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3 years ago

Which elements have similar behavior? barium silicon aluminum strontium osmium beryllium

Physics

2 answers:

tresset_1 [31]3 years ago

8 0

Which elements have similar behavior?
Barium, strontium, beryllium.

Ilya [14]3 years ago

3 0

<u>Answer:</u> The elements which have similar behavior are Barium, strontium and beryllium.

<u>Explanation:</u>

Periodic table is divided into 7 periods and 18 groups.

The elements belonging to the same group will show chemical properties. Groups are the vertical columns in the periodic table.

From the given options:

Barium, strontium and beryllium, all belong to the group 2.

Silicon belong to group 14.

Aluminium belong to group 13 and Osmium belong to group 8.

Hence, the elements which have similar behavior are Barium, strontium and beryllium.

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A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc

ki77a [65]

I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction. Say it's driving North.

a). From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b). From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c). A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south. BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ?

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself ! Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else. And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.

7 0

3 years ago

What is the mass of an object moving with 80N of force and an acceleration of 8 m/s2?

OverLord2011 [107]

Answer:

10 Kg

Explanation:

Force is equal to mass times acceleration

therefore mass is equal to force divided by acceleration

please mark me brainliest

6 0

3 years ago

7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del

GrogVix [38]

Answer:

x = 0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

sin 60 = $T_{y}$ / T

T_{y} = T sin 60

cos 60 = Tₓ / T

Tₓ = T cos 60

we apply the equation

∑ τ = 0

-W L / 2 - w x + T_{y} L = 0

the length of the bar is L = 6m

-Mg 6/2 - m g x + T sin 60 6 = 0

x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

x = (4676 - 5880) / 6860

x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0

3 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$