The electron and positron collide and annihilate each other, converting their total energy (sum of kinetic and rest energy) into energy carried by two photons:
2(1.64MeV + 0.511MeV) = 2
= 1.64MeV + 0.511MeV
= 2.15MeV
Answer:
a) 4 s
b) 36 m/s²
c) 54 m
Explanation:
s = 2t³ – 24t + 6
a) Find t when v = 72 m/s.
v = ds/dt
v = 6t² – 24
72 = 6t² – 24
6t² = 96
t = 4
b) Find a when v = 30 m/s.
a = dv/dt
a = 12t
When v = 30:
30 = 6t² – 24
6t² = 54
t = 3
a = 36
c) Find Δs between t = 1 and t = 4
Δs = (2(4)³ – 24(4) + 6) – (2(1)³ – 24(1) + 6)
Δs = 38 – (-16)
Δs = 54
Answer:
15.2 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S.I unit of acceleration is m/s²
From the question,
F-W = ma......................... Equation 1
Where F = Tension on the line, W = weight of the fish, m = mass of the fish, a = acceleration of the fish.
But,
W = mg........................ Equation 2
Where g = acceleration due to gravity.
Substitute equation 2 into equation 1
F-mg = ma
make a the subject of the equation
a = (F-mg)/m.................... Equation 3
Given: F = 150 N, m = 6 kg,
Constant: g = 9.8 m/s²
Substitute into equation 2
a = [150-(6×9.8)]/6
a = (150-58.8)/6
a = 91.2/6
a = 15.2 m/s²
Hence the minimum acceleration of the fish = 15.2 m/s²
Answer:
There is no options for this answer
Explanation:
Answer:
The fluids speed at a) 
and b) 
are 
and 
respectively
c) Th volume of water the pipe discharges is: 
Explanation:
To solve a) and b) we should use flow continuity for ideal fluids:
(1)
With Q the flux of water, but Q is 
using this on (1) we have:
(2)
With A the cross sectional areas and v the velocities of the fluid.
a) Here, we use that point 2 has a cross-sectional area equal to 
, so now we can solve (2) for 
:
b) Here we use point 2 as 
:
c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is 
, so we can write:
, solving for V:
