In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
- vocational aim
- cultural aim
- spiritual aim
- intellectual aim
Answer:
0.25 m
Explanation:
We can solve the problem by using the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have
f = +20 cm=+0.20 m (the focal length is positive for a converging lens)
q = +1.0 m (the image distance is positive for a real image)
Solving the equation for p, we find
Answer:
The potential energy at point A is 17.1675 J
Explanation:
The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries
In meteorology it is used to describe the level of saturated soil above the water table
Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are
Point A, elevation = 75 cm capillary potential = -100 cm
Point B, elevation = 25 cm capillary potential = -200 cm
The total potential energy at point A is
Elevation above reference - capillary potential =75-(-100) = 175 cm
which gives per unit mass
PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J
Answer
given,
mass of satellite = 545 Kg
R = 6.4 x 10⁶ m
H = 2 x 6.4 x 10⁶ m
Mass of earth = 5.972 x 10²⁴ Kg
height above earth is equal to earth's mean radius
a) satellite's orbital velocity
centripetal force acting on satellite = 
gravitational force = 
equating both the above equation
v = 5578.5 m/s
b) 
T = 14416.92 s
T = 4 hr
c) gravitational force acting
F = 5202 N
