Impulse is the change in momentum
J = 1.5 kg(22 m/s - 15 m/s) = 10.5 kg-m/s
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Answer:
79.5 m
Explanation:
Let t be the time taken to hit the surface of water and x be the horizontal distance traveled.
use II equation of motion in Y axis direction
h = uy t + 1/2 g t^2
- 60 = - 100 Sin 35 x t - 1/2 x 9.8 x t^2
-60 = - 57.35 t - 4.9 t^2
4.9 t^2 + 57.35 t - 60 = 0
By solving we get
t = 0.97 second
The horizontal distance traveled is
x = ux t
x = 100 Cos 35 x 0.97
x = 79.5 m
Answer:
20.6m
Explanation:
The mass is useless in this case.
Kinetic Energy = Potential Energy
g is acceleration due to gravity.
(1/2)mv² = mgh (m's drop out)
(1/2)v² = gh
(1/2)(20.1²) = 9.81(h)
h = (20.1 x 20.1)/9.81 x 2
20.61 meters
