That is the mst best eway to find its solution.
37.4/2.2*10^3 = 0.017 gm/liter or 1.7*10^-2
so we conclude that option b is sorrect
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Answer:
y = 4 Sin (2πt)
Explanation:
Amplitude, A = 4
frequency, f = 1
Wave function is given by
y = A sinωt
where, ω is angular frequency
ω = 2 π f = 2 π x 1 = 2π
So, the desired wave function
y = 4 Sin (2πt)
Multiply it by a fraction equal to ' 1 ', like this:
(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter
Answer: 1.28 sec
Explanation:
Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:
V = ∆x / ∆t
Solving for ∆t, we have:
∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec
