What is true of all particles of gas enclosed within a container?
2 answers:
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Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like
given that
here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2
now from above equation
So force applied by the person must be 150 N
Answer:
X-Positions: Y-Positions
x(0) = 0 y(0) = 0
x(2) = 120 m y(2) = 19.6 m
x(4) = 240 m y(4) = 78.4 m
x(6) = 360 m y(6) = 176.4 m
x(8) = 480 m y(8) = 313 m
x(10) = 600m y (10) = 490 m
Explanation:
X-Positions
- First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
- After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
- So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:
- It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.
Y-Positions
- We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
- As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
- At any moment, it is subject to the acceleration of gravity, g.
- As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:
- Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
- y(2) = 2* 9.8 m/s2 = 19.6 m
- y(4) = 8* 9.8 m/s2 = 78.4 m
- y(6) = 18*9.8 m/s2 = 176.4 m
- y(8) = 32*9.8 m/s2 = 313.6 m
- y(10)= 50 * 9.8 m/s2 = 490.0 m
