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
B. the distance the star is from Earth
Explanation:
The apparent magnitude of star is a function of its distance from the earth. It is one of the physical properties that is used to study a star.
The apparent magnitude of a star or other astronomical bodies is a measure of their brightness as seen from a location on the earth.
The apparent magnitude of a star depends on:
- Distance of the star from the location on earth.
- luminosity of the star
- the particles along the part of the star and earth that cuts off the light the earth receives.
learn more:
Star luminosity brainly.com/question/9084808
#learnwithBrainly
Answer:
a) 588,000 N
b) 294000 N
Explanation:
Given that
Density of water = 1000kg/m3
(g) = 9.8m/s2
volume is given as (V)= 5m*4m*3m
a) force will be equal to weight of water
b) at either end
[A = wh]
F = 294000 N
