Answer:
a)![v=13.2171\,m.s^{-1}](https://tex.z-dn.net/?f=v%3D13.2171%5C%2Cm.s%5E%7B-1%7D)
b)![H=8.9605\,m](https://tex.z-dn.net/?f=H%3D8.9605%5C%2Cm)
Explanation:
Given:
mass of bullet, ![m=4.97\times 10^{-3}\,kg](https://tex.z-dn.net/?f=m%3D4.97%5Ctimes%2010%5E%7B-3%7D%5C%2Ckg)
compression of the spring, ![\Delta x=0.0476\,m](https://tex.z-dn.net/?f=%5CDelta%20x%3D0.0476%5C%2Cm)
force required for the given compression, ![F=9.12 \,N](https://tex.z-dn.net/?f=F%3D9.12%20%5C%2CN)
(a)
We know
![F=m.a](https://tex.z-dn.net/?f=F%3Dm.a)
where:
a= acceleration
![9.12=4.97\times 10^{-3}\times a](https://tex.z-dn.net/?f=9.12%3D4.97%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20a)
![a\approx 1835\,m.s^{-2}\\](https://tex.z-dn.net/?f=a%5Capprox%201835%5C%2Cm.s%5E%7B-2%7D%5C%5C)
we have:
initial velocity,![u=0\,m.s^{-1}](https://tex.z-dn.net/?f=u%3D0%5C%2Cm.s%5E%7B-1%7D)
Using the eq. of motion:
![v^2=u^2+2a.\Delta x](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2a.%5CDelta%20x)
where:
v= final velocity after the separation of spring with the bullet.
![v^2= 0^2+2\times 1835\times 0.0476](https://tex.z-dn.net/?f=v%5E2%3D%200%5E2%2B2%5Ctimes%201835%5Ctimes%200.0476)
![v=13.2171\,m.s^{-1}](https://tex.z-dn.net/?f=v%3D13.2171%5C%2Cm.s%5E%7B-1%7D)
(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,
![u=13.2171\,m.s^{-1}](https://tex.z-dn.net/?f=u%3D13.2171%5C%2Cm.s%5E%7B-1%7D)
∵At maximum height the final velocity will be zero
![v=0\,m.s^{-1}](https://tex.z-dn.net/?f=v%3D0%5C%2Cm.s%5E%7B-1%7D)
Using the equation of motion:
![v^2=u^2-2g.h](https://tex.z-dn.net/?f=v%5E2%3Du%5E2-2g.h)
where:
h= height
g= acceleration due to gravity
![0^2=13.2171^2-2\times 9.8\times h](https://tex.z-dn.net/?f=0%5E2%3D13.2171%5E2-2%5Ctimes%209.8%5Ctimes%20h)
![h=8.9129\,m](https://tex.z-dn.net/?f=h%3D8.9129%5C%2Cm)
is the height from the release position of the spring.
So, the height from the latched position be:
![H=h+\Delta x](https://tex.z-dn.net/?f=H%3Dh%2B%5CDelta%20x)
![H=8.9129+0.0476](https://tex.z-dn.net/?f=H%3D8.9129%2B0.0476)
![H=8.9605\,m](https://tex.z-dn.net/?f=H%3D8.9605%5C%2Cm)
Answer:
a) w = 25.1 rad/s, b) θ = 0.9599 rad
, c) α = 328.1 rad/s² d) t= 0.0765 s
Explanation: Let's work on this exercise with the equations of angular kinematics
a) The angular velocity is
w = 4.00 rev / s (2π rad / 1 rev)
w = 25.1 rad/s
b) let's reduce the angle of degrees to radians
θ = 55 ° (π rad / 180 °)
θ = 0.9599 rad
c) Let's use the initial angular velocity as the system part of the rest is zero
w² = w₀² + 2 α θ
α = w² / 2 θ
α = 25.1²/2 0.9599
α = 328.1 rad / s²
d)
w = w₀ + α t
t = w / α
t = 25.1 / 328.1
t= 0.0765 s
(i) because there's gravity force pulling it downwards.
(ii) because the force on the metal box is gone (either it has landed on the ground or something else happened) that means that the metal box is no longer moving but it has constant velocity/uniform velocity.
I'm sorry if it's wrong but that's how I see it
Answer:
18
Explanation:
we need the to know the extra weight Eric has over Tom. To do that we need to subtract Tom's weight from Eric's, but now we do not know what Eric's weight is or Tom's weight. the only clue we have is that Eric is 4 inches taller than Tom
1. if Eric is 4 inches taller that Tom, then
let
Tom's height be ![T_{h}](https://tex.z-dn.net/?f=T_%7Bh%7D)
and
Eric's height will be 4 inches more
![T_{h} +4](https://tex.z-dn.net/?f=T_%7Bh%7D%20%2B4)
let's plug these two information into the formula given in the question
for the difference in their weight;
Eric's weight will be;
w = ![4.5(T_{h} +4)-154](https://tex.z-dn.net/?f=4.5%28T_%7Bh%7D%20%2B4%29-154)
Tom's weight will be
w = ![4.5h-154](https://tex.z-dn.net/?f=4.5h-154)
taking the difference between the two expressions above
w = ![[4.5(T_{h} +4)-154]-[4.5h-154]](https://tex.z-dn.net/?f=%5B4.5%28T_%7Bh%7D%20%2B4%29-154%5D-%5B4.5h-154%5D)
opening the bracket
w = ![4.5T_{h} +18-154-4.5h+154](https://tex.z-dn.net/?f=4.5T_%7Bh%7D%20%2B18-154-4.5h%2B154)
w = 18 pounds
so Eric's weight is 18 pounds more than Tom's