A 7.5kg block is sliding down a wall with constant velocity. The coefficient of static friction between the block and the wall
1 answer:
Ok, a couple of things have to be accounted for here. First, since the block is moving relative to the wall we have to use the kinetic coefficient of friction, 0.40. The second consideration is that since the block is moving at a constant velocity, the acceleration is zero. This means, by Newton's second Law, that the net force is zero. So the force of gravity must be equal to the friction force of the wall resisting its fall. This friction force is the product of the normal force (which we are seeking) and the kinetic coefficient of friction. We can then set these two forces equal:
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Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor,
The static and dynamic friction coefficient is,
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,
So, the acceleration of the car is . Hence, this is the required solution.
Answer:
40.0⁰
Explanation:
The formula for calculating the magnetic flux is expressed as:
where:
is the magnetic flux
B is the magnetic field
A is the cross sectional area
is the angle that the normal to the plane of the loop make with the direction of the magnetic field.
Given
A = 0.250m²
B = 0.020T
= 3.83 × 10⁻³T· m²
3.83 × 10⁻³ = 0.020*0.250cosθ
3.83 × 10⁻³ = 0.005cosθ
cosθ = 0.00383/0.005
cosθ = 0.766
θ = cos⁻¹0.766
θ = 40.0⁰
<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>