Answer:
Wet surfaces areaA=+25.3ft^2
Explanation:
Using F= K×A× S^2
Where F= drag force
A= surface area
S= speed
Given : F=996N S=20mph A= 83ft^2
K = F/AS^2=996/(83×20^2)
K= 996/33200 = 0.03
1215= (0.03)× A × 18^2
1215=9.7A
A=1215/9.7=125.3ft^2
Answer:
So the acceleration of the child will be 
Explanation:
We have given angular speed of the child 
Radius r = 4.65 m
Angular acceleration 
We know that linear velocity is given by 
We know that radial acceleration is given by 
Tangential acceleration is given by
So total acceleration will be 
Answer: the speed at which it falls toward the Earth.
Explanation:
A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.
Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).
This kind of motion is the result of two independent motions: vertical motion and horizontal motion.
The observed net velocity is the vectorial sum of the vertical and horizontal velocities.
The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).
The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.
Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.
<span>x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v=<span>
<span>-0.13261622 m/s
</span></span>the object moving at 0.13 m/s <span>at time t=0.815 s</span>
