Answer:
The root pass is made with a 5/32” (4.0mm) diameter electrode. A cellulosic electrode (E-XX10) is being used. The root pass is welded with reverse (DC+) polarity.
Explanation:
Answer:
ω₂ = 93.6 rev / min
Explanation:
ω₀ = 260 rev / min
ω₁ = 0.68*ω₀ = 0.68*(260 rev / min) = 176.8 rev / min
ω₂ = ?
t₁ = 1 min
t₂ = 2 min
We can apply the equation:
ω₁ = ω₀ + α*t₁ ⇒ α = (ω₁ - ω₀) / t₁
⇒ α = (176.8 rev / min - 260 rev / min) / 1 min = - 83.2 rev / min²
then we can use the same formula, knowing the angular acceleration:
ω₂ = ω₀ + α*t₂ ⇒ ω₂ = (260 rev / min) + (- 83.2 rev / min²)*(2 min)
⇒ ω₂ = 93.6 rev / min
Answer:6amperes
Explanation:
1/R=1/40 + 1/40
1/R=2/40
Cross multiplying we get
2R=40
Divide both sides by 2
2R/2 =40/2
R=20ohms
Current=voltage ➗ resistance
Current=120 ➗ 20
Current =6amperes
Gravity affet everything and it touches nothing.
Hope this helps!
The horse's position on the ground at time <em>t</em> is
<em>x</em> = (20 m/s) <em>t</em>
The baboon's height from the ground at time <em>t</em> is
<em>y</em> = 3 m - 1/2 <em>g</em> <em>t</em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.
The baboon falls and lands on the horse, so that the two animals meet when the baboon's height is 2 m from the ground, which happens after
2 m = 3 m - 1/2 <em>g</em> <em>t</em>²
1/2 <em>g</em> <em>t</em>² = 1 m
<em>t</em>² = (2 m) / (9.80 m/s²)
<em>t</em> ≈ 0.452 s
In this time, the horse reaches the tree, so its distance from it is
(20 m/s) * (0.452 s) ≈ 9.04 m
