An EMT raceway contains two 3-phase, 480-volt branch circuits. What minimum size THHN branch circuit conductor is required for o
1 answer:
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Answer:
Explanation:
When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.
According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.
As we know that work is given as:
here:
force of gravity on the object (which is acting downwards)
displacement of the object
- Here the work done by the gravity at an instant just before the projectile hits the earth will be negative as the displacement is in the direction opposite to the force of gravity.
Answer:
R = 28.125 ohms
Explanation:
Given that,
The voltage of a bulb, V = 4.5 V
Current, I = 0.16 A
We need to find the resistance of the filament. Using Ohm's law,
V = IR
Where
R is the resistance of the filament
So,
So, the resistance of the filament is equal to 28.125 ohms.
An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let <em>θ</em> denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,
• the normal force with magnitude <em>n</em>
• the car's weight with magnitude <em>w</em>
and the net force points toward the center of the circle made by the turn, with centripetal acceleration
<em>a</em> = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²
Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,
∑ <em>F</em> (⟂) = <em>N</em> + <em>W</em> (⟂) = <em>m</em> <em>a</em> (⟂)
and
∑ <em>F</em> (//) = <em>W</em> (//) = <em>m a</em> (//)
Let the direction of <em>N</em> be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle <em>θ</em> with the banked curve, and <em>W</em> makes the same angle with the negative perpendicular axis, so that the equations above reduce to
<em>N</em> - <em>m g</em> cos(<em>θ</em>) = <em>m</em> <em>a</em> sin(<em>θ</em>)
and
<em>m g</em> sin(<em>θ</em>) = <em>m a</em> cos(<em>θ</em>)
The second equation is all we need at this point to find the ideal <em>θ</em>. The mass <em>m</em> cancels out, and we can solve for <em>θ</em> to get
tan(<em>θ</em>) = <em>a</em>/<em>g</em> ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615
→ <em>θ</em> ≈ 3.52°
