All categories

3 years ago

How could pig xenotransplantation affect humans?

Physics

1 answer:

Korolek [52]3 years ago

5 0

It could cause them to catch animal disease

Explanation:

Xenotransplantation refers to the transplantation of organs or tissues between different species. E.g. Chimpanzee organ (liver) being transplanted into a human body would come under the category of xenotransplantation.

Pigs have come up as prospective organ donors since their organs and tissues are more or less to that of human’s. However, there are issues such as-

Xenotransplantation of pig’s tissue into a human would allow easier passage for all types of zoonotic viruses.
Moreover, if transgenic pigs are used instead of natural pigs it can lead to virus mutation as per the human body demands thus leading to increased susceptibility to human infection.

Thus, engrafting of pig’s tissue in the human body poses a significant risk for humans of catching a zoonotic (animal) disease.

You might be interested in

A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl

frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ = 34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence, the work done by tension before the block goes up the incline = 292.97 J

8 0

2 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

8 0

3 years ago

Read 2 more answers

A total Δν of 15 km/s is required to achieve an interplanetary mission. The proposed rocket has two stages. The first stage alon

AlexFokin [52]

Answer:

102000 kg

Explanation:

Given:

A total Δν = 15 km/s

first stage mass = 1000 tonnes

specific impulse of liquid rocket = 300 s

Mass flow rate of liquid fuel = 1500 kg/s

specific impulse of solid fuel = 250 s

Mass flow of solid fuel = 200 kg/s

First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds

Now,

Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time

Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg

Also,

Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time

Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg

Therefore,

The total jettisoned mass flow of the fuel in first stage

= 90000 kg + 12000 kg

= 102000 kg

3 0

3 years ago

As shown in the diagram, threee equal charges are spaced evenly in a row. The magnitude of each charge is +2e, and the distance

Nataly [62]

Answer: 4.27 x 10^-10 N to the left

Explanation: I just took this quiz

4 0

3 years ago

Read 2 more answers

In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the

sp2606 [1]

<span>I'll tell you how to do it but you must crunch the numbers.

Use Kepler's 3rd Law

T^2 = k R^3

where k = 4(pi)^2/ GM
G =gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2
M = mass of this new planet
pi = 3.14159265
T =3.09 days = 266976 seconds
R = (579,000,000km)/9 = 64333333.3 km

a)
Solve Kepler's 3rd Law for M. Your answer will be in kg

b)
mass of the sun = 1.98892 × 10^30 kilograms
Form the ratio
M(planet)/M(sun) </span>

6 0

2 years ago