The work done on the barbell is -165.62 Nm.
Explanation:
Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.
In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.
Thus, 
Now, the force is -127.4 N and the displacement is 1.3 m.
So, 
So, the work done on the barbell is -165.62 Nm.
Answer:
The water level will drop by about 1.24 cm in 1 day.
Explanation:
Here Mass flux of water vapour is given as
where
is the mass flux of the water which is to be calculated.
- D is diffusion coefficient which is given as
- l is the thickness of the film which is 0.15 cm thick.
is given as
In this
is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
is the air pressure which is given as 0.5 times of
- R is the universal gas constant as
- T is the temperature in Kelvin scale which is
By substituting values in the equation
Converting into 
As 1 mole of water 18 
so
Putting this in the equation of mass flux equation gives
For calculation of water level drop in a day, converting mass flux as
So the water level will drop by about 1.24 cm in 1 day.
Answer:
6N
Explanation:
Given parameters:
Pressure applied by the woman = 300N/m²
Area = 0.02m²
Unknown:
Force applied = ?
Solution:
Pressure is the force per unit area on a body
Pressure = 
Force = Pressure x area
Force = 300 x 0.02 = 6N
Answer:
Answered
Explanation:
1 and 3 are necessary
Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.
If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.
If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.
