(125 kg)*(9.81 m/s^2)*(10 m)
W = 12262.5 J
Answer:
<u>b. 2.27 N</u>
Explanation:
Fe = (k)(Q1)(Q2) / d²
k ≈ 9 × 10⁹ N m² / C² [Coulomb's Constant] or 8.98 × 10⁹ (nearest hundredth).
q = particle charge [Coulombs; C]
d = distance [Meters; m]
Fe = Electrostatic force [Newtons; N]
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Fe = (9×10⁹)(3.5 × 10^(-8)(-4.5 × 10^(-6))/(.25)² N =
(9)(3.5)(-4.5) × (10^(9))(10^(-8))(10^(-6) / (.25)² N =
(9)(3.5)(-4.5) × (10^(9 - 8 - 6)) / (.25)² N =
(-141.75) × 10^(-5) / (.0625) N =
141.75 × 10^(-5) / (.0625) N
[opposite direction] =
(1.4175 × 10^(2)) × 10^(-5) / (.0625) N
[opposite direction] =
1.4175 × 10^(-3) / .0625 N
[opposite direction] =
1.4175 × 10^(-3) / 6.25 × 10^(-2) N =
14.175 / 6.25 N [opposite direction] =
2.268 N ≈ 2.27 N [opposite direction].
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy







= 14.68m/s
Answer:
2.1 cm
Explanation:
The general equation for the graph is:
y = A cos(ωt + ∅)
A: amplitude, is the maximum and minimum value for y(t)