The equation relevant to this is:
S(t) = So + Vot - At²/2 <span>
</span>
<span>Therefore
we can create two equations:
<span>S(t) = 90 = So - 4t - 16.1t² -->
eqtn 1</span>
<span>S(t+2) = 10 = So - 4(t+2) - 16.1(t+2)² --> eqtn 2</span>
</span>
<span>Expanding
eqtn 2:
10 = So - 4t - 8 - 16.1(t² + 4t + 4)
10 = So - 4t - 8 - 16.1t² - 64.4t - 64.4
10 + 8 + 64.4 = So - 68.4t - 16.1t²
<span>82.4 = So - 68.4t - 16.1t² -->
eqtn 3</span></span>
<span>
Subtracting eqtn 1 by eqtn 3:</span>
90 = So - 4t - 16.1t²
82.4 = So - 68.4t - 16.1t²
=> 7.6 = 64.4t
t = 0.118 s
Therefore calculating for initial height So:<span>
<span>82.4 = So - 68.4(0.118) - 16.1(0.118)²
<span>So = 90.7 ft</span></span></span>
Answer:
I think the answer is A
Explanation:
I need this brainliest answer please
Answer:
time period is increased so the clock will become SLOW
Explanation:
As we know that the time period of the simple pendulum is given by the formula
here we know that
L = distance of the pendulum bob from the hinge
g = acceleration due to gravity
now here the bob slide down so that the length of the pendulum is being increased
so time period T of the pendulum is also increased
so here the pendulum will take more time to oscillate or to complete one oscillation
so clock will become SLOW
Answer:
range of movement is 1.49 mm
Explanation:
given data
focal length = 45 mm
distance = 1.4 m
distance from the lens = 35 mm
distance from infinity down = 1.4 m =
to find out
range of movement
solution
we will apply here lens equation that is
1/f = 1/p + 1/q
here f = 45 and p = infinity to 1400 mm
we find here image distance that is q
1/45 = 0 + 1/q ......1
q = 45 mm
and
1/45 = 1/1400 + 1/q ......2
q = 46.49
so range of movement
that is 46.49 - 45
range of movement is 1.49 mm
D) How fast you are moving
