closed and unsecure circuit
(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.
(b) The velocity of the rock after 2 seconds is 7.56 m/s.
(c) The time for the block to hit the surface is 4.03.
(d) The velocity of the block at the maximum height is 0.
<h3>
Velocity of the rock</h3>
The velocity of the rock is determined as shown below;
Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m
v² = u² - 2gh
where;
- g is acceleration due to gravity in mars = 3.72 m/s²
v² = (15)² - 2(3.72)(13.14)
v² = 127.23
v = √127.23
v = 11.28 m/s
<h3>Velocity of the rock when t = 2 second</h3>
v = dh/dt
v = 15 - 3.72t
v(2) = 15 - 3.72(2)
v(2) = 7.56 m/s
<h3>Time for the rock to reach maximum height</h3>
dh/dt = 0
15 - 3.72t = 0
t = 4.03 s
<h3>Velocity of the rock when it hits the surface</h3>
v = u - gt
v = 15 - 3.72(4.03)
v = 0
Learn more about velocity at maximum height here: brainly.com/question/14638187
Answer:
I took 3*sqrt(10/83)= 1.110349815
And rounded to 1.11 Hz
Explanation:
Answer:
It looks like its moving north.
Explanation:
Answer:
Explanation:
Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .
After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4
After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .
So,
1 / 8 the of initial light passes through the last filter .
