Magma forms by partial melting of upper mantle and crust. Partial melt means that only a fraction of the available material forms a melt, and that the remainder stays solid. The partial melt rises because of its lower density and ascends through he crust.
Answer:
a
The hiker (you ) is 200 m below his/her(your) starting point
b
The resultant displacement in the north east direction is
![a = 6562.0 \ m](https://tex.z-dn.net/?f=a%20%20%3D%206562.0%20%5C%20%20m%20)
The resultant displacement in vertical direction (i.e the altitude change )
![b =6503.1 \ m](https://tex.z-dn.net/?f=b%20%3D6503.1%20%5C%20%20m%20)
Explanation:
From the question we are told that
The displacement in the morning is ![S_{morning} = (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)](https://tex.z-dn.net/?f=S_%7Bmorning%7D%20%3D%20%20%282200%20%5Cm%20%2C%20east%29%20%2B%20%284000%5C%20m%5C%20north%29%20%2B%20%28100%20%5C%20m%20%2C%5C%20vertical%29)
The displacement in the afternoon is ![S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)](https://tex.z-dn.net/?f=S%20_%7Bafternoon%7D%3D%20%281300%5C%20m%20%2C%5C%20west%29%20%2B%20%282500%20%5C%20m%20%2C%5C%20north%29%20-%20%28300%5C%20m%20%2C%5C%20vertical%29)
Generally the direction west is negative , the direction east is positive
the direction south is negative , the direction north is positive
resultant displacement is mathematically evaluated as
![(2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east](https://tex.z-dn.net/?f=%282200%20%5Cm%20%2C%20east%29%20%2B%28%20-%201300%5C%20m%20%2C%5C%20west%29%20%3D%20900%20%5C%20m%20%5C%20east)
![(4000\ m\ north) + (2500 \ m ,\ north) = 6500 \ m ,\ north](https://tex.z-dn.net/?f=%284000%5C%20m%5C%20north%29%20%20%2B%20%282500%20%5C%20m%20%2C%5C%20north%29%20%3D%206500%20%20%5C%20m%20%2C%5C%20north)
![(100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m](https://tex.z-dn.net/?f=%28100%20%5C%20m%20%2C%5C%20vertical%29%20-%20%28300%5C%20m%20%2C%5C%20vertical%29%20%3D%20-200%20%5C%20m)
From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position
Generally from Pythagoras theorem , the resultant displacement in the north east direction is
![a = \sqrt{900^2 + 6500^2}](https://tex.z-dn.net/?f=a%20%20%3D%20%20%5Csqrt%7B900%5E2%20%2B%206500%5E2%7D)
=> ![a = 6562.0 \ m](https://tex.z-dn.net/?f=a%20%20%3D%206562.0%20%5C%20%20m%20)
Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )
![b = \sqrt{6500^2 +(-200)^2 }](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%7B6500%5E2%20%2B%28-200%29%5E2%20%20%7D)
=> ![b =6503.1 \ m](https://tex.z-dn.net/?f=b%20%3D6503.1%20%5C%20%20m%20)
Answer:
Is to add all forces, for example either the gf = gravitational force
Ff= force fiction
Fn= normal force
Thus, fg + ff + fn = y will give you results
Well then the forces you use in your exercises or questions.
Answer:
Explanation:650
colour* wavelength (nm) energy (eV)
red 650 1.91
orange 600 2.06
yellow 580 2.14
green 550 2.25
Answer:
x(t) = d*cos ( wt )
w = √(k/m)
Explanation:
Given:-
- The mass of block = m
- The spring constant = k
- The initial displacement = xi = d
Find:-
- The expression for displacement (x) as function of time (t).
Solution:-
- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.
- We apply the Newton's equation of motion in horizontal direction.
F = ma
-kx = ma
-kx = mx''
mx'' + kx = 0
- Solve the Auxiliary equation for the ODE above:
ms^2 + k = 0
s^2 + (k/m) = 0
s = +/- √(k/m) i = +/- w i
- The complementary solution for complex roots is:
x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]
- The given initial conditions are:
x(0) = d
d = [ A*cos ( 0 ) + B*sin ( 0 ) ]
d = A
x'(0) = 0
x'(t) = -Aw*sin (wt) + Bw*cos(wt)
0 = -Aw*sin (0) + Bw*cos(0)
B = 0
- The required displacement-time relationship for SHM:
x(t) = d*cos ( wt )
w = √(k/m)