Physics<span> is a natural science that involves the study of matter and its motion through space time, along with related concepts such as energy and force.</span>
The forces move strongly towards the left by 1N
Given the following
Force towards the right = 4N
Force towards the left = 5N
Note that the force acting towards the left is negative, hence the force acting towards the left is -5N
Take the sum of force
Resultant force = -5N + 4N
Resultant force = -1N
This shows that the forces move strongly towards the left by 1N
Learn more here: brainly.com/question/24629099
<span>Actually the second law of thermodynamics would truly gets violated ie, which means that the entrophy changes of the isolated system can never be negative, which covers the above that if heat were to spontaneously flow between any two objects of equal temperature would be fully violated.</span>
Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is

So the volume flow rate along the pipe is

We can use the similar logic to find the cross-section area at the refinery

The radius of the pipe at the refinery is:



So the diameter is twice the radius = 0.38*2 = 0.754m
Answer:
6.0 ×
W/
Explanation:
From Wien's displacement formula;
Q = e A
Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.
The emissive intensity =
= e
Given from the question that: e = 0.6 and T = 1000K, thus;
emissive intensity = 0.6 × 
= 0.6 × 1.0 × 
= 6.0 ×

Therefore, the emissive intensity coming out of the surface is 6.0 ×
W/
.