Answer:
a)Work done by fireman= 2.15 Btu
b) Time t= 0.86 sec
Explanation:
Given that
Weight = 280 lbf
We know that 1 lbf = 4.44 N
so 280 lbf = 1245.5 N
Weight =1245.5 N
Height h = 60 ft
We know that
1 ft = 0.3048 m
So 60 ft = 18.28 m
h =18.28 m
Power = 3.5 hp
We know that
1 hp =0.74 KW
So 3.5 hp = 2.61 KW
Power = 2.61 KJ/s
So the work done by fireman = Weight x h
Now by putting the values
Work done by fireman= 1245.5 x 18.28 J
Work done by fireman= 2267.74 J
Work done by fireman= 2.26774 KJ
We know that 1 Btu= 1.05 KJ
So 2.266 KJ = 2.15 Btu
Work done by fireman= 2.15 Btu
We know that ,rate of work is called power.
Power x time = work
2.61 x t = 2.26
So t= 0.86 sec
Answer:
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Explanation:
Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.
In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.
Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.
Answer:
P₂ = 1.22 kPa
Explanation:
This problem can be solved using the equation of state:

where,
P₁ = initial pressure = 1 KPa
P₂ = final pressure = ?
V₁ = initial Volume = 1 liter
V₂ = final volume = 1.1 liter
T₁ = initial temperature = 290 k
T₂ = final temperature = 390 k
Therefore,

<u>P₂ = 1.22 kPa</u>
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
Explanation :
It is given that,
Diameter of the coil, d = 20 cm = 0.2 m
Radius of the coil, r = 0.1 m
Number of turns, N = 3000
Induced EMF, 
Magnitude of Earth's field, 
We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :




So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.