YOUR CAR KEYS!!!
2006-08-25 17:11:06 · answer #4 · answered by Aaron M 3 · 0⤊ 1⤋