Answer:
Negative
Explanation:
First law of thermodynamic also known as the law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
The first law relates relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.
The first law is mathematically given as ΔU =
-
= Q - W
Where Q = Quantity of heat
W = Work done
From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.
Analyzing the pistol when it raises in isothermal and when it falls in isobaric state.The following can be said:
In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.
In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.
In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.
Answer:
Part a)

Part b)

So this speed is independent of the mass of the rider
Explanation:
Part a)
By force equation on the rider at the position of the hump we can say

now we will have


now we have



Part b)
At the top of the loop if the minimum speed is required so that it remains in contact so we will have

at minimum speed




So this speed is independent of the mass of the rider
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Answer:
The weights are 1 kg, 3kg, 9kg and 27kg.
Explanation:
The weights are 1 kg, 3kg, 9kg and 27kg.
1+3+9+27= 40
27+9+3= 39
27+9+3-1=38
27+9+1=37
27+9=36
27+9-1=35
27+9+1-3=34
27+9-3=33
27+9-3-1=32
27+3+1=31
27+3=30
27+3-1=29
27+1=28
27
27-1=26
27+1-3=25
27-3=24
27-3-1=23
27+3+1-9=22
27+3-9=21
27+3-9-1=20
Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.

And the distance 'd' is



Through the free-body diagram the tension components are given by


Here we can watch that,

Dividing both expression we have that,

Replacing the values,


PART B) Using the vertical component we can find the tension,



