Here is the genius answer
1. Suppose that you have two samples with the
1 answer:
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The impulse (the variation of momentum of the ball) is related to the force applied by
where
is the variation of momentum, F is the intensity of the force and 
is the time of application of the force.
Using F=1000 N and
, we can find the variation of momentum:
This can be rewritten as
where
and 
are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
from which we find the final velocity of the ball:
where
Using F=1000 N and
This
where
from which we find the final velocity of the ball:
This question is incomplete, the missing image is uploaded along this answer below.
Answer:
the speed of the 50-kg cylinder after it has descended is 3.67 m/s
Explanation:
Given the data in the question and the image below;
relation between velocity of cylinder and velocity of the drum is;
V = ω
× r
----- let this be equ 1
where V is velocity of cylinder, ω
is the angular velocity of drum C and r
is the radius of drum C
Now, Angular velocity of gear B is;
ω = ω
ω = V
/ r
-------- let this equ 2
so;
V / 0.1 m = 10V
Next, we determine the angular velocity of gear A;
from the diagram;
ω( 0.15 m ) = ω
( 0.2 m )
from equation 2; ω = V
/ r
so
ω( 0.15 m ) = (V
/ r
) 0.2 m
substitutive in value of radius r (0.1 m)
ω( 0.15 m ) = (V
/ 0.1 m ) 0.2 m
ω( 0.15 ) = 0.2V
/ 0.1
ω = 2V
/ 0.15
ω = 13.333V
----- let this be equation 3
To get the speed of the cylinder, we use energy conversation;
assuming that the final position is;
T₁ + ∑ = T₂
0 + mgh =
m
V²
+
ω²
+
ω²
so
mgh =
m
V²
+
(m
k
²)(13.333V
)² +
(m
k
²)(10V
)²
we given that; m = 50 kg, h = 2 m, m
= 10 kg, k
125 mm = 0.125 m, m
= 30 kg, k
= 150 mm = 0.15 m.
we know that; g = 9.81 m/s²
so we substitute
50 × 9.81 × 2 = ( × 50 × V
²) +
( 10 × (0.125)² )(13.333V
)² +
( 30 × (0.15)²)(10V
)²
981 = 25V² + 13.888V
² + 33.75V
²
981 = 72.638V²
V² = 981 / 72.638
V² = 13.5053
V = √13.5053
V = 3.674955 ≈ 3.67 m/s
Therefore, the speed of the 50-kg cylinder after it has descended is 3.67 m/s
