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3 years ago

In pea plants, tall height is dominant over short height. How can two tall plants result in a short offspring.

Engineering

1 answer:

MatroZZZ [7]3 years ago

3 0

I think if both the parents are heterozygous. Then if you do a punnet square the last box would be tt. There you go!

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La Patrulla Fronteriza de los Estados Unidos analiza la compra de un helicóptero nuevo para la vigilancia aérea de la frontera d

patriot [66]

Answer:

The Border Patrol of the United States analyzes the purchase of a new helicopter for the aerial surveillance of the border of New Mexico and Texas with the Mexican Republic. 4 years ago a similar helicopter was purchased at a cost of $ 140,000.00. with an interest rate of 7% per year. Calculate the single payment factor and the present value factor with the above data with the table and formula. Draw the flow chart.

Explanation:

7 0

3 years ago

A treatment plant being designed for Cynusoidal City requires an equalization basin to even out flow and BOD variations. The ave

sineoko [7]

Answer:

The size of equalization basin is 6105.6 m³

Explanation:

The average flow is:

flow = ∑flow/n = 9.788/24 = 0.408 m³/s

Where n is the number or observations.

The inflow volume is:

$V_{inflow} =Q*t$

where t is the time interval

$V_{inflow} =0.446(1*3600)=1605m^{3}$

in the same way it is calculated the inflow volume for each observation

The outflow volume is:

$V_{out} =Q_{out} *t=0.4*(1*3600)=1400m^{3}$

The volume of flow is:

$ds=V_{inflow} -V_{out} =1605-1400=205m^{3}$

in the same way it is calculated the volume of flow for each observation. According to the file attach, the highest volume is 6105.6 m³

Download pdf

6 0

3 years ago

While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic

makvit [3.9K]

Answer:

$Given data:\\While walking across campus one windy day\\Frontal area, $A=0.3 m ^{2}$\\Wind speed $V=24 Km / hr$\\The drag coefficient $C_{D, b}=1.2$\\The combined mass $m=75 kg$\\Umbrella diameter, $D=1.22 m$\\Velocity of wind $V=24 \frac{ km }{ hr }$\\The rolling resistance $C_{R}=0.75 \%$$

Solution:

Note: Refer the diagram

$Basic equation:\\'s law of motion: $\sum F_{x}=m a_{x}$\\Lift coefficient, $C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}$\\Drag coefficient, $C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}$$

$From force balance equation:\\$\sum F_{x}=F_{D}-F_{R}=0$\\But $F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}$$F_{R}=C_{R} m g$\\Area of the Umbrella $A_{u}=\frac{\pi D_{u}^{2}}{4}$$A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}$$A_{v}=1.17 m ^{2}$$

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), $C_{D, x}=1.42$

Substituting all parameters,

$\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}$

Then,

$\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}$

And the equation becomes,

$\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}$

Thus the floyds travels at $68.3^{\circ}$ wind speed.

7 0

4 years ago

If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp

victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0

4 years ago

Discuss the trends in reaction forces versus jet velocity. Is the trend consistent with the theory? Does it make sense?

Snowcat [4.5K]

Answer:

The rate of fluid motion(Jet Velocity) exert a force on the object in contact with it. This force is also knowns as reactions forces.

In theory, this is related to Newton Second of motion which States that:

The rate of change of momentum is directly proportional to impressed force.

This makes sense and it is consistent with theory. Detailed explanation below:

Explanation:

A jet which can be illustrated as a moving fluid, in natural or artificial systems, may exert forces on objects in contact with it.

To analyze fluid motion, a finite region of the fluid (control volume) is usually selected, and the gross effects of the flow, such as its force or torque on an object, is determined by calculating the net mass rate that flows into and out of the control volume.

These forces can be determined, as in solid mechanics, by the use of Newton’s second law, or by the momentum equation(Consistent with the theory). The force exerted by a jet of fluid on a flat or curve surface can be resolved by applying the momentum equation. The study of these forces is essential to the study of fluid mechanics and hydraulic machinery.

In practice, Engineers and designers use the momentum equation to accurately calculate the force that moving fluid may exert on a solid body. For example, in hydropower plants, turbines are utilized to generate electricity. Turbines rotate due to force exerted by one or more water jets that are directed tangentially onto the turbine’s vanes or buckets. The impact of the water on the vanes generates a torque on the wheel, causing it to rotate and to generate electricity.

3 0

4 years ago