The best answer is C) total mass of the team.
In tug of war, the mass of each team is a critical factor in determining which side wins. The team with greater mass will require greater force to move, and also is likely able to exert greater force on the other team due to the correlation between strength and mass.
Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
= K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
= K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
Directed down
Answer:
examples of mixture: smoke,fog,sand, potassium nitrate, sulfur and carbon.
examples of compounds: mithane, ammonia,salt and nitrous oxide.
examples of elements: hydrogen , helium, oxygen and neon.
Explanation:
there are more elements
By the work-energy theorem, the total work done on the mass as it swings is
<em>W</em> = ∆<em>K</em> = 1/2 (18 kg) (17 m/s)² = 153 J
No work is done by the tension in the string, since it's directed perpendicular to the mass at every point in the arc. Similarly, the component of the mass's weight <em>mg</em> pointing perpendicular to the arc also performs no work.
If we ignore friction/drag for the moment, the only remaining force is the parallel component of weight, which performs <em>mgh</em> = (176.4 N) <em>h</em> of work, where <em>h</em> is the vertical distance between points A and B.
Now, if <em>w</em> is the amount of work done by friction/air resistance, then
(176.4 N) <em>h</em> - <em>w</em> = 153 J
If you know the starting height <em>h</em>, then you can solve for <em>w</em>.
