Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π 
f= 1/2π 
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
Answer:
1.566 x 10^2
Move the decimal to where the number being multiplied by 10^x is greater than 1 but less than 10. Then multiply it by 10^x
X is the number of times you moved the decimal, so in this case it would be 10^2
A hypothesis is an educated prediction that can be tested.
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
