Answer:
If a vertical line extending down from an object's CG extends outside its area of support, the object will topple
Explanation:
We can understand better this situation using a diagram with the forces acting on it.
In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.
Answer:
a
The speed of the quarterback backward is
b
Known are
Unknown
Explanation:
From the question we are told that
The mass of the quarterback is
The mass of the ball is
The speed of the ball is
The law of momentum conservation can be mathematically represented as
Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball
So
=>
substituting values
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°
