Answer: 6.36
Explanation:
Given
Radius of grindstone, r = 4 m
Initial angular speed of grindstone, w(i) = 8 rad/s
Final angular speed of the grindstone, w(f) = 12 rad/s
Time used, t = 4 s
Angular acceleration of the grinder,
α = Δw / t
α = w(f) - w(i) / t
α = (12 - 8) / 4
α = 4/4 = 1 rad/s²
Number of complete revolution in 4s =
Δθ = w(i).t + 1/2.α.t²
Δθ = 8 * 4 + 1/2 * 1 * 4²
Δθ = 32 + 1/2 * 16
Δθ = 32 + 8
Δθ = 40 rad/s
40 rad/s = 40/2π rpm = 6.36 rpm
Therefore, the grindstone does 6.36 revolutions during the 4 s interval
Explanation:
Only few supernova are observed in our galaxy -
Type II supernovae ( i.e. the explosions of the massive stars ) occurred in the Milky Way, and they might be hidden by the intervening dust if they are located in the more distant parts of our Galaxy .
Type Ia supernovae , which need a white dwarf star in the binary star system , are brighter than the type II supernovae , but some of them could also happen in the older parts of Galaxy which are hidden due to the buildup of the dust and gas .
<span>The atom becomes positively charged.
When you add electrons to a neutral atom, it is no longer a neutral atom, it has a negative change and is an anion. When you take away electrons from a neutral atom, it is no longer a neutral atom- it becomes a positive atom, and is a cation.</span>
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
The acceleration is 3 m/s per minute, or 0.05 m/s per second.
