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2 years ago

Ionic compounds form crystal structures. molecular structures. pure elements. neutral elements

Physics

2 answers:

cestrela7 [59]2 years ago

5 0

Answer:

ionic compounds are the compounds whose atoms are combined together by transfer of electron

Explanation:

Naddik [55]2 years ago

3 0

Answer: Option (a) is the correct answer.

Explanation:

Ionic compounds are the compounds whose atoms are combined together by transfer of electrons.

An ionic compound has partial positive charge on the cation and partial negative charge on anion.

For example, NaCl is an ionic compound whose atoms are arranged orderly due to the opposite charges on its atoms.

So, an ionic compound forms a neutral compound but not a neutral element.

Thus, we can conclude that ionic compounds form crystal structures.

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ioda

I believe the answer is C. Positive, since it is a sustainable plan to use a renewable source hope this helps :)

4 0

2 years ago

Light travels fastest when moving through

vitfil [10]

A vaccum unlike sound,light can travel through any matter including a great vacuum of nothing (space)

4 0

2 years ago

A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.

Sedbober [7]

Answer:

The work done by the car is 363 kJ

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W = 1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

Given: m=1500 kg, v=22 m/s

Substituting these values into equation 1

W = 1/2(1500)(22)²

W = 750 × 484

W = 363000 J

W = 363 kJ

Thus the work done by the car is 363 kJ

8 0

2 years ago

Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p

Bad White [126]

Answer:

Diode Lasers

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is

4. The optical gain bandwidth (as measured between half intensity points) will

normally depend on the pumping current (diode current) but for this problem

assume that it is 2 nm.

(a) What is the mode integer m of the peak radiation?

(b) What is the separation between the modes of the cavity? Please express your

answer as Δλ.

(d) What is the reflection coefficient and reflectance at the ends of the optical

cavity (faces of the InGaAsP crystal)?

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the

beam is a Gaussian beam with the waist located at the output. And the beam

waist size is approximately the x-y dimensions of the cavity.)

Solution:

(a) The wavelength λ of a cavity mode and length L are related by

λ = , where m is the mode number, and n is the refractive index.

So the mode integer of the peak radiation is

1290

1055.1

10250422

= ×

××× == −

−

m .

(b) The mode spacing is given by nL

c f 2

=Δ . As

c f = , λ

Δ−=Δ 2

c f .

Therefore, we have nm

nL f

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .

bandwidth could fit in two possible modes.

For mode integer of 1290, nm

nL 39.1550

1290

10250422 6

= ××× ==

−

Take m = 1291, nm

nL 18.1549

1291

10250422 6

= ××× ==

−

Or take m = 1289, nm

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0

3 years ago

A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a

RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2$

$F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N$

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0

3 years ago