Answer:
K = 0.076 J
Explanation:
The height of the target, h = 0.860 m
The mass of the steel ball, m = 0.0120 kg
Distance moved, d = 1.50 m
We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.
Put all the values,
The velocity of the ball is :
The kinetic energy of the ball is :
So, the required kinetic energy is 0.076 J.
Answer:
E=15.3*10¹³ N/C : approximate
Explanation:
We use the following formula to calculate the electric field due to a disk with uniform surface charge at a point P that is along the central perpendicular axis of the disk and at a distance x from the center of the disk:
E= 2*π*k*σ*F Fórmula ( 1 )
E: Electric field at point P (N/C)
σ: surface charge density (C/m²)
R: disk radio (m)
x :distance from the center of the disk to the point P located on the axis of the disk (m)
K: Coulomb constant ( N*m²/C)
Equivalences
1cm = 10⁻²m
1mm = 10⁻³m
Data
R =2.6 cm= 2.6*10⁻²m = 0.026m
x=3.7 mm = 3.7* 10⁻³m = 0,0037 m
Q=+6.71C.
k= 8.98774 * 10⁹ N* m²/C
Calculation of surface charge density (σ )
σ= Q/A
Q: uniformly distributed charge (C)
A: disk area (m²) = π*R²
σ= +6.71C/π*(2.6*10⁻²)²m² = 3159.56C/m²
Calculation of the electric field at point P
We apply formula (1) and replace data
E= 2*π*3159.56*8.98774*10⁹ *F
E=15.3*10¹³ N/C : approximate