Deadpool is doing a superhero landing from a 21 meter tall building what would his velocity be right before he hits the ground?
1 answer:
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Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Explanation:
First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C
a) I(t) = -ωqosin(ωt+φ)
⇒Io= ωqo
⇒ω= Io/qo
also we know that T= 2π/ω
⇒T=
now putting the values we get
=
= 1.71×10^{-3} sec
b) note that the time it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Answer:
the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Explanation:
This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Answer:
a. speed, v = 0.97 c
b. time, t' = 20.56 years
Given:
t' = 5 years
distance of the planet from the earth, d = 10 light years = 10 c
Solution:
(a) Distance travelled in a round trip, d' = 2d = 20 c = L'
Now, using Length contraction formula of relativity theory:
(1)
time taken = 5 years
We know that :
time =
5 = (2)
Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):
Squaring both the sides and Solving above eqution, we get:
v = 0.97 c
(b) Time observed from Earth:
Using time dilation:
Solving the above eqn:
t'' = 20.56 years