A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t
1 answer:
Mass of the block = 1.4 kg
Weight of the block = mg = 1.4 × 9.8 = 13.72 N
Normal force from the surface (N) = 13.72 N
Acceleration = 1.25 m/s^2
Let the coefficient of kinetic friction be μ
Friction force = μN
F(net) = ma
μmg = ma
μg = a
μ =
μ =
μ = 0.1275
Hence, the coefficient of kinetic friction is: μ = 0.1275
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Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ =
So,
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
i. 0.34
ii. 0.4
iii. 1700 w/m²
iv. 2211.36 w/m²
Explanation:
Given that
Irradiation of the plate, G = 2500 w/m²
Reflected rays, p = 500 w/m²
Emissive power, E = 1200 w/m²
See attachment for calculations
