Answer:
Normal force, 
Explanation:
Let 
is the weight of suitcase. A man attempts to pick up his suitcase by pulling straight up on the handle. The weight of the suitcase in downward direction. The normal force is acting in upward direction. Let F is the force with which it is pulled straight up.
So, the normal force is given by :
N = mg - F
mg is the weight of the suitcase.
Hence, this is the required solution.
Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of
<em>F</em> = <em>m a</em> → <em>a</em> = <em>F </em>/ <em>m</em>
so that the cart's final speed is
<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>
<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>
<em />
If we force is halved, so is the accleration:
<em>a</em> = <em>F</em> / <em>m</em> → <em>a</em>/2 = <em>F</em> / (2<em>m</em>)
So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give
(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>
- Mass=1167kg
- Initial velocity=u=10m/s
- Acceleration=a=4m/s^2
- Work done=105J=W
- Final velocity=v=?
- Force=F
- Distance=d
Apply Newton's second law
Now
Now
According to third equation of kinematics
Hi,
My best answer would be Gravity. Is it a multiple choice question? Or is it an essay question.
We know that either team is winning which means that the cumulative forces (sum of each girl force) on both teams are equal.
This means,
being x the force of the fourth girl on the east team.
Solving to x, you get
The fourth girl on the east is pulling with a force of 220N.
