Answer:
P_(pump) = 98,000 Pa
Explanation:
We are given;
h2 = 30m
h1 = 20m
Density; ρ = 1000 kg/m³
First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,
Thus, it can be expressed as;
P_(tank)+ P_(pump) = P_(nozzle)
Now, the pressure would be given by;
P = ρgh
So,
ρgh_1 + P_(pump) = ρgh_2
Thus,
P_(pump) = ρg(h_2 - h_1)
Plugging in the relevant values to obtain;
P_(pump) = 1000•9.8(30 - 20)
P_(pump) = 98,000 Pa
Answer:
Half life of S = 3.76secs
Explanation:
The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.
For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564
for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615
Simply say = 1/3.9 + 1/104 = 0.266015
So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active
The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle

between the direction of the force and the direction of the displacement:

In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so

Therefore, the work done is
Answer:
There is 148.35 Joules of heat is released in the process.
Explanation:
Given that,
Heat capacity of the object, 
Initial temperature, 
Final temperature, 
We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Let the mass of the object is 10 g or 0.01 kg
So,

Q = 148.35 Joules
So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.