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Grace [21]
2 years ago
9

1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer

depend on whether the charges were of the same magnitude or different? How does this relate to Newton’s 3rd law?
Physics
1 answer:
Aloiza [94]2 years ago
7 0

Answer:

Following are the solution to the given question:

Explanation:

Its strength from both charges is equivalent or identical. The power is equal. And it is passed down

F=\frac{kq_1q_2}{r^2}

Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.

|F_{12}| = |F_{21}|

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Red Riding Hood sat down to rest and placed her 1.20-kg basket of goodies beside her. A wolf came along, spotted the basket, and
anygoal [31]

Answer:

  117.6°

Explanation:

The vertical component of a force directed at some angle α from the vertical is ...

  F·cos(α)

We want the vertical components of the wolf's force (Fw) and Red's force (Fr) to total zero. So for some angle from vertical α, Red's force will satisfy ...

  Fw·cos(25°) + Fr·cos(α) = 0

  cos(α) = -Fw/Fr·cos(25°) ≈ -(6.4 N)/(12.5 N)·0.906308 ≈ -0.464030

  α ≈ arccos(-0.464030) ≈ 117.6°

Red was pulling at an angle of about 117.6° from the vertical.

_____

<em>Additional comment</em>

That's about 27.6° below the horizontal.

3 0
2 years ago
What is the pressure exerted by a column of fresh water 10m high<br>(p=1000kg/m3 and g=10m/s)​
alexira [117]
  • Height (h) = 10 m
  • Density (ρ) = 1000 Kg/m^3
  • Acceleration due to gravity (g) = 10 m/s^2
  • We know, pressure in a fluid = hρg
  • Therefore, the pressure exerted by a column of fresh water
  • = hρg
  • = (10 × 1000 × 10) Pa
  • = 100000 Pa

<u>Answer</u><u>:</u>

<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>

Hope you could understand.

If you have any query, feel free to ask.

8 0
2 years ago
A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one
Fittoniya [83]
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
8 0
3 years ago
Read 2 more answers
How do physicists distinguish one type of electromagnetic wave from another?
GenaCL600 [577]
A wave is characterized by the cyclic occurrences of crests and troughs. Wavelengthis defined as the distance between two consecutive troughs or two crests and the Frequency is defined as the number of cycles that pass through a point per second
7 0
3 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
2 years ago
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