The distance between the charges is 13.86 X 10⁴m
<u>Explanation:</u>
Given:
Force, F = 1.2N
Charge, q₁ = 1.602 X 10⁻¹⁹ C
k = 8.987 X 10⁹ Nm²/C²
Distance, d = ?
According to Coulomb's law:
Substituting the value in the formula we get:
Therefore, the distance between the charges is 13.86 X 10⁴m
Answer:
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Answer:
θ = 29.38°
Explanation:
The centripetal force is given by the formula;
F_c = F_n(sin θ) = mv²/r
Now, the vertical component of the normal force is; F_n(cos θ)
Now, this vertical component is also expressed as; F_n(cos θ) = mg
Thus, the slope is;
F_n(sin θ)/F_n(cos θ) = (mv²/r)/mg
tan θ = v²/rg
v² = rg(tan θ)
The initial speed will be gotten from the relation;
(v_o)² = μ_s(gr)
Plugging rg(tan θ) for (v_o)², we have;
μ_s(gr) = rg(tan θ)
rg will cancel out to give;
μ_s = (tan θ)
Thus, θ = tan^(-1) μ_s
μ_s is coefficient of static friction given as 0.563
θ = tan^(-1) 0.563
θ = 29.38°
Answer:
C 350W
Explanation:
Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s
m = 70kg and g =9.8m/s².
x = horizontal distance covered
Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.
E = mgh + 300t
Where t is the time taken to cover the distance
x = vt and h = 0.05vt
So
E = mg×0.05×vt + 300t
Substituting respective values
E = 70×9.8×0.05×1.4t +300t = 348t
P = E/t = 348W ≈ 350W.
Answer:
Explanation:
First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:
Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.
Since 
and 
, we can rewrite the first equation as:
Now, we solve for 
and calculate it:
This means that the crate's coefficient of kinetic friction on the floor is 0.18.
