Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
Answer: the answer is d
Explanation: there are not more than 10 violations within a twelve month period hope this helps
<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)
cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0
therefore, wmax=1.2rad/s
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)
adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)
adn=r*w^2=200*1.2^2=288
ad= square root of adt^2+adn^2 = 288mm/s^2</span>
Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the <span>friction force keeping on table is given by the following formula:</span>
<span>Ff = u*(1-f)*m*g </span>
and we need to know aso that <span>gravity force pulling off the table Fg is given by this other formula:</span>
<span>Fg = f*m*g </span>
What you need to do is <span>Equate the two and solve for f: </span>
<span>f*m*g = u*(1-f)*m*g </span>
<span>=> f = u*(1-f) = u - uf </span>
<span>=> f + uf = u </span>
=> f = u/(1+u) = fraction of rope
With that you can find the answer
