Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
Answer:
E = 1440 kJ
Explanation:
It is given that,
Power of a cooker oven is 800 W
Voltage at which it is operated is 230 V
Time, t = 30 minutes = 1800 seconds
We need to find the electrical energy used by the cooker oven. The product of power and time is equal to the energy consumed. So,

So, electrical energy of 1440 kJ is consumed by the cooker oven.
Answer:
103239.89 days
Explanation:
Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)
where,
a is the distance of the semi-major axis in a.u
T is the orbit time in days
Converting the mean distance of the new planet to astronomical unit (a.u.)
1 a.u = 9.296 × 10⁷ miles

Substituting the values into Kepler's third law equation;
(days)²

T = 103239.89 days
An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days
Answer:
Explanation:
ignore air resistance
Let t be the time of fall for the dropped stone.
½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²
4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)
4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716
0 = 21.56t - 71.148
t = 71.148/21.56 = 3.3 s
h = ½(9.8)3.3² = 53.361 = 53 m
or
h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m
Answer:A
Explanation:
In R-L circuit current is given by
![i=i_0\left [ 1-e^{\frac{-t}{L/R}}\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-e%5E%7B%5Cfrac%7B-t%7D%7BL%2FR%7D%7D%5Cright%20%5D)
where i=current at any time t

R=resistance
L=Inductance
at t=0
approaches to 1
therefore ![i=i_0\left [ 1-1\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-1%5Cright%20%5D)
i=0
when t approaches to
,
approaches to zero
thus 
thus we can say that initially circuit act as broken wire with zero current
and it increases exponentially with time and act as ordinary connecting wire