Answer:
1.6 ft/min
Explanation:
Since trough is 10 ft long and water is filled at the rate of 12ft3/min. We can calculate the rate of water filled with respect to area:
= 12 / 10 = 1.2ft2/min
As the water level rises, so does the water surface, or the bottom side of the isosceles triangles. In fact we can calculate the bottom side when the trough is half foot deep:
= 3 / 2 = 1.5 ft
The rate of change in water level would be the same as calculating the height of the isosceles triangles knowing its base
= 1.2 * 2 / 1.5 = 1.6 ft/min
The atomic number is the same as the proton number so the answer would be D) 10
Answer
given,
length of the swing = 26.2 m
inclined at an angle = 28°
let, the initial height of the Tarzan be h
h = L (1 - cos θ)
a) initial velocity v₁ = 0 m/s
final velocity of Tarzan = v_f
law of conservation of energy
PE_i + KE_i = PE_f + KE_f
= 7.75 m/s
the speed tarzan at the bottom of the swing
v_f = 7.75 m/s
b)initial speed of the = 3 m/s
v_f= 11.29 m/s
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
