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Aneli [31]
3 years ago
12

Why must indirect evidence be used to study the structure of atoms?

Physics
1 answer:
Sloan [31]3 years ago
8 0
Indirect evidence must be used to study the structure of atoms, because the direct evidence affects the position and energy of the particles of the atomic structure.

For example, imagine you use a light ray to find the location of an electron in an atomic structure, the current location of the electron you are now seeing is as a result of the light ray, and the electron has changed it's initial position to a new position as a result of the effect of the light ray.

So indirect means would probably give more accurate answers about the atomic structure than that of the direct means.

An example of the indirect means would be for example setting up a mathematical equation or  model of the atomic structure and trying to provide solution for the model or equation.
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12. A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s. If the runner has a mass of
Margaret [11]
1/2 m/s2
30 Na




I hope this helps!
4 0
3 years ago
Someone please help
saul85 [17]

Based on the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • The location will correspond to any point on the same latitude as A

<h3>What are lines of longitude?</h3>

Lines of longitude are imaginary lines which run along the earth from the North pole. to the South pole.

Longitude lines divide the earth into semi-circles.

Longitude lines are known as meridians and each meridian measures one arc degree of longitude.

Considering the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • the location will correspond to any point on the same latitude as A

In conclusion, longitude lines are imaginary lines and run from North to South on the earth.

Learn more about lines of longitude at: brainly.com/question/1939015

#SPJ1

8 0
2 years ago
A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the
shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

Force, F = ma

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

a = 0.47 * 9.8 m/s^2

a = 4.606m/s^2

We know,

v^2 - u^2 = 2as

where, v is the final velocity

           u is the initial velocity

           a is the acceleration

           s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s

The car travels at a speed of 25m/s.

6 0
3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
Một chất điểm khối lượng m=200g chuyển
MrMuchimi
Can you please translate to English?
8 0
2 years ago
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