Assuming that acceleration due to gravity is g = 9.8 m/s^2.
mass of block = 5 kg
<span>μ = 0.2
</span>applied force = 9 N
The force of gravity, Fg, on this block is given by Fg = m * g --> F = 5 kg * 9.8m/s^2 = 49 N.
The normal force on this block exerted by the ground is Fn = Fg --> 49 N. (It points in direction opposite to Fg, but we don't need to worry about that here.)
The <em>maximal </em>force of friction on the block that resists its movement is Ff = Fn * <span>μ --> Ff = 49 N * 0.2 = 9.8 N
So, the max frictional force is 9.8 N.
Since the applied force is only 9 N, the block does not move.
Since we deduced that the block is not moving, then the block must be in equilibrium (all external forces must equal zero). So, the frictional force must match the applied force of 9 N.</span>
You measure the open distance between the floor and the bottom surface of the gas pedal. Then you press the gas pedal down 1/2 of that distance.
2) Higher frequency and higher pitch.
Due to the Doppler effect, sounds emitted at a distance have a higher frequency and therefore a higher pitch.
Complete Question
A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton
Answer:
The vector position is 
Explanation:
From the question we are told that
The position of the proton is
Generally the vector location of the proton is mathematically represented as

So substituting values

Answer:
the correct answue are B, A, C, C, B
Explanation:
1) The electric field is requested, let's approximate the membrane by a parallel plate with surface charge density
E =
E =
E = 5.65 10⁵ N / C
the correct answer is B
2) A calcium ion has two positive charges, so the force applied by each side of the membrane (plate)
F = q E
F = 2 1.6 10⁻¹⁹ 5.65 10⁵
F = 1.8 10⁻¹³ N
the total force is the sum of the force of each membrane and the two forces go to the same side
F = total = 2 F
F_total = 3.6 10⁻¹³ N
the correct answer is A
3) the field and the electric potential are related
ΔV = - E s
ΔV = - 5.65 10⁵ 10 10⁻⁹
ΔV = - 5.65 10⁻³ V
the correct answer is C
4) In the exercise they indicate that the outer wall has a positive charge, therefore, as they indicate that we approximate the system to a capacitor, the inner wall must be negatively charged.
The electric field goes from the positive to the negative charge, which is why it goes from the outer wall to the inner wall
the correct answer is C
5) For this part we use conservation of energy
starting point. On the inside wall, brown
Em₀ = U = qV
final point. On the outside
Em_f = K
energy is conserved
Em₀ = Em_f
q V = K
K = 3 10⁻¹⁵ 5.65 10⁻³
K = 1.7 10⁻¹⁷ J
the correct answer is B