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4 years ago

Whar are the phenotypes for FF Ff ff

Physics

2 answers:

kondor19780726 [428]4 years ago

7 0

Answer:

Explanation:

brainly.com/question/14753236

Bezzdna [24]4 years ago

5 0

Answer:

F, F, f (if I'm understanding the question correctly)

Explanation:

Phenotypes are the physical trait shown. In FF, Ff, ff a capital letter means that the gene is dominant and therefore always shows when paired with either another of itself or a recessive (lowercase). So, for FF, you see F as the phenotype shown, and for Ff, you see F as the phenotype because F is dominant over the recessive f. In ff, however, since you have two recessives, only then can you see f as the phenotype because you have no dominant traits.

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---------is a gas law that describes the direct relationship between the volumes of a gas and the number of gas particles it con

Ad libitum [116K]

Answer:

The answer is Avogadro's Law.

Explanation:

Avogardo's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." The derivation of Avogadro's law follows directly from the ideal gas law is:

$P*V=n*R*T$

where ;

V is the volume of the gas;

n is the amount of substance of the gas (measured in moles);

k is a constant for a given temperature and pressure

P is pressure of the environment

T is temperature of the environment

3 0

3 years ago

The strings in a compound bow behave approximately like a

Pepsi [2]

The speed at which the arrow would be launched is 133.42 m/s

The work-energy theorem asserts that the net work done applied by the forces on a particular object is equivalent to the change in its kinetic energy.

The equation for the work-energy theorem can be computed as:

$\mathbf{W =\Delta K.E}$

$\mathbf{F\Delta x =\dfrac{1}{2} mv^2}$

where;

Force (F) = 267 N
distance Δx = 0.60 m
mass (m) = 18 g
speed (v) = ???

From the above equation, let make speed(v) the subject of the formula:

∴

$\mathbf{v = \sqrt{\dfrac{2(F \Delta x)}{m}} }$

$\mathbf{v = \sqrt{\dfrac{2(267 \times 0.60)}{0.018}} }$

v = 133.42 m/s

Learn more about the work-energy theorem here:

brainly.com/question/17081653

7 0

2 years ago

A concrete block is hung from an ideal spring that has a force constant of 100 N/m . The spring stretches 0.129 m .A- What is th

madam [21]

Answer:

1.31498 kg

0.72050 s

Explanation:

m = Mass of block

g = Acceleration due to gravity = 9.81 m/s²

k = Spring constant = 100 N/M

x = Displacement = 0.129 m

The force balance is

$mg=kx\\\Rightarrow m=\dfrac{kx}{g}\\\Rightarrow m=\dfrac{100\times 0.129}{9.81}\\\Rightarrow m=1.31498\ kg$

The mass of the block is 1.31498 kg

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{1.31498}{100}}\\\Rightarrow T=0.72050\ s$

The period of oscillations is 0.72050 s

The time period does not depend on the acceleration due to gravity. It varies with the mass and the spring constant.

Hence, the time period would be the same

8 0

3 years ago

Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)

stepladder [879]

Answer:

$m(t)=20e^{-0.5t}$

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope $(t_{1/2})$ = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

$m(t)=m_0e^{-kt}$

Where,

$m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year$

So, the equation is: $m(t)=20e^{-kt}$

At half-life, the mass is reduced to half of the initial value.

So, at $t=t_{1/2},m(t)=\frac{m_0}{2}$ . Plug in these values and solve for 'k'. This gives,

$\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5$

Hence, the equation for the mass remaining is given as:

$m(t)=20e^{-0.5t}$

8 0

3 years ago

If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?

joja [24]

Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: $g=-9.81~m/s^2$ , and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
$v_f^2 = v_i^2 + 2gS$
where $v_i=4~m/s$ is the initial vertical velocity of the athlete, $v_f=0$ is the vertical velocity of the athlete at the maximum height (and $v_f=0~m/s$ at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
$S= \frac{v_f^2-v_i^2}{2g}=0.82~m$

3 0

3 years ago