It refers to a base substitution<span> when the change in nucleotide changes the amino acid coded for by the </span>affected<span> codon</span>
The kinetic energy and gravitational potential energy changes during its movement from ground to the top height.
<h3>What happens to kinetic and potential energy while motion?</h3>
When the ball moves upward, its gravitational potential energy is increases and kinetic energy begins to decrease but when the ball falls towards the earth, its gravitational potential energy is transformed into kinetic energy. When the ball collides with the ground, the kinetic energy is transformed into other forms of energy.
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Answer:
Mechanical advantage = 2.875
Explanation:
Given:
A diagram is shown below for the above scenario.
Length of ramp (Effort arm) = 4.6 m
Height of truck bed ( Resistance length) = 1.6 m
Mechanical advantage (MA) is the ratio of effort arm and resistance length.
So, mechanical advantage is given as,
![MA=\frac{\textrm{Effort arm}}{\textrm{Resistance length}}= \frac{4.6}{1.6}=2.875](https://tex.z-dn.net/?f=MA%3D%5Cfrac%7B%5Ctextrm%7BEffort%20arm%7D%7D%7B%5Ctextrm%7BResistance%20length%7D%7D%3D%20%5Cfrac%7B4.6%7D%7B1.6%7D%3D2.875)
Answer: 22.6 hours
Explanation:
The power is the measure of the rate of energy.
In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as
![P=IV](https://tex.z-dn.net/?f=P%3DIV)
where
I is the current
V = 12.0 V is the voltage of the battery
So the energy delivered by the battery can be written as
![E=Pt=VIt](https://tex.z-dn.net/?f=E%3DPt%3DVIt)
Where
![It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s](https://tex.z-dn.net/?f=It%3D51.0%20A%5Ccdot%20h%20%3D%2051.0%20A%20%5Ccdot%203600%20s%2Fh%3D183%2C600%20A%5Ccdot%20s)
So the energy delivered is
![E=(12.0)(183,600)=2.2\cdot 10^6 J](https://tex.z-dn.net/?f=E%3D%2812.0%29%28183%2C600%29%3D2.2%5Ccdot%2010%5E6%20J)
At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:
![t=\frac{2.2\cdot 10^6 J}{27.0 W}=81481 s = 22.6 h](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B2.2%5Ccdot%2010%5E6%20J%7D%7B27.0%20W%7D%3D81481%20s%20%3D%2022.6%20h)
The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
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