In genetic traits, p and q represent the relative probabilities of the two alleles manifesting. If these two are the only options (ex. a dominant one and a recessive one), then the probabilities of both must sum up to 1. In this case, since we are given that q = 0.4, then p + q = 1, p + 0.4 = 1, and p = 0.6.
Answer:
Explanation:
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Answer:
a. 0.18Hz
b. 0.56m/s
Explanation:
From the question we can deduct the following parameters
The wavelength, λ is define as the distance between two successful crest or trough and from the question we conclude that wavelength is 3.17m.
Also the period of the wave T can be computed as
T=22.6/4
T=5.65secs.
a. To compute the frequency, recall that frequency, F=1/period.
Hence,
F=1/5.65
F=0.18Hz
b. Next we compute the wave speed.
Wave speed=frequency *wavelength
Wave speed =0.18*3.17
Wave speed =0.56m/s
Answer:
3360 N
Explanation:
In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.
The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.
The torques are equal:
Fr = Fr
(1440 N) (3.5 m) = F (1.5 m)
F = 3360 N
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
