Does anyone know how to perform better/quicker in odysseyware english 4/12? i’ve got less than a month to finish this

Does anyone know how to perform better/quicker in odysseyware english 4/12? i’ve got less than a month to finish this class and my reading has significantly slowed down, so i’m having a problem completing all my work. i’m pretty sure there’s no way to find all the answers on the internet.

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  1. First let's find the acceleration required in the barrel to speed the ball up from 0 to 83 m/s in a distance of 2.17 m. We know the force the cannon exerts on the cannonball is 20000 N; if we can find this acceleration then we can use F = ma to find the mass. We can find the acceleration using one of the kinematic equations of motion. We have: u = initial speed = 0 m/s v = final speed = v0 = 83 m/s d = distance = 2.17 m a = acceleration = ? v² = u² + 2ad. Since u = 0, this reduces to v² = 2ad and rearranges to a = v²/2d = 83²/2*2.17 = 83²/4.34 = 1587.327 m/s². Now F = ma, so m = F/a = (20000N)/(1587.327 m/s²) = 12.6 kg. For part 2, use the Range Equation: If R is the horizontal distance the cannonball travels, v = v0 = the initial velocity = 83 m/s g = acceleration due to gravity - 9.8 m/s² x the launch angle relative to the horizontal, then R = (v²sin(2x))/g. So R = (83²sin(2*37))/9.8 = (6889sin74)/9.8 = 676 m. So the target ship is 676 m away.

  2. $1,848 because you’d multiply 1,650 by .12 which is the percentage which equals to 198 which is the 12%. Then you just add 1650 and 198

  3.  I have until the end of this month to complete two classes so I feel your stress. What program are you using? I'm on invirtual.agilixbuzz  which don't have any set schedule on when to complete specific assignments, but the class ends whether you have finished everything or not. I'd be glad to help with anything, I believe I have an A+ currently in English, and I was in English-H at my public high school

  4. The EMF or electromotive force is the energy supplied by a battery or a cell per coulomb (Q) of charge passing through it. The magnitude of emf is equal to V (potential difference) across the cell terminals when there is no current flowing through the circuit. (byjus)


    [tex]How can we calculate the e.m.f of the battery?.[/tex]

  5. Option D is the correct choice.

    Step-by-step explanation:

    Let m be the number of marbles per pack. Since we are told that total mass of each marble is  [tex]4\frac{1}{2}[/tex], the mass of all marbles in one pack will be  [tex]4\frac{1}{2}*m[/tex].

    There are such 6 packs so mass of all marbles in 6 packs will be [tex]6*(4\frac{1}{2} m)[/tex].

    We are also given that mass of each packaging is [tex]\frac{2}{3}[/tex]g we can find mass of all 6 packaging as [tex]6*(\frac{2}{3})[/tex] .

    Now we will write an equation using this information.

    [tex]6(4\frac{1}{2}*m+\frac{2}{3} )=629[/tex]

    We can see that option D matches with our answer, therefore option D is the correct choice.

  6. Explanation:

    The emf is equal to the work done on the charge per unit charge (ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt (1V=1J/C).

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