The free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.
When an object falls from up to the ground, the object falls under in the influence of acceleration due to gravity.
The vertical component of the force on the apple as it falls trough the air is given as;
∑Fy = 0
Fₙ - W = 0
Fₙ = W
where;
- <em>Fₙ is the frictional force on the apple acting upwards</em>
- <em>W is the weight of the apple acting downwards</em>
The free-body diagram of the apple is represented as follows;
↑ Fₙ
Ο
↓ W
Thus, the free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.
Learn more here:brainly.com/question/18770265
The acceleration of this car is equal to 5
.
<u>Given the following data:</u>
- Initial velocity = 0 m/s (assuming it's starting from rest).
To determine the acceleration of this car:
<h3>How to calculate acceleration.</h3>
In Science, the acceleration of an object is calculated by subtracting the initial velocity from its final velocity and dividing by the time.
Mathematically, acceleration is given by this formula:
![a = \frac{V\;-\;U}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BV%5C%3B-%5C%3BU%7D%7Bt%7D)
<u>Where:</u>
- U is the initial velocity.
- is the time measured in seconds.
Substituting the given parameters into the formula, we have;
![a = \frac{10\;-\;0}{2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B10%5C%3B-%5C%3B0%7D%7B2%7D)
Acceleration, a = 5 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Read more on acceleration here: brainly.com/question/24728358
Answer:
![6.63\times 10^8\ N/C](https://tex.z-dn.net/?f=6.63%5Ctimes%2010%5E8%5C%20N%2FC)
Explanation:
Given that,
The magnitude of magnetic field, B = 2.21
We need to find the magnitude of the electric field. Let it is E. So,
![\dfrac{E}{B}=c\\\\E=Bc](https://tex.z-dn.net/?f=%5Cdfrac%7BE%7D%7BB%7D%3Dc%5C%5C%5C%5CE%3DBc)
Put all the values,
![E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C](https://tex.z-dn.net/?f=E%3D2.21%5Ctimes%203%5Ctimes%2010%5E8%5C%5C%5C%5C%3D6.63%5Ctimes%2010%5E8%5C%20N%2FC)
So, the magnitude of the electric field is equal to
.