Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × 
s
speed = 0.8 c = 0.8 × 3 × 
to find out
expected distance over
solution
we know c is speed of light in air is 3 × m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × × 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
Answer:
a) W = - 318.26 J, b) W = 0
, c) W = 318.275 J
, d) W = 318.275 J
, e) W = 0
Explanation:
The work is defined by
W = F .ds = F ds cos θ
Bold indicate vectors
We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces
Let's use trigonometry to break down weight
sin θ = Wₓ / W
Wₓ = W sin 60
cos θ = Wy / W
Wy = W cos 60
X axis
How the body is going at constant speed
fr - Wₓ = 0
fr = mg sin 60
fr = 15 9.8 sin 60
fr = 127.31 N
Y Axis
N - Wy = 0
N = mg cos 60
N = 15 9.8 cos 60
N = 73.5 N
Let's calculate the different jobs
a) The work of the force of gravity is
W = mg L cos θ
Where the angles are between the weight and the displacement is
θ = 60 + 90 = 150
W = 15 9.8 2.50 cos 150
W = - 318.26 J
b) The work of the normal force
From Newton's equations
N = Wy = W cos 60
N = mg cos 60
W = N L cos 90
W = 0
c) The work of the friction force
W = fr L cos 0
W = 127.31 2.50
W = 318.275 J
d) as the body is going at constant speed the force of the tape is equal to the force of friction
W = F L cos 0
W = 127.31 2.50
W = 318.275 J
e) the net force
F ’= fr - Wx = 0
W = F ’L cos 0
W = 0
The two forces should be equal therefore:
2.10 * Fa = Fa + 2 * F * cos 18
simplifying the right side:
2.10 * Fa = Fa + 1.902 * F
1.10 Fa = 1.902 F
<span>F / Fa = 0.578</span>
Answer:
The ratio (U₁/U₂) = 6
Explanation:
U, the potential energy is given as
U = kqQ/r
k = Coulomb's constant
q = charge we're concerned about
Q = charge of the negative plate of the capacitor
r = distance of q from the negative plate of the capacitor.
For charge q₁
U₁ = kq₁Q/s
U₂ = kq₂Q/2s
But q₂ = q₁/3
U₂ becomes U₂ = kq₁Q/6s
U₁ = kq₁Q/s
U₂ = kq₁Q/6s
(U₁/U₂) = 6
