Answer:

t'=1.1897 μs
Explanation:
First we will calculate the velocity of micrometeorite relative to spaceship.
Formula:

where:
v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).
u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)
u' is the relative velocity of micrometeorite with respect to spaceship.
In order to find u' , we can rewrite the above expression as:


u'=0.9806c
Time for micrometeorite to pass spaceship can be calculated as:

(c = 3*10^8 m/s)


t'=1.1897 μs
Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g