Answer:
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Explanation:
Answer:
A) 5.2 x 10³ N
B) 8.8 x 10³ N
Explanation:
Part A)
= weight of the craft in downward direction = tension force in the cable when stationary = 7000 N
= Tension force in upward direction
= Drag force in upward direction = 1800 N
Force equation for the motion of craft is given as
-
-
= 0
7000 - 1800 -
= 0
= 5200 N
= 5.2 x 10³ N
Part B)
= weight of the craft in downward direction = tension force in the cable when stationary = 7000 N
= Tension force in upward direction
= Drag force in downward direction = 1800 N
Force equation for the motion of craft is given as
-
-
= 0
- 7000 - 1800 = 0
= 8800 N
= 8.8 x 10³ N
Answer:
Plants are a good starting point when looking at the carbon cycle on Earth. Plants have a process called photosynthesis that enables them to take carbon dioxide out of the atmosphere and combine it with water. Using the energy of the Sun, plants make sugars and oxygen molecules.
Anything times zero is zero
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671