Answer:
I don't speak spanish oso sorry
Explanation:
Answer:
Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension
ex = бx/E
бx = Fx/A = Fx/π
Using both equation and solving for the modulus of elasticity E
E = бx/ex = Fx / π
ex
E = 
Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius
ey =
(бy - v (бx + бz)) =
бx
=
= 
Finally
ey = Δr / r
Δr = ey * r = 10 * 
Δd = 2Δr = 
Explanation:
Answer:
Explanation:
mass of object, m = 3 kg
spring constant, K = 750 n/m
compression, x = 8 cm = 0.08 m
angle of gun, θ = 30°
(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.
(b) Let v be th evelocity of ball at the tim eof launch.
by using the conservation of energy
1/2 Kx² = 1/2 mv²
750 x 0.08 x 0.08 = 3 x v²
v = 1.265 m/s
By use of the formula of maximum height


h = 0.02 m
h = 2 cm
Answer:
on an average, <em>2</em><em>0</em><em>H</em><em>z</em><em> </em><em>to</em><em> </em><em>2</em><em>0</em><em>k</em><em>H</em><em>z</em>