Answer:
A. when the mass has a displacement of zero
Explanation:
The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:
where
k is the spring constant
x is the displacement of the mass with respect to the equilibrium position of the spring
m is the mass
v is the velocity of the mass
Since the total energy E must remain constant, we can notice the following:
- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum
- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0
Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.
Answer:
58.5 m
Explanation:
First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.
The initial vertical velocity of the ball is
where
u = 21.5 m/s is the initial speed
is the angle
Substituting,
The vertical position of the ball at time t is given by
where
h = 13.5 m is the initial heigth
is the acceleration of gravity (negative sign because it points downward)
The ball reaches the water when y = 0, so
Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.
The horizontal velocity of the ball is
And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:
3.
a)
r = distance of each mass in each hand from center = 0.6 m
m = mass of each mass in each hand = 2 kg
v = linear speed = 1.1 m/s
L = combined angular momentum of the masses = ?
Combined angular momentum of the masses is given as
L = 2 m v r
L = 2 (2) (1.1) (0.6)
L = 2.64 kg m²/s
b)
v' = linear speed when she pulls her arms = ?
r' = distance of each mass from center after she pulls her arms = 0.15 m
Using conservation of momentum , angular momentum remains same, hence
L = 2 m v' r'
2.64 = 2 (2) (0.15) v'
v' = 4.4 m/s
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Q = magnitude of charge on each of the two point charge = 3.60 mC = 3.60 x 10⁻³ C
r = distance between the two point charges = 9.3 cm = 0.093 m
k = constant = 9 x 10⁹ Nm²/C²
F = magnitude of the force between the two point charges = ?
according to coulomb's law , force between two charges is given as
F = k Q²/r²
inserting the values
F = (9 x 10⁹) (3.60 x 10⁻³)²/(0.093)²
F = 1.35 x 10⁷ N
