A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t
he center of mass of the bowling pin is moving upward with a speed vi of
10 The maximum height of the center of mass of the bowling pin is most nearly
10 The maximum height of the center of mass of the bowling pin is most nearly
1 answer:
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Complete question:
A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Answer:
The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Explanation:
Given;
charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C
length of the conductor, L = 50 m
inner radius, r₁ = 1.304 mm
outer radius, r₂ = 9.249 mm
The magnitude of the electric field halfway between the two cylindrical conductors is given by;
Where;
λ is linear charge density or charge per unit length
r is the distance halfway between the two cylindrical conductors
The magnitude of the electric field is now given as;
Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Answer:
Final speed of the train is 7.5 m/s
Explanation:
It is given that,
Uniform acceleration of the train is, a = 1.5 m/s²
It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :
Here, train starts from rest so, u = 0
v = 7.5 m/s
So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.