Answer:
1) Addition of a catalyst
2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)
Explanation:
1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.
You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.
<span>Answer: Force = 81.6 N
Explanation:
According to Newton's Second law:
F = ma --- (1)
Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2
Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>
Answer:
<h3>The answer is 0.59 m/s²</h3>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
f is the force
m is the mass
From the question we have
We have the final answer as
<h3>0.59 m/s²</h3>
Hope this helps you
-6.98 × 10-^7 is the answer <3
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
