Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
Answer:
body position 4 is (-1,133, -1.83)
Explanation:
The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by
x_cm = 1 /M ∑
m_{i}
y_cm = 1 /M ∑ y_{i} mi
Where M is the total mass of the body, mi is the mass of each element
give us the mass and position of this masses
body 1
m1 = 2.00 ka
x1 = 0 me
y1 = 0 me
body 2
m2 = 2.20 kg
x2 = 0m
y2 = 5 m
body 3
m3 = 3.4 kg
x3 = 2.00 m
y3 = 0
body 4
m4 = 6 kg
x4=?
y4=?
mass center position
x_cm = 0
y_cm = 0
let's apply to the equations of the initial part
X axis
M = 2.00 + 2.20 + 3.40
M = 7.6 kg
0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)
x4 = -6.8 / 6
x4 = -1,133 m
Axis y
0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)
y4 = -11/6
y4 = -1.83 m
body position 4 is (-1,133, -1.83)
I don’t think I will have any time to go
Idk what is growing but if it’s a free than c
Answer:
speed of golf ball is 1.15 ×
m/s
and % of uncertainty in speed = 2.07 ×
%
Explanation:
given data
mass = 45.9 gram = 0.0459 kg
speed = 200 km/hr = 55.5 m/s
uncertainty position Δx = 1 mm =
m
to find out
speed of the golf ball and % of speed of the golf ball
solution
we will apply here heisenberg uncertainty principle that is
uncertainty position ×uncertainty momentum ≥
......1
Δx × ΔPx ≥
here uncertainty momentum ΔPx = mΔVx
and uncertainty velocity = ΔVx
and h = 6.626 ×
Js
so put here all these value in equation 1
× 0.0459 × ΔVx = 
ΔVx = 1.15 ×
m/s
and
so % of uncertainty in speed = ΔV / m
% of uncertainty in speed = 1.15 ×
/ 55.5
% of uncertainty in speed = 2.07 ×
%