Answer:
C. N/m (newtons/meter)
Explanation:
Since the equation for potential energy for a spring is PE = 1/2kx², after you know <em>k </em>and <em>x</em>, you will get an answer in Joules.
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Answer:
The transverse component of acceleration is 26.32
where as radial the component of acceleration is 8.77 
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s

So the transverse component of acceleration are given as

Here


So

The transverse component of acceleration is 26.32 
The radial component is given as

Here

So

The radial component of acceleration is 8.77 
Well the trivial answer is zero, since there is indeed a "zero vector". Assuming you aren't allowed to use the zero vector you would need at least two. They would be antiparallel and of equal magnitude. (That is be pointing in opposite directions and have the same length)
Answer:
m₁ / m₂ = 1.3
Explanation:
We can work this problem with the moment, the system is formed by the two particles
The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero
p₀o = m₁ v₁ + m₂ v₂
pf = 0
m₁ v₁ + m₂ v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂= - (-6.2) / 4.7
m₁ / m₂ = 1.3
Another way to solve this exercise is to use the mass center relationship
Xcm = 1/M (m₁ x₁ + m₂ x₂)
We derive from time
Vcm = 1/M (m₁ v₁ + m₂v₂)
As they say the velocity of the center of zero masses
0 = 1/M (m₁ v₁ + m₂v₂)
m₁ v₁ + m₂v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂ = 1.3
Answer:
Fractional error = 0.17
Percent error = 17%
F = 112 ± 19 N
Explanation:
Plug in the values to find the force:
F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N
Find the fractional error:
ΔF/F = Δm/m + 2Δv/v + Δr/r
ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5
ΔF/F = 0.17
Multiply by 100% to find the percent error:
ΔF/F × 100% = 17%
Solve for the absolute error:
ΔF = 0.17 × 112 N = 19 N
Therefore, the force is:
F = 112 ± 19 N