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3 years ago

A ____ is another name for a group of elements with similar properties

1 answer:

7 0

The answer is “Family”

Explanation- There are 18 numbered groups in the periodic table. The elements in a group have similar physical or chemical characteristics and some groups have a specific name, for example group 17 (the halogens).

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If it were possible to move a star towards the earth then its apparent magnitude number would ______ while its absolute magnitud

ArbitrLikvidat [17]

If it were possible to move a star towards the earth then its apparent magnitude number would decrease while its absolute magnitude number would stay the same.

Definition of apparent magnitude:

The luminosity of a celestial body (such as a star) as observed from the earth compare absolute magnitude.

So for example, the apparent magnitude of the Sun is -26.7 and is the brightest celestial object we can see from Earth. However, if the Sun were 10 parsecs away, its apparent magnitude would be +4.7, only about as bright as Ganymede appears to us on Earth.

Definition of absolute magnitude:

Absolute magnitude is a measure of the luminosity of a celestial object on an inverse logarithmic astronomical magnitude scale.

To learn more about apparent magnitude here

brainly.com/question/2949443

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5 0

2 years ago

Plz guys answer my question

Sophie [7]

Answer:

Explanation:

whats your question?

6 0

3 years ago

Read 2 more answers

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

Assume that the polymer material has a constant refractive index of 1.5. For light of 600nm wavelength at normal incidence, what

yaroslaw [1]

Answer:

Minimum thickness will be 100 nm

Explanation:

We have given refractive index is n = 1.5

Wavelength of the light incidence $\lambda$ = 600 nm

We have to find the smallest thickness of the film so that there will be minimum light reflect

For minimum thickness of non reflecting film

$t=\frac{\lambda }{4n}$ , here t is thickness, $\lambda$ is wavelength and n is refractive index

Putting all values $t=\frac{600}{4\times 1.5}=100nm$

So minimum thickness will be 100 nm

8 0

3 years ago