Inertia= inactivity
The factor involved in increasing inertia would be "mass". The more mass something has, the more inertia.
Answer:
net torque on the system is 2.5 Nm
Explanation:
As we know that net torque on the system is given as
now we have
now we have
Since the system itself is giving off heat, this is a
reduction in the internal energy.
heat = - 25,000 J
Since work is being done on the system, therefore it is
an additional energy to the system. Work is given as:
work = - P dV
work = - 1.50 atm (6 L – 12 L)
work = 9 L atm
Since it is given that 1 L atm is equivalent to 101.3 J,
therefore the total energy added is:
energy due to work = 9 L atm (101.3 J / 1 L atm)
energy due to work = 911.7 J
Therefore the total change in internal energy is the sum
of heat and energy due to work:
Change in internal energy = - 25,000 J + 911.7 J
Change in internal energy = - 24,088.3 J
<span>Therefore, approximately 24.1 kJ of energy is lost by the
system in the total process.</span>
<span>
</span>
<span>Answer:</span>
<span>-24.1 kJ</span>
Answer:
D. 0.9
Explanation:
Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.
Resolving into x component:
∑
------(1)
Resolving into y component:
∑
------(2)
Static frictional force, 
μ 
------(3)
substituting 
from equation (1) and from equation (2) into equation (3)
μ 
μ 
μ 
μ 
The angle the vehicles make with the horizontal α = 42°
μ ≥ tan 42°
μ ≥ 0.9
Answer:
We know that force applied per unit area is called pressure.
Pressure = Force/ Area
When force is constant than pressure is inversely proportional to area.
1- Calculating the area of three face:
A1 = 20m x 10 m =200 Square meter
A2 = 10 mx 5 m = 50 Square meter
A3 = 20m x 5 m = 100 Square meter
Therefore A1 is maximum and A2 is minimum.
2- Calculate pressure:
P = F/ A1 = 30 / 200 = 0.15 Nm⁻² ( minimum pressure)
P = F / A2 = 30 / 50 = 0.6 Nm⁻² ( maximum pressure)
Hence greater the area less will be the pressure and vice versa.
