Answer:
v = 6295.55 m/s
Explanation:
Given that,
The radius of a planet, 
The free fall acceleration of the planet, a = 7.45 m/s²
We need to find the tangential speed of a person standing at the equator.
Also, the centripetal acceleration was equal to the gravitational acceleration at the equator.
We know that,
Centri[etal acceleration,

So, the tangential speed of the person is equal to 6295.55 m/s.
Answer:
165.8 V/m
Explanation:
The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)
Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)
V = Q/C
V = Q ÷ 2πε₀L/㏑(b/a)
V = Q㏑(b/a)/2πε₀L
So, the potential difference per unit length V' is
V' = V/L = Q㏑(b/a)/2πε₀
Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have
V' = Q㏑(b/a)/2πε₀
V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)
V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)
V' = 0.16575 × 10³ V/m
V' = 165.75 V/m
V' ≅ 165.8 V/m
All except C, feelings are not physical.
Answer:
The force exerted is 6189.4 N
Solution:
As per the question:
Area of the book, A = 
Mass of the book, m = 3 kg
Pressure of the book, P = 
Now,
Pressure, P = 
Force, F = PA = 
Force on mass of 3 kg, F' = mg = 
Now, the force at the bottom is given by:
F'' = F + F' = 6160 + 29.4 = 6189.4 N