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
The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let 
be the masses of the three fragments.
Let 
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,
So the x- component of the velocity of the m2 fragment after the explosion is,
∴ 
When you touch an object and heat flows OUT of it, INTO your finger, you say the object feels hot.
When you touch an object and heat flows INTO it, OUT of your finger, you say the object feels cold.
If the object has the same temperature as your finger ... <em>around the mid-90s</em> ... then no heat flows in or out of your finger when you touch the object, and the object doesn't feel hot or cold.
Answer:
Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation
there will be some initial kinetic energy but after collision kinetic energy will zero
on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the 
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by
Where
g = Gravitational acceleration
t = time
As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.
Both bodies will end with the same thermal speed.
