Answer:
Yes
Explanation:
The cart will definitely behave differently because more weight is added to it.
In order to successfully push a cart, the applied force must exceed the force of the weight of the cart along with its constituents and the frictional force between the tyres and the floor.
<em>When more weights are added (the 2 packs of water bottles), it means that more force would be needed to overcome the opposing forces to the movement of the cart in order to successfully push the cart again.</em>
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
Answer:
most likely 0m/s bc it was dropped, so it had to be still before
Explanation:
Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg
The second law states that the total entropy can never decrese over time for an isolated system