The answer is 21m because the motion is in one dimension with constant acceleration.
The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>
Plugging in our known values, we have
<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>
Slow-twitch<span> muscles help enable long-endurance feats such as distance running.
</span>fast-twitch<span> muscles fatigue </span>faster<span> but are used in powerful bursts of movements like sprinting.</span>
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
<u>α = - 1.047 rad/s²</u>
B.
We can use the second equation of motion to find the angular distance:
<u>θ = 14.1 rad</u>
C.
θ = (14.1 rad)(1 rev/2π rad)
<u>θ = 2.24 rev</u>
Communicating the findings is part of a scientific investigation! The last option
Answer:
v=1.295
Explanation:
What we are given:
a=5÷(3s^(1/3)+s^(5/2)) m/s^2
Start by using equation a ds = v dv
This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:
<em>a=2</em>
<em>b=1</em>
<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>
<em>dx=dv</em>
Integrate the left side the standard method.
<em>a=v</em>
<em>b=0</em>
<em>dx=dv</em>
<em>Integrating</em>
=v^2/2
Use Simpson's rule for the right site.
<em>a=b</em>
<em>b=a</em>
<em>x=f(x)</em>
f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)
If properly applied. you should now have the following equation:
v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]
=0.8376
Solve for v.
v=1.295