Answer:
Explanation:
Given
Wooden board is pivoted at center and
Older child of mass 
is sitting at a distance of L from center
if two child of mass 
is sitting at a distance 
and 
(say) from pivot then net torque about pivot is zero
i.e.
as 
Therefore
Therefore another child is sitting at a distance of
Answer:
1.31×10¯⁶ N
Explanation:
From the question given above, the following data were obtained:
Mass of John (M₁) = 81 Kg
Mass of Mike (M₂) = 93 Kg
Distance apart (r) = 0.620 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The gravitational force between the two students, John and Mike, can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 81 × 93 / 0.62²
F = 6.67×10¯¹¹ × 7533 / 0.3844
F = 1.31×10¯⁶ N
Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N
Answer:
a = 0.009 J
b = 0.19 m/s
c = 0.005 J and 0.004 J
Explanation:
Given that
Mass of the object, m = 0.5 kg
Spring constant of the spring, k = 20 N/m
Amplitude of the motion, A = 3 cm = 0.03 m
Displacement of the system, x = 2 cm = 0.02 m
a
Total energy of the system, E =
E = 1/2 * k * A²
E = 1/2 * 20 * 0.03²
E = 10 * 0.0009
E = 0.009 J
b
E = 1/2 * k * A² = 1/2 * m * v(max)²
1/2 * m * v(max)² = 0.009
1/2 * 0.5 * v(max)² = 0.009
v(max)² = 0.009 * 2/0.5
v(max)² = 0.018 / 0.5
v(max)² = 0.036
v(max) = √0.036
v(max) = 0.19 m/s
c
V = ±√[(k/m) * (A² - x²)]
V = ±√[(20/0.5) * (0.03² - 0.02²)]
V = ±√(40 * 0.0005)
V = ±√0.02
V = ±0.141 m/s
Kinetic Energy, K = 1/2 * m * v²
K = 1/2 * 0.5 * 0.141²
K = 1/4 * 0.02
K = 0.005 J
Potential Energy, P = 1/2 * k * x²
P = 1/2 * 20 * 0.02²
P = 10 * 0.0004
P = 0.004 J
F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N
Well depending on the speed of both of those things is were the rock will be placed and it also determines how fast can an environment change
<span />
