A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her sw
1 answer:
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Answer:
2.63 Hz
Explanation:
m = mass of the object = 8.0 kg
x = stretch in the spring = 3.6 cm = 0.036 m
k = spring constant of the spring
using equilibrium of force
Spring force = weight of object
k x = m g
k (0.036) = (8) (9.8)
k = 2177.78 N/m
frequency of oscillation is given as
= 2.63 Hz
Answer:
U = 1794.005 × 10⁶ J
Explanation:
Data provided;
Capacitance of the original capacitor, C = 1.27 F
Potential difference applied to the original capacitor, V = 59.9 kV
= 59.9 × 10³ V
Now,
The Potential energy (U) for the capacitor is calculated as:
Potential energy of the original capacitor, U = × C × V²
on substituting the respective values, we get
U = × 1.27 × ( 59.9 × 10³ )²
or
U = 1794.005 × 10⁶ J
Explanation:
Given that,
Angular velocity = 0.240 rev/s
Angular acceleration = 0.917 rev/s²
Diameter = 0.720 m
(a). We need to calculate the angular velocity after time 0.203 s
Using equation of angular motion
Put the value in the equation
The angular velocity is 0.426 rev/s.
(b). We need to calculate the tangential speed of the blade
Using formula of tangential speed
Put the value into the formula
The tangential speed of the blade is 0.963 m/s.
(c). We need to calculate the magnitude at of the tangential acceleration
Using formula of tangential acceleration
Put the value into the formula
The tangential acceleration is 2.074 m/s².
Hence, This is required solution.