Answer:
1.24 m/s
Explanation:
Metric unit conversion:
9.25 mm = 0.00925 m
5 mm = 0.005 m
The volume rate that flow through the single pipe is

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

So the flow speed of each of the narrower pipe is:

Newton´s thrid law of motion states that:
Every action has an equal and opposite reaction.
c. equal, opposite.
Answer:
0.466 (3 sig. fig.)
Explanation:
Frictional force acting on the box = 5.00×10^2xsin25
Normal force acting on the box = 5.00×10^2xcos25
coefficient of friction = 0.466 (3 sig. fig.)
Answer:
True The net force must be zero for the acceleration to be zero
Explanation:
In order to analyze the statements of this problem we propose your solution.
First let's look at Newton's first, which stable that every object is at rest or with constant speed unless something takes it out of this state (acceleration)
Now let's look at the second postulate, which says that force is related to the product of the mass of a body and its acceleration.
As a result of these two laws, for a body is a constant velocity the summation force on it must be zero.
Now we can analyze the statements given.
True The net force must be zero for the acceleration to be zero
False. If the force is different from zero, there is acceleration that changes the speeds
False. There may be forces, but the sum of them must be zero
False. If a force acts, the acceleration is different from zero and the speed changes
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁