<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem.
The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So:
0.30 * 2N = 0.6N
And if you look at your options, you'll see that option "B" matches exactly.</span>
The answer is true. I hope that this helps you out!!
F = q₁q₂C / r²
F force
q charge
C Coulomb constant
r separation between charges
If Juan used a Celsius thermometer, it would tell him the Celsius temperature.
If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.
Answer:
a= 0.22 m/s²
Explanation:
Given that
M = 3.5 kg
θ = 30°
m = 1 kg
μ= 0.3
The force due to gravity
F₁= M g sinθ
F₁=3.5 x 10 x sin 30
F₁= 17.5 N
F₂ = m g
F₂ = 1 x 10 = 10 N
The maximum value of the friction force on the incline plane
Fr = μ M g cosθ
Fr = 0.3 x 2.5 x 10 cos30°
Fr= 6.49 N
Lets take acceleration of the system is a m/s²
F₁ - F₂ - Fr = (M+m) a
17.5 - 10 - 6.49 = (3.5+1)a
a= 0.22 m/s²