Answer:
A block device is a computer data storage device that supports reading and (optionally) writing data in fixed-size blocks, sectors, or clusters. These blocks are generally 512 bytes or a multiple thereof in size
Take the tiny bit of carbon dioxide and the tiny bit of water vapor out of the air,
and the rest of what you're breathing right now is a mixture of elements.
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía
Answer:
1.97 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
Solving the above equation we get
So, the time the package was in the air is 1.97 seconds
Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
Let,
m
A
= Mass of ball A
m
B
= Mass of ball B
u
A
= initial velocity of ball A
u
B
= initial velocity of ball B
v
A
= Velocity after the collision of ball A
v
B
= Velocity after the collision of ball B
F
ab
= Force exerted by A on B
F
ba
= Force exerted by B on A
Now,
Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision
= m
A
v
A
−m
A
u
A
Rate of change of momentum A= Change in momentum of A/ time taken
=
t
m
A
v
A
−m
A
u
A
Force exerted by B on A (F
ba
);
F
ba
=
t
m
A
v
A
−m
A
u
A
........ [i]
In the same way,
Rate of change of momentum of B=
t
m
b
v
B
−m
B
u
B
Force exerted by A on B (F
ab
)=
F
ab
=
t
m
B
v
B
−m
B
u
B
.......... [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
F
a
b=−F
b
a [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]
Using [i] and [ii] , we have
t
m
B
v
B
−m
B
u
B
=−
t
m
A
v
A
−m
A
u
A
m
B
v
B
−m
B
u
B
=−m
A
v
A
+m
A
u
A
Finally we get,
m
B
v
B
+m
A
v
A
=m
B
u
B
+m
A
u
A
This is the derivation of conservation of linear momentum.
