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Answer:
1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>
2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>
3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>
4. The magnitude of the force on the object is a maximum on <em>the Top.</em>
Explanation:
<em>1. Because the change in position delta X is zero.</em>
<em>2. Because of delta X.</em>
<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>
<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>
Answer:
B = 62.9 N
Explanation:
This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid
B = ρ g V
write the equilibrium equation
T + B -W = 0
B = W- T (1)
use the density to write the weight
ρ = m / V
m = ρ V
W = ρ g V
substitute in 1
B = m g -T
B = 
g V - T
To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³
calculate
B = 7850 9.8 1.20 10⁻³ - 29.4
B = 92.3 - 29.4
B = 62.9 N
<span>Two plastic balls suspended by strings are placed close to each other. If they have the same charge then they will repel each other.</span>
Answer:
d = 10.076 m
Explanation:
We need to obtain the velocity of the ball in the y direction
Vy = 24.5m/s * sin(35) = 14.053 m/s
To obtain the distance, we use the formula
vf^2 = v0^2 -2*g*d
but vf = 0
d = -vo^2/2g
d = (14.053)^2/2*(9.8) = 10.076 m
For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m
400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.
