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:
B. Trial 2
Explanation:
Trial 2, because the student’s finger applied the largest force to the sensor.
Because the trial 2 student finger applied to largest force.
Answer:
7.3 kg m/s
Explanation:
First of all, let's calculate the gravitational potential energy of the stone as it reaches its highest point:
For the law of conservation of energy, this is equal to the initial kinetic energy of the stone at ground level (where the potential energy is zero), just after the stone leaves your hand:
From this equation we can find the velocity of the stone as it leaves your hand:
The initial velocity of the stone (before leaving your hand) is zero:
The impulse received by the stone is equal to its change in momentum, so:
Answer:
27 °C
Explanation:
BY the statement of the question it is clear that it is about an ideal gas - and hence if change in KE is about zero - then there will be no change of temperature.
So, answer is 27 °C
