ANSWER:
What effect does a catalyst have on a system in equilibrium?
The system is unaffected.
~batmans wife dun dun dun....
Answer:
d_2 = 4d_1
Explanation:
The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by
R = U²sin2θ/g
Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be
d_1 = U²sin2θ/g
Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be
d_2 = V²sin2θ/g
= (2U)²sin2θ/g
= 4U²sin2θ/g
= 4d_1 (since d_1 = U²sin2θ/g)
So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.
A ball kept on 3rd floor of a building.
A pendulum bob kept at 3m height
A stone thrown vertically upward.
A pressed spring.
A squashed spunge ball.
Answer:
<em>The period of the motion will still be equal to T.</em>
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Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as

where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>