Answer:
72.98 km
Explanation:
Her displacement is simply the distance from her final position to her initial position.
Now, I've drawn and attached a triangle diagram to depict this her movement.
Point O is her initial starting point.
Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.
From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.
Thus;
x² = 62² + 26² - 2(62 × 26)cos 120
x² = 4520 + 806
x² = 5326
x = √5326
x = 72.98 km
Answer:
Vr = 20 [km/h]
Explanation:
In order to solve this problem, we have to add the relative velocities. We must remember that velocity is a vector, therefore it has magnitude and direction. We will take the sea as the reference measurement level.
Let's take the direction of the ship as positive. Therefore the boy moves in the opposite direction (Negative) to the reference level (the sea).
![V_{r}=30-10\\V_{r}=20 [km/h]](https://tex.z-dn.net/?f=V_%7Br%7D%3D30-10%5C%5CV_%7Br%7D%3D20%20%5Bkm%2Fh%5D)
Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
A “real” image occurs when light rays actually intersect at the image, and become inverted, or turned upside down. ... In flat, or plane mirrors, the image is a virtual image, and is the same distance behind the mirror as the object is in front of the mirror. The image is also the same size as the object.