What is a resistance training method that involves performing a set with light load followed by increasing the load and decreasi
1 answer:
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Answer:
<em>63.44 rad/s</em>
<em></em>
Explanation:
mass of bullet = 3.3 g = 0.0033 kg
initial velocity of bullet = 250 m/s
final velocity of bullet = 140 m/s
loss of kinetic energy of the bullet =
==> = 70.785 J
this energy is given to the stick
The stick has mass = 250 g =0.25 kg
its kinetic energy = 70.785 J
from
KE =
70.785 =
566.28 =
= 23.79 m/s
the stick is 1.5 m long
this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m
The angular speed will be
Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>
Answer:
The horizontal component of the vector ≈ -16.06
The vertical component of the vector ≈ 19.15
Explanation:
The magnitude of the vector, = 25 units
The direction of the vector, θ = 130°
Therefore, we have;
The horizontal component of the vector, Rₓ = × cos(θ)
∴ Rₓ = 25 × cos(130°) ≈ -16.06
<em>The horizontal component of the vector, Rₓ ≈ -16.06</em>
The vertical component of the vector, R =
× sin(θ)
∴ R = 25 × sin(130°) ≈ 19.15
<em>The vertical component of the vector, R</em><em> ≈ 19.15</em>
(The vector, R = Rₓ + R
= Rₓ·i + R
·j
∴ ≈ -16.07·i + 19.15j)