Answer:
Explanation:
Did you mean: V = d/t a = (V - Vit Average = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Showing results for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = "-9.81" m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Search instead for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
The answer is A cause gas clouds hold elements in it
Answer:Noble gases:
are highly reactive.
react only with other gases.
do not appear in the periodic table.
are not very reactive with other elements.
Explanation:Noble gases:
are highly reactive.
react only with other gases.
do not appear in the periodic table.
are not very reactive with other elements.
Answer:
I believe that it is the 2nd option.
Explanation:
My reasonings are because C4H10O has 7 isomers. In which 4 are alcohol and the other 3 are ether.
The first option is ethers, specifically ethoxyethane.
The third option is ethers, specifically 1-methoxypropane.
The fourth option is an alcohol, specifically 1- butanol.
Therefore, leads us to the 2nd option that it is NOT an isomer of C4H10O
<span>the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
molar mass of Na</span>₂SO₄<span> is - 142 g/mol.
1 mol of </span>Na₂SO₄<span> has a mass of 142 g.
In 1 mol of </span>Na₂SO₄<span> the mass of Na is 23 g/mol x 2 = 46 g.
Mass of Na in 1 mol of </span>Na₂SO₄ is - 46 g
mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
mass of Na is 37.72 g
