The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
- The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.
Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, <em>C</em> represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
The steepness of the slope is therefore;
Where;
= Half the width of the truck = 
= 1.2 m
= The elevation of the center of gravity above the ground = 2.2 m
The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.
Learn more here:
brainly.com/question/20793607
<span>4.5 m/s
This is an exercise in centripetal force. The formula is
F = mv^2/r
where
m = mass
v = velocity
r = radius
Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop.
Let's determine the force we get from gravity.
0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N
Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N
Plug known values into formula.
F = mv^2/r
13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m
6.88064 kg m^2/s^2 = 0.34 kg V^2
20.23717647 m^2/s^2 = V^2
4.498574938 m/s = V
Rounding to 2 significant figures gives 4.5 m/s
The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
Answer:
the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
Explanation:
This exercise asks to describe the inflation situation of a spherical fultball.
Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.
When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.
Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
Rotational kinetic energy <span>is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.</span>
A) The lifter's and the barbell's force are equal and opposite in direction
B) The lifter's force is greater than that of the barbell and in the upward direction
C) The barbell's force is greater than that of the lifter and in the downward direction
