Answer:
If iron, nickle, or cobalt is an answer choice, those are the three metals that are purely magnetic.
Answer:
a) 1.75s b) 17.2 m/s (down)
Explanation:
d1= 15m d2= 0m (because it hits ground)
a= -9.81 m/s^2 t=???
Equation
the triangle means change in so d2-d1
Δd= v1 * t + 1/2 * a * t^2
0m-15m= v1*t + 1/2 a t^2
-15 m= 0m/s*t (goes away) + 1/2* a *t^2
-15mx2= t^2
-15mx2/a= t^2
Square root (-30/-9.81m/s^2)
t=1.75 s
b) now v2!!
Im going to use v2= v1 + a*t
v2= 0m/s + -9.81 x 1.75s
v2 = -17.2 m/s or you can say 17.2 m/s down!!!
1045
if this is actually if this is right tell me
Answer:
Say a 14 year old girl was at a construction site and she was asked to move something like a 10,000 pound brick( one brick). She would be acting on it as the unbalanced force but they would still not change their position.
so to say the girl would be doing everything she could to move that brick but the brick would still be in that same spot so the unbalanced force (the girl) would be acting on the thing that was at rest but it wouldn't move.
so the unbalanced force would not really be acting on the thing at rest; even though the unbalanced force was doing something to the brick.
( just think about it and you will eventually get it...just imagine in your head...)
Explanation:
Answer:
After 4 s of passing through the intersection, the train travels with 57.6 m/s
Solution:
As per the question:
Suppose the distance to the south of the crossing watching the east bound train be x = 70 m
Also, the east bound travels as a function of time and can be given as:
y(t) = 60t
Now,
To calculate the speed, z(t) of the train as it passes through the intersection:
Since, the road cross at right angles, thus by Pythagoras theorem:
![z(t) = \sqrt{x^{2} + y(t)^{2}}](https://tex.z-dn.net/?f=z%28t%29%20%3D%20%5Csqrt%7Bx%5E%7B2%7D%20%2B%20y%28t%29%5E%7B2%7D%7D)
![z(t) = \sqrt{70^{2} + 60t^{2}}](https://tex.z-dn.net/?f=z%28t%29%20%3D%20%5Csqrt%7B70%5E%7B2%7D%20%2B%2060t%5E%7B2%7D%7D)
Now, differentiate the above eqn w.r.t 't':
![\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600](https://tex.z-dn.net/?f=%5Cfrac%7Bdz%28t%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7B1%7D%7Bsqrt%7B3600t%5E%7B2%7D%20%2B%204900%7D%7D%5Ctimes%202t%5Ctimes%203600)
![\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t](https://tex.z-dn.net/?f=%5Cfrac%7Bdz%28t%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7Bsqrt%7B3600t%5E%7B2%7D%20%2B%204900%7D%7D%5Ctimes%203600t)
For t = 4 s:
![\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s](https://tex.z-dn.net/?f=%5Cfrac%7Bdz%284%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7Bsqrt%7B3600%5Ctimes%204%5E%7B2%7D%20%2B%204900%7D%7D%5Ctimes%203600%5Ctimes%204%20%3D%2057.6%5C%20m%2Fs)