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2 years ago

HELP ME PLSSS What are the possible genotypes

Physics

2 answers:

riadik2000 [5.3K]2 years ago

7 0

Answer:

the answer is b i think

seraphim [82]2 years ago

7 0

Answer:Within an individual organism, the specific combination of alleles for a gene is known as the genotype of the organism, and (as mentioned above) the physical trait associated with that genotype is called the phenotype of the organism.

Explanation:

The phenotype is the trait those genes express. ... Looking at the possible offspring, each box (or possible offspring) has two copies of the dominant gene. This means there is a 100% chance of the offspring having brown eyes, or being BB. It's important to note here that each box represents a possible offspring.

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If you use the same force to push a motorcycle as you would push a bike which one would have more acceleration and why explain u

zaharov [31]

Answer:

The bike would have more acceleration

Explanation:

Accourding to newtons first law a force is equal to its mass multiplied by its acceleration (f=ma) therefore an object with a higher mass compared to an object with a lower mass would experience less acceleration.

Eg.

F=50N

Motorbike M=200kg

F=ma

50=200 x a

50/200=a

0.25m/s/s =a

Bike M=35kg

F=ma

50=35 x a

50/35= a

1.43m/s/s=a

6 0

2 years ago

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A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some sola

irakobra [83]

C I’m pretty sure!!!!!

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2 years ago

What flows in electricity? A. Electrons B. Protons C. Neutrons

nordsb [41]

Answer:

Explanation:

Electrons

6 0

2 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

Given:

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

Assume:

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago

Un coche inicia un viaje de 450 km a las ocho de la mañana con una velocidad media de 90 km/h. ¿A qué hora llegará a su destino?

Artyom0805 [142]

Answer:

Llegara a su destino a la 1:00 pm

Explanation:

Si el coche va a 90 km/h buscamos un numero q al multiplicarlo por 90 nos de 450. Entonces 90×5 = 450, si hacemos la cuenta desde las ocho de la mañana mas las 5 horas del viaje terminaria llegando a su destino a la 1:00 pm.

5 0

2 years ago