Answer:
<em>The distance the car traveled is 21.45 m</em>
Explanation:
<u>Motion With Constant Acceleration
</u>
It occurs when an object changes its velocity at the same rate thus the acceleration is constant.
The relation between the initial and final speeds is:
![v_f=v_o+at\qquad\qquad [1]](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat%5Cqquad%5Cqquad%20%5B1%5D)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is given by:
![\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Dv_o.t%2B%5Cfrac%7Ba.t%5E2%7D%7B2%7D%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for a:
Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:
The distance is now calculated with [2]:
x = 21.45 m
The distance the car traveled is 21.45 m
i'm stuck on that question also
Oooooo there's a spongy bone? that's cool! Lol okay okay, I will research it and help you out.
Here's what I found:
Cancellous bone<span>, also known as </span>spongy<span> or </span>trabecular bone<span>, is one of the </span>two<span> types of </span>bone<span> tissue found in the human body. ... It is very porous and contains red </span>bone<span>marrow, where blood cells are made.</span>
no, work is = force * distance or displacement
Answer:
I should be active for 15 hours to meet the physical activity requirement.
Explanation:
Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.
So, t = t₀/√(1 - β²)
t = 9/√(1 - (v/c)²)
= 9/√(1 - (0.8c/c)²)
= 9/√(1 - (0.8)²)
= 9/√(1 - (0.64)
= 9/√0.36
= 9/0.6
= 15 hr
So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.
