Answer:
The speed of the block is 8.2 m/s
Explanation:
Given;
mass of block, m = 2.1 kg
height above the top of the spring, h = 5.5 m
First, we determine the spring constant based on the principle of conservation of potential energy
¹/₂Kx² = mg(h +x)
¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)
0.03125K = 118.335
K = 118.335 / 0.03125
K = 3786.72 N/m
Total energy stored in the block at rest is only potential energy given as:
E = U = mgh
U = 2.1 x 9.8 x 5.5 = 113.19 J
Work done in compressing the spring to 15.0 cm:
W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J
This is equal to elastic potential energy stored in the spring,
Then, kinetic energy of the spring is given as:
K.E = E - W
K.E = 113.19 J - 42.6 J
K.E = 70.59 J
To determine the speed of the block due to this energy:
KE = ¹/₂mv²
70.59 = ¹/₂ x 2.1 x v²
70.59 = 1.05v²
v² = 70.59 / 1.05
v² = 67.229
v = √67.229
v = 8.2 m/s
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Answer:
n = 1,875
Explanation:
The speed of light in vacuum is constant (c) and in a material medium it is
v = d / t
The refractive index of a material is defined by
n = c / v
Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised
These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,
v = d / t
v = 1 / 6.25 10⁻⁹
v = 1.6 10⁸ m / s
we calculate the refractive index
n = 3 10⁸ / 1.6 10⁸
n = 1,875
<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>
