Answer:
20 Yards
Explanation:
|---20----|
| |
| 50 |50
|---D--->|
Start End
Total displacement(D) 20 yards (East).
Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force, 
= m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F = = m·v²/r = 16 N
(1/2) × r × = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r × = (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.
Answer:
I think it would be A
Explanation:
Air, water, and warmth
Excuse my mistake if I get it wrong.
Answer:
Approximately 
to the right (assuming that both astronauts were originally stationary.)
Explanation:
If an object of mass 
is moving at a velocity of 
, the momentum 
of that object would be 
.
Since momentum of this system (of the astronauts) conserved:
.
Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be 
since the velocity of both astronauts was 
.
Therefore:
.
The final momentum of the first astronaut (
, 
to the left) would be 
to the left.
Let 
denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be 
.
.
Hence, 
. In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude (
) but opposite in direction (to the right versus to the left.)
Rearrange the equation to obtain an expression for velocity in terms of momentum and mass: 
.
.
Hence, the velocity of the astronaut in question (
) would be to the right.
Answer:
Explanation:
⁶₃Li will have 3 protons and 3 neutrons .
mass of proton in amu = 1.00727 amu
mass of neutron in amu = 1.00866 amu
mass of lithium nucleus in amu = 6.01512 amu
mass defect = 3 ( 1.00727 + 1.00866 ) - 6.01512 amu
= .03267 amu
Binding energy = mass defect in amu x 931 Mev
= 30.41 MeV
binding energy per nucleon
no of nucleon = 3 + 3 = 6
binding energy per nucleon = 30.41 / 6 Mev
= 5.068 MeV .
