Answer: I would choose options C and D
Explanation:
The answer would be 60 cuz a watetfall is a waterfall
Answer:
The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
Explanation:
Given;
distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m
current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively
The magnitude of the magnetic field halfway between the wires can be calculated as;
where;
B is magnitude of the magnetic field halfway between the wires
I₁ is current in the first wire
I₂ is current the second wire
μ₀ is permeability of free space
r is distance half way between the wires
Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
230 Newton
Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
F = k(q1 q2)/ r^2
Distance between protons = d = 10⁻¹⁵ m
charge of proton = q = 1.6 × 10⁻¹⁹ C
Here q1=q2
electric force = F =230N
Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.
2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.
Find more about electric force of repulsion between nuclear protons
brainly.com/question/8404637
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Answer:
Explanation:
"the velocity of the block is constant, so the net force acting on the block must be zero. Thus the normal force equals the weight and the force of friction equals the applied force" The only thing which is wrongly stated here is that the normal force equals the weight. Normal force is not always equal to weight. If external force is applied at some angle with the horizon , normal force may be more or less than weight. Even in case of an object placed on an inclined plane , weight is not equal to normal force .
All other parts of the statement are correct.
