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.
As it is given that the air bag deploy in time

total distance moved by the front face of the bag

Now we will use kinematics to find the acceleration




now as we know that

so we have

so the acceleration is 400g for the front surface of balloon
Answer:
Explanation:
Speed is defined as the rate at which an object covers a particular distance. So the formula for determining speed is given as the ratio of distance to time taken for covering that distance.
Speed = Distance/Time
As here the distance is given in km units and time in s units, so the units of any one parameter should be changed. Since we know that speed of sound is always about 300 m/s. So it is better to convert the unit of distance from km to m.
Hence, now the distance traveled by the noise is 2000 m and time taken is 5.8 s.
So the speed of noise = Distance/Time = 2000/5.8=345 m/s.
Thus, the speed of noise is slightly greater than the speed of sound and it is found to be 345 m/s.
(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then
(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>
==> <em>v</em> ≈ 6.2 m/s
(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:
<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>
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Answer:
just guys
Explanation:
and if not i need how old you are sorry just trying to be safe