Answer:
PART A)
External force will be 75 N
PART B)
distance moved will be 1.125 m
Explanation:
PART A)
Given that net force on the mower is

now we also know that friction force due to ground is given as

now we have



so external force will be 75 N
PART B)
deceleration due to friction when external force is removed from it


now we can find the distance by kinematics



so the distance moved will be 1.125 m
Answer:

Explanation:
Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(
) and inversely proportional to the square of the distance(d) that separates them.
Replacing the given values, where k is the Coulomb constant:

No, because superconductivity cannot occur if there is resistance
In addition to explaining electrical resistance, equilibrium distance theory also foretells the existence of superconductivity. According to its postulates, electrical resistivity decreases with distance from the equilibrium. There is only superconductivity at zero distance, with no resistance
<h3>What is Superconductivity ?</h3>
The ability of some materials to transmit electric current with virtually little resistance is known as superconductivity.
- This ability has intriguing and maybe beneficial ramifications. Low temperatures are necessary for a material to exhibit superconductor behaviour. H. K. made the initial discovery of superconductivity in 1911.
- Aluminum, magnesium diboride, niobium, copper oxide, yttrium barium, and iron pnictides are a few well-known examples of superconductors.
Learn more about Superconductivity here:
brainly.com/question/17166152
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Answer:
(a) 42.28°
(b) 37.08°
Explanation:
From the principle of refraction of light, when light wave travels from one medium to another medium, we have:
= sinθ
/sinθ
In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.
For critical angle θ
= θ
, θ
= 90°; 
(a) 
= sinθ
/sin90°
0.6728 = sinθ![_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584](https://tex.z-dn.net/?f=_%7Bc%7D%3C%2Fp%3E%3Cp%3E%CE%B8%5Btex%5D_%7Bc%7D%20%3D%20sin%5E%28-1%29%200.6728%20%3D%2042.28%C2%B0%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%28b%29%20%5Btex%5Dn_%7Ba%7D%20%3D%201.6584)
= sinθ
/sin90°
0.60299 = sinθ[tex]_{c}
θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°