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Answer:
block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s
Explanation:
We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.
Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)
Before the crash
p₀ = m v₀ + 0
After the crash
= (m + M) v
p₀ =
m v₀ = (m + M) v (1)
Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring
Initial
Em₀ = K = ½ m v2
Final
E = Ke = ½ k x2
Emo = E
½ m v² = ½ k x²
v² = k/m x²
Let's look for the spring constant (k), with Hook's law
F = -k x
k = -F / x
k = - 0.75 / -0.25
k = 3 N / m
Let's calculate the speed
v = √(k/m) x
v = √ (3/8.00) 0.15
v = 0.09186 = 9.18 10⁻² m/s
This is the spped of the block plus bullet rsystem right after the crash
We substitute calculate in equation (1)
m v₀ = (m + M) v
v₀ = v (m + M) / m
v₀ = 0.09186 (0.008 + 0.992) /0.008
v₀ = 11.5 m / s
Answer:
<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>
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Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia =
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
therefore,
for the first case, the K.E. is given as
and for the second case, the K.E. is given as
<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>
<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>
Answer:
Explanation:
In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:
In this case:
Therefore, using the previous equation and the data provided, let's solve for :
Hence, the number of loops in the secondary is approximately 41667.