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3 years ago

What organs are being affected from liver cancer and how?

1 answer:

5 0

The liver, because its liver cancer.. lol
The liver filters your blood, without it, your blood will stay 'dirty' and cannot do its jobs like it usually should be

You might be interested in

Which would you expect to freeze faster pure water or salt solution why?

krek1111 [17]

Pure water.

A salt solution contains impurities whereas pure water will not contain any impurities.

Impurities increase the boiling point (freezing point) of a substance.

Thus, I would expect the pure water solution to freeze faster than the salt solution.

6 0

3 years ago

What is the magnitude and direction (right or left) of the

Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

$Fy_{net}=F_{n}+F_{g}$ (1)

Where $F_{n}=12 N$ is the Normal force, directed upwards and $F_{g}=-12 N$ is the weight of the box (gravity force), directed downwards.

$Fy_{net}=12 N-12 N$ (2)

$Fy_{net}=0 N$ (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

$Fx_{net}=F_{left}+F_{right}$ (4)

Where $F_{left}=-3 N$ and $F_{right}= 15 N$

$Fx_{net}=-3 N + 15 N$ (5)

$Fx_{net}=12 N$ (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0

3 years ago

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

7 0

2 years ago

Two people stand facing each other at roller skating rink then push off each other

9966 [12]

a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

The momentum of an object is a vector quantity given by:

$p=mv$

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

$p=p_1 + p_2$

Which can be rewritten as

$p=m_1 u_1 + m_2 u_2$

where $m_1,m_2$ are the masses of the two people and $u_1,u_2$ their initial velocities.

However, the two people are initially at rest, so

$u_1 = 0\\u_2 = 0$

Therefore the total momentum is

$p=0+0=0$

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

In this problem,

$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

where:

$m_1 = 30 kg$ is the mass of the girl

$v_1 = -5 m/s$ is her velocity (she moves backward, so the negative sign)

$m_2 = 50 kg$ is the mass of the boy

$v_2$ is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

$0=m_1 v_1 + m_2 v_2$

And solving for v2 we find the velocity of the boy:

$v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s$

and the positive sign means he is moving forward.

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

$\Delta p = F\Delta t$