Answer:

Explanation:
Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

now we can say that


now electric potential is given as


now work done to bring a small charge from infinite to the surface of this sphere is given as


here we know that

now the total energy of the sphere is given as




Pressure is the amount of force per unit area. In formula it
is,
P = F ÷ A
P = 7000 N ÷ 0.4 m2
P = 17,500 N/m2
The amount of pressure the truck exerts on the piston is
17,500 N/m2
☁️ Answer ☁️
annyeonghaseyo!
Your answer is:
True.
Several simple machines change the direction of the applied force. These include lever, fulcrum and the pulley.
Hope it helps.
Have a nice day hyung/noona!~  ̄▽ ̄❤️
Answer:
the human brain is responsible for involuntary activities, such as regulating heartbeat, breathing, and blinking. Although the brain controls both voluntary and involuntary activities, different regions of the brain are devoted to each type of task.
Both involuntary and voluntary actions are controlled by the same parts of the brain. Hindbrain and midbrain control involuntary actions like salivation, vomiting, etc. All the body's voluntary actions are controlled by the motor cortex in the frontal lobe of the cerebrum.
Explanation:
Answer:
System inputs and outputs fluctuate around a stable average so the system does not move far from its average condition.
Explanation:
Steady-state equilibrium can also be called dynamic equilibrium. The main difference between the two is the type of system we are considering. When a system is closed, and it attains equilibrium, there is no transfer of energy. In the case of an open system, even if the system achieves equilibrium there will be some transfer of energy but it will not deviate far from it's equilibrium point, that is, it will be in a steady-state.