How do you write a letter to the editor?
Open the letter with a simple salutation. ...
Grab the reader's attention. ...
Explain what the letter is about at the start. ...
Explain why the issue is important. ...
Give evidence for any praise or criticism. ...
State your opinion about what should be done. ...
Keep it brief. ...
Sign the letter.
First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:
2as = v² - u²
Because the final velocity v is 0 in such cases
s = -u²/2a; because both u and a are downwards, the negative sign cancels
s = 14.5² / 2*9.81
s = 10.72 meters
Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m
We will use the formula
s = ut + 0.5at²
to find the time taken with the initial velocity u = 0.
55.72 = 0.5 * 9.81 * t²
t = 3.37 seconds
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
Answer:
E. all of these
Explanation:
The designation of a point in space all the points that necessary
- reference point
- a direction
- fundamental units
- a direction
- motion
all are necessary to designate a point in space. Hence option E is correct.
For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.
