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:
R = 2481 Ω
L= 1.67 H
Explanation:
(a) We have an inductor L which has an internal resistance of R. The inductor is connected to a battery with an emf of E = 12.0 V. So this circuit is equivalent to a simple RL circuit. It is given that the current is 4.86 mA at 0.725 ms after the connection is completed and is 6.45 mA after a long time. First we need to find the resistance of the inductor. The current flowing in an RL circuit is given by
i = E/R(1 -e^(-R/L)*t) (1)
at t --> ∞ the current is the maximum, that is,
i_max = E/R
solve for R and substitute to get,
R= E/i_max
R = 2481 Ω
(b) To find the inductance we will use i(t = 0.940 ms) = 4.86 mA, solve (1) for L as,
Rt/L = - In (1 - i/i_max
)
Or,
L = - Rt/In (1 - i/i_max
)
substitute with the givens to get,
L = -(2481 Si) (9.40 x 10-4 s)/ In (1 - 4.86/6.45
)
L= 1.67 H
<u><em>note :</em></u>
<u><em>error maybe in calculation but method is correct</em></u>
The kinetic energy for a large vehicle is different from that of a smaller vehicle, assuming that the vehicles are travelling at the same speed and stopping the same distance. This is because for a larger vehicle the kinetic energy is higher, as the mass for a larger vehicle, is more than the smaller vehicle.
Answer:
M = 1.38 10⁵⁹ kg
Explanation:
For this problem we will use the law of universal gravitation
F = G m₁ m₂ / r²
Where G is the gravitation constant you are value 6.67 10⁻¹¹ N m2 / kg2, m are the masses and r the distance
In this case the mass of the planet is m = 3.0 10²³ kg and the mass of the start is M
Let's write Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
The speed module is constant, so we can use the kinematic relationship
v = d / t
The distance remembered is the length of the circular orbit and the time in this case is called the period
d = 2π r
a = 2π r / T
Let's replace Newton's second law
G m M / r² = m (4π² r² / T²) / r
G M = 4 π² r³ / T²
M = 4 π² r³ / T² G
Let's calculate
M = 4 π² (3.0 10²³)³ / (3.4 10¹¹)² 6.67 10⁻¹¹
M = 13.82 10⁵⁸ kg
M = 1.38 10⁵⁹ kg