Answer:
Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.
Again item 2 refers to bonds between atoms and is eliminated.
This leaves only item 1.
Nuclear forces are very short range forces between components of the nucleus.
Weak nuclear forces are trillions of times smaller than strong forces.
Gravitational forces are much much smaller than the weak nuclear force.
Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
Explanation:
Given:
v₀ = 0 m/s
a = 9.8 m/s²
t = 4.7 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²
Δy ≈ 110 m
