Answer:
what is the question? there isn't one
Answer: -25.4
Explanation:
Acellus don’t forget the negative sign
Answer / Explanation
It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.
So for proper understanding, the two parts of the question will be highlighted.
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.
a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?
b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.
Solve for Vmax
Step Process
a) Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]
Where V₀ = 6m/s,
We have,
h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]
= 0.411 m
Where V₀ = 9.5m/s
We have,
h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]
= 1152 m
b) From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.
Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.
Therefore Vₙ = V₀max = 9.8 m/s
Answer:
Explanation:
Given the magnitude of the forces, 7N and 2N, the minimum combining force acting on the forces are when the force's is acting in opposite direction.
Magnitude of the force in opposite direction is 7N - 5N = 2N
The maximum combining force occurs when they act in the same direction. Magnitude of the force in the same direction is 5N+7N = 12N
Hence the range of magnitude requires is 2N≤F≤12N
Second hand:
1 rev per minute = (2π radians/minute) x (1 min/60sec) = π/30 rad/sec
Minute hand:
1 rev per hour = (2π radians/hour) x (1 hr/3600 sec) = π/1800 rad/sec
Hour hand:
1 rev per 12 hours = (2π rad/12 hr) x (1 hr/3600 sec) = π/21,600 rad/sec
As long as the clock is in good working order, and the hands are turning steadily at their normal rate, there is no angular acceleration.
