Answer:
Gravity acts to pull the object down.
The object’s inertia carries it forward.
The path of the object is curved.
Explanation:
The motion of a projectile consists of two separate motions:
- A uniform motion along the horizontal direction, where the velocity is constant; since there are no forces along this direction, the velocity does not change, and so the object continues its motion for inertia --> so, the statement "The object’s inertia carries it forward" is true.
- A uniformly accelerated motion along the vertical direction, with a constant downward acceleration (g=9.8 m/s^2, acceleration due to gravity). So, the vertical velocity changes, due to the presence of the gravity that acts to pull the object down.
- As a result of the combination of these two motions, the object follows a curved path (in particular, it is a parabolic path).
Answer:
The period is ![T = 0.00255 \ s](https://tex.z-dn.net/?f=T%20%20%3D%20%200.00255%20%5C%20%20s)
Explanation:
From the question we are told that
The frequency is ![f = 392 \ Hz](https://tex.z-dn.net/?f=f%20%20%3D%20%20392%20%5C%20Hz)
Generally the period is mathematically represented as
![T = \frac{1}{f}](https://tex.z-dn.net/?f=T%20%20%3D%20%5Cfrac%7B1%7D%7Bf%7D)
=> ![T = \frac{1}{ 392}](https://tex.z-dn.net/?f=T%20%20%3D%20%5Cfrac%7B1%7D%7B%20392%7D)
=> ![T = 0.00255 \ s](https://tex.z-dn.net/?f=T%20%20%3D%20%200.00255%20%5C%20%20s)
Answer:
Promotes Cardiovascular Health. ...
Burns Calories. ...
Builds Bone Strength. ...
Boosts The Immune System. ...
Provides Strength Training. ...
Boosts Mental Development. ...
Develops Better Coordination And Motor Skills. ...
Develops Self-Discipline And Concentration.
Explanation:
Answer:
(2.5,0)
Explanation:
The particle can be described by the following equations:
![x=Rsin(-\omega t)+2.5\\y=Rcos(-\omega t)\\\frac{dx}{dt}=-\omega Rcos(-\omega t)\\\frac{dy}{dt}=\omega Rsin(-\omega t)\\\frac{d^2x}{dt^2}=-\omega^2Rsin(-\omega t)\\\frac{d^2y}{dt^2}=-\omega^2Rcos(-\omega t)](https://tex.z-dn.net/?f=x%3DRsin%28-%5Comega%20t%29%2B2.5%5C%5Cy%3DRcos%28-%5Comega%20t%29%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%3D-%5Comega%20Rcos%28-%5Comega%20t%29%5C%5C%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Comega%20Rsin%28-%5Comega%20t%29%5C%5C%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%3D-%5Comega%5E2Rsin%28-%5Comega%20t%29%5C%5C%5Cfrac%7Bd%5E2y%7D%7Bdt%5E2%7D%3D-%5Comega%5E2Rcos%28-%5Comega%20t%29)
For R = 2.5, ω = 2 and t = 0:
![x=2.5\\y=2.5\\\\\frac{dx}{dt}=-5\\ \frac{d^2y}{dt^2}=-10](https://tex.z-dn.net/?f=x%3D2.5%5C%5Cy%3D2.5%5C%5C%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%3D-5%5C%5C%20%5Cfrac%7Bd%5E2y%7D%7Bdt%5E2%7D%3D-10)
The center of the circle would be at point (2.5,0)
Newton 3rd law:
F=ma
1600=85*a
1600/85=a
a=18.82 m/s^2