Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,
1 answer:
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First, let's find the speed 
of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
After the collision, the two blocks stick together and so now they have mass
and they are moving with speed :
For conservation of momentum
So we can write
From which we find
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force
. The work done by the frictional force to stop the two blocks is
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
From which we can find the value of the coefficient of kinetic friction:
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
After the collision, the two blocks stick together and so now they have mass
For conservation of momentum
So we can write
From which we find
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
From which we can find the value of the coefficient of kinetic friction:
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost = = 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J