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wolverine [178]
2 years ago
7

On the planet Abby, there is an homozygous lethal condition known as Gumball. Individuals with this condition chew gum till thei

r jaws hurt. A study of 1000 heterozygote Abbers (Gg) shows that 600 show the Gumball phenotype. Homozygous wild-type (gg) Abbers never show the phenotype. If we cross of a wild-type gg male with an affected female, what is the probability that an arbitrarily selected F1 will be wild-type?
A. 0.6
B. 0.0
C. 1.00
D. 0.70
E. 0.3
Physics
1 answer:
Ipatiy [6.2K]2 years ago
6 0

Answer:

D. 0.7 would be wild-type healthy

E. 0.3 would be affected Gg

Explanation:

Using the Punnett square we cross an homozygous male (gg), that is always healthy vs an heterozygous female (Gg). The result is 2 gg individuals, and 2 Gg individuals. Both gg individuals are healthy.

In a population of 1000 Gg individuals, 600 show the Gumball phenotype (unhealthy). It means that from the two heterozygous Gg (from the Punnett square), 1.2 show the unhealthy phenotype.

Again, from the Punnett square we have 4 individuals, that represent the 100%. 1.2 are unhealthy Gg, and represent 30% in F1.

This 30%, is equivalent to 0.3, that is the option E.

100 - 30 % = 70 %.  This is 0.7 for healthy wild-type, option D.

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The amount of pressure required to move a 6800 lb force with a 6" d piston is ___ psi.
Katena32 [7]
The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19  sq inches
Thus, the pressure= 6800/21.19
                              = 320.91 PSI

7 0
2 years ago
A fan is to accelerate quiescent air to a velocity of 12.5 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
Reika [66]

Answer:

= 829.69 Watt

≅ 830 Watt

Explanation:

Given that,

Velocity of air flow = 12.5m/s

Rate of flow of air = 9m³/s

Density of air = 1.18kg/m³

power by kinetic energy = 1/2(mv²)

mass = density × volume

m = 1.18 × 9

  = 10.62 kg/s

power = 1/2 mV²

           = 1/2 (10.62 × 12.5²)

           = 829.69 Watt

           ≅ 830 Watt

Flow rate  

u

=

9

 

m

3

/

s

Velocity of the air  

V

=

8

 

m/s

Density of the air  

ρ

=

1.18

 

kg

/

m

3

5 0
2 years ago
100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak
Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

Q_m=AC_o\sqrt{2\rho g_cP_g}

Here, Q_m is the mass flow rate through the leak of the pipeline. A is the area of the hole, C_o is the discharge rate, \rho is the fluid density, g_c is the gravitational constant and P_g is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft

Calculate the area (A) of the hole. The area of the hole is.

A=\frac{\pi d^2}{4}

Substitute 3.14 for \pi and 8.33\times 10^{-3}ft for d and calculate A.

A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is 62.4\frac{Ib_m}{ft^3}. Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

P_g=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate P_g.

Upstream x distance from upstream pressure end

P_g=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig

Convert the pressure from psig to Ib_f/ft^2

P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute 5.45\times 10^{-5}ft^2 for A. 0.61 for C_o, 54.9\frac{Ib_m}{ft^3} for \rho, 32.17\frac{ft.Ib_m}{Ib_f.s^2} for g_c, and 6,379.2\frac{Ib_f}{ft^2} for P_g and calculate Q_m

Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}

The mass flow rate of benzene through the leak in the pipeline is 0.158\frac{Ib_m}{s}

8 0
2 years ago
The boom hoisting sheave must have pitch diameters of no less than _______times the nominal diameter of the rope used.
alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

7 0
1 year ago
A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t
choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25
8 0
2 years ago
Read 2 more answers
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