The answer to the first one is sublimation.
The gravitational acceleration at any distance r is given by

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.
The Earth's radius is
, so the meteoroid is located at a distance of:

And by substituting this value into the previous formula, we can find the value of g at that altitude:

Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle, 
Coefficient of kinetic friction, 
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:





Now, the speed is given by the third eqn of motion with initial velocity being zero:

where
u = initial velocity = 0
Thus


Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory