Answer: 0
Explanation: Initial velocity is 0.
Answer: 1.04N
Explanation:
Given
q1 = 2*10^-6C
q2 = 3.6*10^-6C
r = 0.25m
k = 9*10^9
Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."
F =(kq1q2) / r²
F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²
F = 0.0648/0.0625
F = 1.04N
The type of electrostatic force between the charges is the repulsive force
NH4OH is the answer. Hope this helps you.
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
Explanation:
It is given that,
Weight of the person on Earth, W = 818 N
Weight of a person is given by the following formula as :
g is the acceleration due to gravity on earth
m = 83.46 kg
The mass of an object is same everywhere. It does not depend on the location.
Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N
g' is the acceleration due to gravity on that planet. So,
So, the acceleration due to gravity on that planet is 
. Hence, this is the required solution.
