Answer:
The act of using senses or tools to gather information is called <em>Obser</em><em>vation</em><em>.</em>
Answer:

Explanation:
We are given that







We have to find the exit temperature.
By steady energy flow equation



Substitute the values




Answer:
26b) 66.7%
27) 500 N
Explanation:
26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.
26.b) Mechanical advantage is the ratio between the forces:
MA = load force / effort force
Efficiency is the ratio between the work:
e = work done on load / work done by effort
Work is force times distance.
e = (F load × d load) / (F effort × d effort)
Rearranging:
e = (F load / F effort) × (d load / d effort)
e = MA × (d load / d effort)
In a two pulley system, the load moves half the distance of the effort. So the efficiency is:
e = (4/3) × (1/2)
e = 2/3
e = 66.7%
27) In a three pulley system, the load moves a third of the distance of the effort.
e = (F load / F effort) × (d load / d effort)
0.40 = (600 N / F) × (1/3)
F = 500 N
I used to wish that I can fly
Answer:
The magnitude of electric force is 
Explanation:
Coulomb's Law:
The force of attraction or repletion is
- directly proportional to the products of charges i.e

- inversely proportional to the square of distance i.e


[ k is proportional constant=9×10⁹N m²/C²]
There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C
Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C
and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C
Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.
If we draw a line from q₁ to Q .
The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.
Let hypotenuse = r
Therefore, 
we know,


Total force 


[ r=5]
N
The magnitude of electric force is 