Answer:
k = 26.25 N/m
Explanation:
given,
mass of the block= 0.450
distance of the block = + 0.240
acceleration = a_x = -14.0 m/s²
velocity = v_x = + 4 m/s
spring force constant (k) = ?
we know,
x = A cos (ωt - ∅).....(1)
v = - ω A cos (ωt - ∅)....(2)
a = ω²A cos (ωt - ∅).........(3)
now from equation (3)
k = 26.25 N/m
hence, spring force constant is equal to k = 26.25 N/m
I believe all of these would be known as specific phobias.
Answer:
The magnitude of the angular acceleration ∝ = ![\frac{rxF}{2.8[tex]r^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BrxF%7D%7B2.8%5Btex%5Dr%5E%7B2%7D)
}[/tex]
Explanation:
The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.
<span>Jun 16, 2012 - Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius? –73°C 27°C 327°C 673°C.</span><span>
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