Answer:
P-waves travel through liquids and solid while S-waves only travel through solids.
Explanation:
Scientists are able to use the fact that P-waves travel through both solids and liquids and waves travel through only solids to determine what makes the different layers of the Earth.
Answer:
7.48 x 10⁵ m
Explanation:
g = 7.86 N/kg
M = 5.97 x 10²⁴ kg, R = 6.37 x 10⁶ m.
Find height h
g = GM/(R + h)²
(R + h)² = GM/g = 6.67 x 10⁺¹¹ x 5.97 x 10²⁴ /7.86 = 5.066 x 10¹³
R + h = 7.12 x 10⁶ m
so
h = 7.12 x 10⁶ - 6.37 x 10⁶ = 7.48 x 10⁵ m
Hi there!
(a)
Recall that:
![W = F \cdot d = Fdcos\theta](https://tex.z-dn.net/?f=W%20%3D%20F%20%5Ccdot%20d%20%3D%20Fdcos%5Ctheta)
W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.
![W =248(56)cos(30) = 12027.36 J](https://tex.z-dn.net/?f=W%20%3D248%2856%29cos%2830%29%20%3D%2012027.36%20J)
To the nearest multiple of ten:
![W_A = \boxed{12030 J}](https://tex.z-dn.net/?f=W_A%20%3D%20%5Cboxed%7B12030%20J%7D)
(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:
![\boxed{W_g = 0 J}](https://tex.z-dn.net/?f=%5Cboxed%7BW_g%20%3D%200%20J%7D)
(c)
Similarly, the normal force is perpendicular to the displacement, so:
![\boxed{W_N = 0 J}](https://tex.z-dn.net/?f=%5Cboxed%7BW_N%20%3D%200%20J%7D)
(d)
Recall that the force of kinetic friction is given by:
![F_{f} =\mu_k mg](https://tex.z-dn.net/?f=F_%7Bf%7D%20%3D%5Cmu_k%20mg)
Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
![W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J](https://tex.z-dn.net/?f=W_f%20%3D%20-%5Cmu_k%20mgd%5C%5CW_f%20%3D%20-%20%280.1%29%2856%29%289.8%29%2856%29%20%3D%20-3073.28%20J)
In multiples of ten:
![\boxed{W_f = -3070 J}](https://tex.z-dn.net/?f=%5Cboxed%7BW_f%20%3D%20-3070%20J%7D)
(e)
Simply add up the above values of work to find the net work.
![W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J](https://tex.z-dn.net/?f=W_%7Bnet%7D%20%3D%20W_A%20%2B%20W_f%20%5C%5C%5C%5CW_%7Bnet%7D%20%3D%2012027.36%20%2B%20%28-3073.28%29%20%3D%208954.08%20J)
Nearest multiple of ten:
![\boxed{W_{net} = 8950 J}}](https://tex.z-dn.net/?f=%5Cboxed%7BW_%7Bnet%7D%20%3D%208950%20J%7D%7D)
(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
![F_{net} = F_{Ax} - F_f](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_%7BAx%7D%20-%20F_f)
![W = F_{net} \cdot d = (F_{Ax} - F_f)](https://tex.z-dn.net/?f=W%20%3D%20F_%7Bnet%7D%20%5Ccdot%20d%20%3D%20%28F_%7BAx%7D%20-%20F_f%29)
![W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J](https://tex.z-dn.net/?f=W%20%3D%20%28F_Acos%2830%29%20-%20%5Cmu_k%20mg%29d%5C%5CW%20%3D%20%28248cos%2830%29%20-%200.1%2856%29%289.8%29%29%20%2A%2056%20%5C%5C%5C%5CW%20%3D%208954.08%20J)
Nearest multiple of ten:
![\boxed{W_{net} = 8950 J}](https://tex.z-dn.net/?f=%5Cboxed%7BW_%7Bnet%7D%20%3D%208950%20J%7D)
Im sorry may you please retake the picture then i will answer