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 gas is ignited (I think) and combustion happens where the gasoline turns into gas (the state of being) and expands, pushing something and making the blades turn so
from stationary to explosive so potentioal to kenetic
Answer:
it would be option C
Explanation:
Speed of light = 3×10^8m/s
Planck's constant = 6.626×10^-34 Js
Wavelength = 8 x 10^-9 m
Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9
Energy = [19.878×10^(8-34)] / 8 x 10^-9
Energy = 2.48475 × 10^(-26+9)
Energy = 2.48×10^-17 J
The question is incomplete. Complete question is attached below:
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Correct Answer is : 54 131Xe
Reason:
Emission of beta particle from radioactive material increases atomic number by one, but atomic mass number donot change.In present case, the reactant is 53 131 I. When, iodine emits beta particle i.e. electron, it atomic number increases by one. Hence, it gets converted into Xe. Atomic number of Xenon is 54.
