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3 years ago

How are metals identified i the periodic table??

Physics

2 answers:

zaharov [31]3 years ago

8 0

Answer:

okay here is a thing I learned when I was younger in my middle school:

Explanation:

my teacher would tell me that metals are considered a weak metals are on the left side and the good metals are located on the right side because the only way I remembered was the right means it is really strong and the left is weak and not that supportive. but I think that's how I still think it is or other people may have their own opinions. but hope this helped out with your question!

Svetllana [295]3 years ago

4 0

Answer:

Metals are at the bottom left of the Periodic Table and non-metals are at the top right. IDK if you were asking for this

Explanation:

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Technician A says that you can usually depressurize a brake accumulator by turning the ignition switch ""off"" to disable the el

Rzqust [24]

Answer:

Both Technician A and Technician B

Explanation:

In order to gain a better understanding of the solution above let define some terms

Break Accumulator

We can define a break accumulator as storage that that helps generate the required pressure in order for the breaking system to respond faster this accumulator is charged by turning the steering wheel slowly at once from lock to lock now this build the pressure in the accumulator and one way to depressurize is it is by turning the ignition switch ""off""

Now a scan tool is a device that can interface with a car it can also be used to diagnose a car an get the diagnostic information to help in the cars diagnoses and also be used to reprogram a car

3 0

2 years ago

Which of the following describes a role of gravity in the formation of our solar system?

Black_prince [1.1K]

I think the answer is A because it is gravity that caused the planets to form and start orbiting around the nearest star which is the sun

5 0

3 years ago

Read 2 more answers

How do gamma rays use electromagnetic waves to function....PLS HELP

BabaBlast [244]

Answer:

Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions.

Explanation:

Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

Hope this helps!

Brain-LIst?

4 0

3 years ago

a car company wants to ensure its newest model can stop in 50 m when traveling at 30 m/s (which is about 108 km/h). if we assume

Brut [27]

A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.

<h3>How to calculate deceleration ?</h3>

While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.

We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.

The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.

velocity of car = 30 m/s

car need to stop in 50m

Deceleration a = v^2 – u^2 / 2s

= 0^2 - 50^2 / 2*30

= 11.56

Deceleration of the care = 11.56 ms−2

To learn more about deceleration refer :

brainly.com/question/75351

#SPJ4

8 0

8 months ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago