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:
Try B or C if I'm wrong sorry
Explanation:
Answer:
Oxygen cycle
Explanation:
The components of the reservoirs of oxygen that are exchange in our environment is the oxygen cycle
It suggests the movement of oxygen between the living and non-living parts.
- The cycle does not account for oxygen that is trapped and cannot be exchanged in nature.
- Oxygen is important component of the atmosphere.
- Gaseous exchange between living organisms and atmosphere involves oxygen to a very large extent.
Answer:
537 N
Explanation:
The force due to gravity of a planet is:
F = GMm / r²
where G is the universal gravitational constant
M is the mass of the planet
m is the mass of the object
and r is the distance between the object and the center of the planet
On Earth, you weigh 716 N, so:
716 N = GMm / r²
On planet X:
F = G (3M) m / (2r)²
F = 3/4 GMm / r²
F = 3/4 (716 N)
F = 537 N
The reasoning for this is false
