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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.

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A person is sitting with one leg outstretched and stationary, so that it makes an angle of θ = 27.5° with the horizontal, as the

dangina [55]

here we will use the torque balance about the knee joint

here we can say that

$\tau_g = \tau_m$

here torque due to weight is given as

$\tau_g = 40.1 cos\theta*(l_1 + l_2)$

$\tau_g = 40.1 cos27.5*(0.105 + 0.150)$

$\tau_g = 9.07 Nm$

now torque due to applied force of muscle

$\tau_m = M*sin\alpha * l_1$

$\tau_m = M*sin30* 0.105$

now by torque balance we will have

$9.07 = M*0.5*0.105$

$M = 173 N$

so here the magnitude of m will be 173 N

7 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

4 0

3 years ago

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Brut [27]

Answer:

The gravity does a work of - 117.6 Joules.

The tension does not do work as the force is perpendicular to the direction of motion at any point in the trajectory.

Explanation:

The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:

$W_g = - \Delta E_p = - (mgh_f - m g h_i)$

where m is the mass of the ball, g is the acceleration due to gravity, $h_f$ is the final height and $h_i$ is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

$W_g = - mg (h_f - h_i)$

$W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m$

$W_g = - 117.6 Joules$

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

$dW = \vec{F} d\vec{r} = 0$

6 0

3 years ago

Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on

rosijanka [135]

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

$s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)$

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

$T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}$

s₁ = 1.1 m

Hence, the correct option is:

4 0

3 years ago

A body accelerates by 25m/s 2 when it applied by 40n forces.what would be acceleration if it is applied by 80n forces

il63 [147K]

Answer:

50m/s²

Explanation:

cross muliplyy write down the values acceleration and force

6 0

2 years ago

Read 2 more answers