Complete Question
A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).
J
(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)
MW
Answer:
The heat transferred is
The power is
Explanation:
From the question we are told that
Mass of the water per second is
The initial temperature of the water is
The boiling point of water is
The final temperature
The latent heat of vapourization of water is
The specific heat of water
The specific heat of stem is
Generally the heat needed each second is mathematically represented as
Then substituting the value
The power required is mathematically represented as
From the question
So
Answer:
The average net force on the truck is 375 Newtons.
Explanation:
Using Newton's 3rd equation of motion, we have :
×a×s
where, v = final velocity = 25 m/s
u = initial velocity = 20 m/s
a = acceleration
s = distance traveled = 300 m
Using these values in the above equation, we get acceleration = 0.375 m/
Using Newton's second law, we have:
F=m×a
where m = mass = 1000 kg
a= acceleration = 0.375 m/
Putting values we have F=375 N
Answer:
What is the speed of the mountain lion as it leaves the ground?
9.98m/s
At what angle does it leave the ground?
50.16°
Explanation:
This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.
To solve this, you will need use two kinematic equations and SOHCAHTOA:
With these formulas, we can derive formulas for everything you need:
Things you need to remember:
First we need to take your given:
10.0 m long (horizontal) and maximum height of 3.0m (vertical).
What your problem is looking for is the initial velocity and the angle it left the ground.
Vi = ? Θ =?
Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.
Let's deal with the vertical components first:
We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.
Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:
From here we can derive the formula of time:
Now we just plug in what we know:
Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.
Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.
So we have our first initial vertical velocity:
Viy = 7.66m/s
Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:
But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.
T= 2t
T = 2 (0.782s)
<em>T = 1.564s</em>
<em>So we use this in our formula:</em>
<em></em>
Vix=6.39m/s
Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)
To get the diagonal, you just use the Pythagorean theorem:
c²=a²+b²
Using it in the context of our problem:
The lion leapt at 9.98m/s
Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:
The lion leapt at an angle of 50.16°.
