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Answer:
0.63 s
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 50 g
Extention (e) = 10 cm
Period (T) =?
Next, we obtained 50 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
50 g = 50 g × 1 Kg / 1000 g
50 g = 0.05 kg
Next, we shall convert 10 cm to m. This is illustrated below:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass = 0.05 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 0.05 × 9.8
F = 0.49 N
Next, we shall determine the spring constant of the spring.
Extention (e) = 0.1 m
Force (F) = 0.49 N
Spring constant (K) =?
F = Ke
0.49 = K × 0.1
Divide both side by 0.1
K = 0.49 /0.1
K = 4.9 N/m
Finally, we shall determine the period as follow:
Mass = 0.05 Kg
Spring constant (K) = 4.9 N/m
Pi (π) = 3.14
Period (T) =?
T = 2π√(m/k)
T = 2 × 3.14 × √(0.05 / 4.9)
T = 6.28 × √(0.05 / 4.9)
T = 0.63 s
Thus, the period of oscillation is 0.63 s
Answer:
the cannonball’s velocity parallel to the ground is 86.6m/S
Explanation:
Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.
For this case we must draw the velocity triangle and find the component in X(see atached image).
V= Initial velocity=100M/S
V= Initial velocity=100M/S
Vx=cannonball’s velocity parallel to the ground
Solving for Vx
Vx=Vcos30
Vx=(100m/S)(cos30)=86.6m/s
the cannonball’s velocity parallel to the ground is 86.6m/S
Answer:
Explanation:
(a) When the plate starts to spin:
Its angular velocity increases, so the angular acceleration is non zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.
(b) When the plate rotates at constant angular velocity:
Its angular velocity is constant so the angular acceleration is zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.
(c) When the plate sows to halt:
Its angular velocity decreases, so the angular acceleration is non zero( but negative). As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.
