Answer:
a)
125.6 rad/s
b)
25.12 rad/s²
Explanation:
a)
t = time required by the fan to get up to final operating speed = 5 sec
w = final operating rotational speed = 1200 rpm
we know that :
1 revolution = 2π rad
1 min = 60 sec
w = 
w = 
w = 125.6 rad/s
b)
w₀ = initial angular speed = 0 rad/s
α = angular acceleration
using the equation
w = w₀ + α t
125.6 = 0 + α (5)
α = 25.12 rad/s²
Answer:
P = 2 pi R / v period of space station
F / m = v^2 / R centripetal force per unit of mass
So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2
Also, F / m = 9.8 m/s^2 earth's gravitational attraction
So 9.8 = 4 pi^2 R / P^2 or R = 9.8 P^2 / 4 * pi^2) = 195 m
Or D = 2 R = 390 m the diameter required
Hope this shows! It has all the equations for all of the problems u asked in the comments
Answer: B = 1380T
Explanation: please find the attached file for the solution
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
