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Maurinko [17]
2 years ago
13

The atomic number of an element is also the number of

Physics
1 answer:
lilavasa [31]2 years ago
8 0
Answer: A, protons

Why: it just is
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An automobile steering wheel is shown. What is the ideal mechanical advantage? If the AMA is 8, what is the efficiency of the st
Mars2501 [29]

1. Ideal Mechanical Advantage (IMA): 9

Explanation:

For a wheel and axle system like the steering wheel, the IMA is given by:

IMA=\frac{r_w}{r_a}

where

r_w is the radius of the wheel

r_a is the radius of the axle

For the steering wheel of the problem, we see that r_w = 18 cm and r_a=2 cm, so the IMA is

IMA=\frac{18 cm}{2 cm}=9


2. Efficiency: 88.9%

Explanation:

The efficiency of a system is defined as the ratio between the AMA (actual mechanical advantage) and the IMA:

\eta=\frac{AMA}{IMA}\cdot 100

In this problem, AMA=8 and IMA=9, so the efficiency is

\eta=\frac{8}{9}\cdot 100=88.9\%


6 0
3 years ago
The behavior of which objects are better explained by Einstein’s general theory of relativity than Newton’s universal law of gra
liq [111]
If I had to go with any of those answers, It would be A maybe D, But im not too sure on how to decide between them. Because Einstein mentioned the sun in his theory which has a very large mass <span> 1.989 x 10 with a exponent of 30 to be exact. Hope this helped though.</span>
7 0
3 years ago
Read 2 more answers
What is the voltage of the electrochemical cell written as: Cu(s) | Cu2+(aq) || Mg2+(aq) | Mg(s)?
erma4kov [3.2K]
Thank you for posting your question here at brainly. Feel free to ask more questions.  
<span><span>The best and most correct answer among the choices provided by the question is  </span>B.-2.71 V.</span>  
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V  

Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V  

Since Cu is acting as the anode, the equation needs to be reversed.  

Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V  

Ecell= -2.37 V+ (- 0.34 V) = -2.71 V   <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span> <span> </span>
4 0
3 years ago
Which of the following renaissance scientists made improvements to the telescope?
LenaWriter [7]
Michelaneglo DDDDDDDDDDDDDDDDDDDDDDDDD

7 0
3 years ago
Please someone help, I’m very confused and it’s due soon, thanks
Anit [1.1K]

Answer:

  1. 1 s
  2. 19.6 m
  3. 2 s
  4. 0.8 m/s^2
  5. 28 m/s
  6. 79 m/s
  7. 0.37 s
  8. 26 m/s
  9. 242 m/s
  10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0
3 years ago
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