Answer:
a) 0m
b) 6m
Explanation:
First, we need to remember:
Displacement: Difference between final and initial position.
Distance traveled: Total distance traveled.
a) If the final position is the same as the initial position, then:
final position = initial position
And we know that:
displacement = final position - initial position = 0
Then the displacement of the book is zero.
b)
We can assume that the book traveled along the perimeter of the table.
The table is a rectangle of width 1.2m and length 1.8m
Remember that for a rectangle of width W and length L, the perimeter is:
P = 2*L + 2*W
Then the perimeter of the table is:
P = 2*1.2m + 2*1.8m = 6m
This means that the distance traveled by the book is 6 meters.
The force required to push the box upward is 145.3N and the force to pus the box downward is -109.3N
Data given;
- mass = 15kg
- angle = 30 degree
- acceleration = 1.2 m/s^2
- acceleration due to gravity = 9.8 m/s^2
<h3>Force against gravity</h3>
To move the plane upward, the box will move against gravity.

Let's solve for F

<h3>Force towards gravity</h3>
When the force pushes the box down the inclined plane, it moves towards gravity.

The force required to push the box upward is 145.3N and the force required to push the box downward is -109.3N
Learn more on force across an inclined plane here;
brainly.com/question/11888124
Answer:
0.1m/s²
Explanation:
Using the equation of motion
V = u+at
V is the final speed = 0.9m/s
Initial speed u = 0.5m/s
Time = 4secs
Get the acceleration
0.9 = 0 5+4a
0.9-0.5 = 4a
0.4 = 4a
a = 0.4/4
a = 0.1m/s²
Hence the acceleration is 0.1m/s²
Stop and look. always treat a railroad crossing with no lights or gates like a stop sign.
Answer
given,
mass of the solid door = 22 Kg
dimension of door = 220 cm x 91 cm
moment of inertia about the hinge

r is the distance from the one edge which is equal to 91 cm or 0.91 m


Moment of inertia about center for rectangular gate is equal to

moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge




