Answer:
The stitches and dimples around a baseball and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.
Explanation:
The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.
A smooth ball with no stitches or dimples has more air drag that opposes the motion.
A golf ball is smooth ball with dimples to create a thin turbulent boundary layer of air that clings to the ball's surface. This allows the smoothly flowing air to follow the ball's surface a little farther around the back side of the ball, thereby decreasing the size of the wake, and allowing the ball to travel more easily.
The air pressure inside the balloon increases as the number of particles increases.
Answer:
4.7 m³
Explanation:
We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2
* Givens :
P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K
( We must always convert the temperature unit to Kelvin "K")
* What we want to find :
V2 = ?
* Solution :
101 × 2 / 300.15 = 40 × V2 / 283.15
V2 × 40 / 283.15 ≈ 0.67
V2 = 0.67 × 283.15 / 40
V2 ≈ 4.7 m³
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )