Answer
Ceres, Pluto, and Eris are categorized as DWARF PLANETS.
A) Remaining planetesimals formed within the frost line are referred to as ASTEROIDS.
B) METEORITES are fragments of asteroids that have landed on Earth.
C) COMETS are celestial objects that are often visible with their long tails.
D) COMETS are also planetesimals that were left over and originated in the region of the solar system dominated by the jovian planets.
E) Meteor showers are linked to debris from COMETS.
Response:
C. vx
F. ax
G. ay
Clarification:
The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.
Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.
The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.
Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.
<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack.
centripetal force = weight of the ball
m v^2 / r = m g
v^2 / r = g
v^2 = g r
v = sqrt { g r }
v = sqrt { (9.80~m/s^2) (0.7 m) }
v = 2.62 m/s
Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>
assuming north-south is along the Y-axis and east-west along the X-axis
X = total X-displacement
from the graph, total displacement in the X-direction is computed as
X = 0 - 20 + 60 Cos45 + 0
X = 42.42 - 20
X = 22.42 m
Y = total Y-displacement
from the graph, total displacement in the Y-direction is computed as
Y = 40 + 0 + 60 Sin45 + 50
Y = 90 + 42.42
Y = 132.42 m
To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding
magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m
Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east
To tackle this question, we know the following:
1 Albert equals 88 meters.
1 A = 88 m.
Initially, we square both sides of the equation:
(1 A)^2 = (88 m)^2
1 A^2 = 7,744 m^2
<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>
1 A^2 / 7,744 = 7,744 m^2 / 7,744
(1 / 7,744) A^2 = 1 m^2
Then multiply both sides by 4,050.
(4050 / 7744) A^2 = 4050 m^2
0.523 A^2 = 4050 m^2
<span>Thus, one acre is approximately 0.52 square alberts.</span>
