A bullet with a mass of 0.00658 kg is fired through a 1.95 kg block of wood on a frictionless surface. The initial speed of the
1 answer:
The final velocity of the block is 1.29 m/s
Explanation:
We can solve this problem by using the principle of conservation of momentum: in fact, the total momentum of the system must be conserved before and after the collision. Therefore, we can write:
where:
is the mass of the bullet
is the initial velocity of the bullet
is the final velocity of the bullet
is the mass of the block
is the initial velocity of the block
is the final velocity of the block
Re-arranging the equation and solving for v2, we find:
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Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
Answer:
f = q
Explanation:
In the attachment we can see a diagram of the parallel rays.
The dotted line represents the normal to the mirror surface
These rays when reflected using the constructor equation
where p and q are the distance to the object and the image respectively.
Since the rays are parallel P = inf
1 / f = 1 / inf + 1 / q
f = q
this means that all the rays focus on one focal point.
