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3 years ago

Which compound is an example of a binary ionic compound?

2 answers:

7 0

C, Since binary ionic compound is only of 2 elements. Mg is ionic and so i F so thats it. H20 is covalent B is more than 1 element, D looks dodgy, nah its dat ionic compunds wouldnt form big ones like SF (6)

pogonyaev3 years ago

4 0

The answer is C) MgF2

You might be interested in

Takumi works in his yard for 45 minutes each Saturday. He works in the morning, and he wears sunscreen and a hat each time he wo

MrRa [10]

Explanation :

Takumi wears sunscreen and a hat each time he works in the yard. This is to protect himself with the strong radiation coming from the sun. UV rays that are coming from the sun are the main cause of skin cancer.

Stochastic effects are the effects that are caused by chance. Cancer is one of the main stochastic effects.

So, the correct option is (b) "the severity of stochastic effects, such as cancer".

7 0

4 years ago

Read 2 more answers

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

☆ Correct and rewrite the following statements:

il63 [147K]

Answer:

<h3>a) </h3><h3>Question:-</h3>

Rubber is a good conductor of electricity.

<h3>Answer:-</h3>

It's incorrect sentence.
As we know that, rubber are an insulators of electricity, and they do not not have free ions which can help in conducting electricity.

So correct sentence is:

Rubber is a bad conductor of electricity.

<h3>b)</h3><h3>Question:-</h3>

Electroplating wastes are useful to human health and environment.

<h3>Answer:-</h3>

It's incorrect sentence. 
As we know that, electroplating of substances releases harmful and toxic substances into the environment causing serious health issues in human, so it's waste substances released into environment can have very harmful effects.

So correct sentence is:

Electroplating wastes are harmful to human health and environment.

<h3>c)</h3><h3>Question:-</h3>

In an LED bulb, the shorter lead is connected to the positive terminal of the battery.

<h3>Answer:-</h3>

It's incorrect sentence. 
As we know that, in an LED bulb their are two terminals that is the positive and the negative. So here the lead connected to the positive terminal is always slightly longer than the lead connected to negative terminal of the bulb.

So correct sentence is:

In an LED bulb, the longer lead is connected to the positive terminal of the battery.

<h3>d)</h3><h3>Question:-</h3>

Pure water conducts electricity.

<h3>Answer:-</h3>

It's incorrect sentence. 
As we know that, pure water is a form of H2O where Hydrogen (H) is positive ion and Oxygen (O) is a negative ion which can conduct electricity when they are in individuals but as in pure water they are in combined form of compound H2O in which they become neutral and can't conduct any electricity.

So correct sentence is:

Pure water can't conduct electricity.

3 0

2 years ago

One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist

Svetach [21]

<h3>Answer:</h3>

(a) i₁ = 0.03818 A = 38.18 mA

(b) downward

(c) i₂ = 0.01091 A = 10.91 mA

(d) rightward

(e) i₃ = 0.02727 A = 27.27 mA

(f) leftward

(g) Eₐ = 3.818 Volts

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0

E₂ - i₃R₃ - i₁R₁ = 0

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

i₃ = 0.02727 A

using this value in eqn of i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

i₂ = 0.01091 A

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

i₁ = 0.03818 A

(a)

i₁ = 0.03818 A = 38.18 mA

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will downward

(c)

i₂ = 0.01091 A = 10.91 mA

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will rightward

(e)

i₃ = 0.02727 A = 27.27 mA

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will leftward

(g)

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

Eₐ = 3.818 Volts

3 0

4 years ago

PLEASE HELPPP!!! I'LL GIVE BRAINLEST FOR CORRECT ANSWERS

babymother [125]

Answer:

Explanation:

Weight will be highest at the pole where there is no tangential velocity requiring centripetal acceleration. Also, due to the slight bulge of the earth at the equator, the distance from the surface to the center of mass of earth is slightly shorter there meaning gravity is slightly stronger. Fg = GmM/R²

Weight will be lowest at the equator where the object is moving about the earth axis at an angular velocity of 2π/(24(3600)) rad/s ( about 7.27e-5 rad/s)

This means that some of the (already weaker, see above) gravity there is required to supply the needed centripetal acceleration to keep the object on the ground.

4 0

3 years ago