True, an object at rest stays and rest and an object in motion stays in motion
Answer:
The relative uncertainty gives the uncertainty as a percentage of the original value. Work this out with: Relative uncertainty = (absolute uncertainty ÷ best estimate) × 100%. So in the example above: Relative uncertainty = (0.2 cm ÷ 3.4 cm) × 100% = 5.9%. The value can therefore be quoted as 3.4 cm ± 5.9%.
Explanation:
hope it helps :)
They both provide a range of years of an object. I think. They’re just 2 different ways to tell the age of fossils or rocks
Answer:
![t=12.25\ seconds](https://tex.z-dn.net/?f=t%3D12.25%5C%20seconds)
Explanation:
<u>Diagonal Launch
</u>
It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.
The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is
![x=v_ocos\theta t](https://tex.z-dn.net/?f=x%3Dv_ocos%5Ctheta%20t)
![\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2Bv_osin%5Ctheta%20%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Where vo is the magnitude of the initial velocity,
is the angle, t is the time and g is the acceleration of gravity
The maximum height the object can reach can be computed as
![\displaystyle t=\frac{v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7Bv_osin%5Ctheta%7D%7Bg%7D)
There are two times where the value of y is
when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making ![y=y_o](https://tex.z-dn.net/?f=y%3Dy_o)
![\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_o%3Dy_o%2Bv_osin%5Ctheta%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Removing
and dividing by t (t different of zero)
![\displaystyle 0=v_osin\theta-\frac{gt}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%3Dv_osin%5Ctheta-%5Cfrac%7Bgt%7D%7B2%7D)
Then we find the total flight as
![\displaystyle t=\frac{2v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7B2v_osin%5Ctheta%7D%7Bg%7D)
We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is
![\boxed{t=24.5/2=12.25\ seconds}](https://tex.z-dn.net/?f=%5Cboxed%7Bt%3D24.5%2F2%3D12.25%5C%20seconds%7D)
Answer:
Answer is A. damage to buildings.
Explanation:
I hope it's helpful!