Answer:
1.66 kg
Explanation:
Given that a 0.83-kg block is hung from and stretches a spring that is attached to the ceiling.
From Hook's law
F = Ke
But F = mg
Substitute mg for force in the Hook's law
Mg = ke
0.83 × 9.8 = ke
Make K the subject of formula
8.134 = Ke
K = 8.134 /e
Given that a second block is attached to the first one, and the amount that the spring stretches from its unstretched length triples.
That is
(0.83 + M) × 9.8 = K (3e)
Substitutes K into the above equation
(0.83 + M) × 9.8 = 8.134 / e (3e)
The e will cancel out
(0.83 + M) × 9.8 = 24.402
0.83 + M = 24.402/9.8
0.83 + M = 2.49
M = 2.49 - 0.83
M = 1.66 kg
Therefore, the mass of the second block is 1.66kg
Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
Answer:
the branch of mechanics concerned with the interaction of electric currents with magnetic fields or with other electric currents.
Explanation:
