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scoray [572]
3 years ago
5

a sample of steam at 15 bar pressure and 400°c temperature is first expanded at a constant enthalpy to 6 bar and then expanded i

sentropically where the dryness fraction of the steam is 0.90. find the change of enthalpy, the pressure of isentropically expanded steam and change of entropy.
Physics
1 answer:
kow [346]3 years ago
3 0

Answer:

Answer.

Explanation:

Explanation.

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You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi
True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0
3 years ago
100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.
FinnZ [79.3K]
A) The power delivered to the lines is
P_{in}= 100 kW=1 \cdot 10^5 W
And the voltage at which the lines work is
V=13014.1 V
Since the power delivered is the product between the voltage and the current:
P=VI
We can find the current flowing in the lines:
I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A

b) The voltage change along each line can be found by using Ohm's law:
\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V

c) The power wasted as heat along each line is given by:
P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W
And since we have 2 lines, the total power wasted as heat in both lines is
P_d = 2 \cdot 590 W=1180 W
6 0
3 years ago
Positive electric charges are always attracted to ________ charges.
Georgia [21]

Answer:

Negative electric charges

4 0
3 years ago
In uniform circular motion, which of the following quantities are constant?
Sindrei [870]

Answer:

speed

Explanation:

When a body is in uniform circular motion, its speed remains constant but its velocity, angular acceleration, angular velocity changes due to change in its direction.

6 0
2 years ago
Compare the momentum of a softball and a basketball if they are both moving at the same velocity
kondor19780726 [428]
Softball weigh around 180 grams or that would be 0.18 Kg. Basketball usually weighs around 22 ounce or 0.6 Kg. To compare the momentum of the two ball we have to use the formula of momentum which is p = mv. m for mass and v for velocity. let us assume both ball are travelling 2 m/s).

Softball. p = mv = (0.18 Kg)(2 m/s) = 0.36 Kg.m/s ( smaller momentum)

Basketball p = mv = (0.6 Kg)(2 m/s) = 1. 2 Kg.m/s (larger momentum
6 0
3 years ago
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