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Doss [256]
3 years ago
9

At

Physics
1 answer:
AnnyKZ [126]3 years ago
4 0

Explanation:

One common property with all halogens in group 7 is that they are all non-metals.

Fluorine, chlorine, Bromine and Iodine are classified as non-metallic elements and they have a high electronegativity.

  • In chemical reactions, they are very reactive because they require just one electron to complete their octet configuration and be isoelectronic with noble gases.

The most prominent observation from halogens is that they all non-metallic in nature.

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What is the value of a conversion factor ratio?<br>A.1<br>B.3<br>C.10<br>D.12
Alecsey [184]
A is the correct answer :D

4 0
4 years ago
Use the periodic table to answer the question
True [87]

Answer: I hope this helps you

Explanation:

8 0
3 years ago
Read 2 more answers
PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
AleksAgata [21]
22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
Using v^2 = u^2 - 2as 
0^2 - u^2 + 2 (-7.00) (92) 
initial velocity, u = sqrt (1288) = 35.9 m/s 
This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment
Andrei [34K]

Answer:

  C. her moment of inertia increases and her angular speed decreases

  D. her moment of inertia increases and her angular speed decreases

Explanation:

The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.

The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.

This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.

When the person extends her arms, her moment of inertia increases and her angular speed decreases.

_____

<em>Note to those looking for a letter answer</em>

Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.

3 0
4 years ago
Imagine a star very similar to our own Sun in both size and mass. This star moves very slightly back and forth in the sky once e
Inessa05 [86]

Answer:

(c) The planet must have a mass about the same as the mass of Jupiter,

(d) The planet must be closer to the star than Earth is to the Sun.

Explanation:

Astrometry is the ideal method to detect high-mass planets that are close to their star. That is because the gravitational effect that it will have the planet over its host star will be greater. This effect can be seen as a wobble in the star as a consequence of how they orbit a common center of mass¹. The center of mass will be closer to the most massive object, So, in the case of an extrasolar planet with masses like Jupiter (Jovian), this point will be a little bit farther from the star, making the wobble more notable than in a system with a low-mass planet.          

Key terms:

Astrometry: study of the position of the stars over time in the sky.

¹Center of mass: a geometrical point in which the mass from a whole system is summed.

6 0
3 years ago
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