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Rudiy27
3 years ago
7

What is the relationship between dominant and recessive genes

Physics
1 answer:
Triss [41]3 years ago
7 0
Dominant genes and recessive genes are both given to a parents offspring. However, not both can be expressed causing the difference of dominant and recessive. Dominant genes are more likely to be expressed and recessive genes are more likely to be repressed.
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Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M
horsena [70]

Answer:

sin\theta - \mu_k cos\theta = \frac{m}{M}

sin\theta - \mu_k cos\theta = 1

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

F_f = \mu Mgcos\theta

now if it is moving down the inclined plane at constant speed

so we will have

Mgsin\theta - T - \mu mgcos\theta = 0

on other side the mass "m" will go up at constant speed

so we have

T - mg = 0

so we have

Mgsin\theta = \mu Mgcos\theta + mg

so we have

sin\theta - \mu_k cos\theta = \frac{m}{M}

for special case when m = M

then we have

sin\theta - \mu_k cos\theta = 1

5 0
3 years ago
The figure shows the original channel of a river, as well as its current channel. Which of the following forces MOST LIKELY chan
mash [69]

Answer: bbbb

Explanation:

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3 years ago
What to do if the patient stops breaching during Fainting ?​
oksian1 [2.3K]

Answer:

unloosen the tight attire parts,open all windows for better air circulation,if he/she does not react place your hands on the chest and press gently three times per interval while giving them air through the mouth CPR then when they react place them to lie horizontal face sideways

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What is the common size for air mass?
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Thecorrect answer is B. 

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3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

5 0
3 years ago
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