Exercise is the activity and fitness is a lifestyle and done with time
Answer: F = 2.1 x 10^-4N
Explanation: Question is incomplete.
The complete question is; A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth’s magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east.
Given parameters; l = 2.5m, I = 1.5A, B = 0.55 guass = 0.55 x 10^-4 Tesla , theta = 90 (from West to East), F = ?
F = BILsin(theta)
F = 0.55 x 10^-4 x 1.5 x 2.5 x sin 90
F = 2.1 x 10^-4 N.
According to right hand rule, it's direction is upward.
When the net forces equal 0 Newtons, they are a balanced force.
Answer:
The total normal force acting on the system is approximately 58.8 N
Explanation:
The masses arranged in the stack are;
3 kg, 2 kg, and 1 kg
The mass of the stack system, m = 3 kg + 2 kg + 1 kg = 6 kg
Weight = The force of gravity on an object = m·g
Where;
m = The mass of the object
g = The acceleration due to gravity ≈ 9.8 m/s²
∴ The weight of the stack system, W ≈ 6 kg × 9.8 m/s² ≈ 58.8 N
The direction of the weight force = Perpendicular to the surface (acting downwards)
From Newton's third law of motion, the normal force acts perpendicular to the plane and it is equal in magnitude to the force acting perpendicular to the plane
∴ The magnitude of the total normal force acting on the system = The magnitude of the weight of the system ≈ 58.8 N
The (magnitude of the) total normal force acting on the system ≈ 58.8 N