Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
P = S.G*p_w*g*h
Where, h = P / S.G*p_w*g
- Input the values given:
h = 101.325 KPa / 0.739*9.81*997
h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.
Answer
given,
current (I) = 16 mA
circumference of the circular loop (S)= 1.90 m
Magnetic field (B)= 0.790 T
S = 2 π r
1.9 = 2 π r
r = 0.3024 m
a) magnetic moment of loop
M= I A
M=
M=
M=
b) torque exerted in the loop
Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J
Solution :
(1) The equation used is,
where,
= final internal energy
= initial internal energy
q = heat energy
w = work done
(2) The known variables are, q, w and
initial internal energy = = 2000 J
heat energy = q = 1000 J
work done = w = 500 J
(3) Now plug the numbers into the equation, we get
(4) By solving the terms, we get
(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J
Answer:
F. 25.82 s
Explanation:
Given:
Δy = 90 m
v₀ = 0 m/s
a = 0.27 m/s²
Find: t
Δy = v₀ t + ½ at²
90 m = (0 m/s) t + ½ (0.27 m/s²) t²
t = 25.82 s
