Electric force is present between proton and electron of an atom.
<h3>What is the attractive force present between a proton and electron?</h3>
The attractive force between the electrons and the proton of the nucleus is called the electric force. This force attract electron towards the nucleus so that it can't escape from the atom easily so this force is very important for the atom.
<h3>How an atom of this element changes when it forms an ionic bond?</h3>
An atom of this element changes when it forms an ionic bond because the number of electron decreases due to losing of electron in order to get stability. We know that ionic bond forms between atoms when electron is lost or gain by the atom.
So we can conclude that electric force is present between proton and electron of an atom.
Learn more about force here: brainly.com/question/12970081
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Answer:
292796435 seconds ≈ 300 million seconds
Explanation:
First of all, the speed of the car is 121km/h = 33.6111 m/s
The radius of the planet is given to be 7380 km = 7380000 m
From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec
If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have
w(vehicle) = 9.78 x w(planet)
w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec
To find the period of the planet's rotation; we use the equation
w(planet) = 2π÷T
Where w(planet) is the angular velocity of the planet and T is the period
From the equation T = 2π÷w = 2×(22/7) ÷ 4.66 x 10⁻⁷ = 292796435 seconds
Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds
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.
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<em>Additional comment</em>
That's about 27.6° below the horizontal.
