Why do cells undergo mitosis?

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter st

devlian [24]

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

4 0

3 years ago

What is not a factor affecting impact in a collision; vehicle weight, speed, kinetic energy, centrifugal force

navik [9.2K]

I believe the answer is vehicle weight

6 0

3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$

So, the y-component of the resultant force is

$F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N$

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}$

Where in this problem, we have:

$F_x=-7.14 N$ is the x-component

$F_y=2.70 N$ is the y-component

And substituting, we find:

$F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N$

6 0

3 years ago

Read 2 more answers

Un'automobile di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

sesenic [268]

Answer:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

Explanation:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

3 0

2 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (the train must see the car advancing at a lower speed), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (relative to the ground):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, we have to do all the same but using $v_{AT}=v_A+v_T$ (the train must see the car advancing at a faster speed), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago