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The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
The given parameters;
- initial horizontal velocity, vₓ = 16 m/s
- initial vertical velocity,
- time interval 1 seconds
The components of the velocity can be horizontal or vertical velocity.
The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.
The vertical component of the velocity is calculated as;
The horizontal component of the velocity is constant since it is not affected by gravity.
The horizontal component of the velocity = 16 m/s
Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
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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
Explanation:
Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their
dipole - dipole attractions and hence hydrogen bonds are formed.
However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.
The ozone layer that is inside the stratosphere blocks UV radiation.
The ozone layer contains high concentrations of ozone relative to other parts of the atmosphere. This was discovered by Charles Fabry and Henri Buisson who are both French Physicists.
The ozone in the earth's stratosphere is created through ultraviolet light striking a group of ordinary oxygen molecules containing two oxygen atoms, subsequently splitting them into individual oxygen atoms and finally these said atomic oxygen then combines with unbroken O2 to create ozone (O3).
