let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer:
F n = 0.2 N
Explanation:
given,
you are exerting force of 10 N on the ball.
mass of the ball = 1 kg
acceleration due to gravity = 9.8 m/s²
normal force on the ball = ?
normal force is force exerted by the object to counteract the force from other object.
normal force acting on the ball will be
F n = F - mg
F n = 10 - 1 × 9.8
F n = 10 -9.8
F n = 0.2 N
Hence, normal force acting on the ball is equal to 0.2 N
Answer:
Explanation:
A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.
In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.
When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.
