When you ask for help with a crossword clue, don't forget to tell us how many letters it has. For this one, see if "temperature" fits the space.
Answer:72,000
Explanation:
I believe the answer is a. Because the formula of kinetic energy is 1/2(m)•(v^2)
Answer:
I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.
Explanation:
In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:
R = V₀² Sin 2θ/g
where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.
The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:
V₀ = √(V₀ₓ² + V₀y²)
where,
V₀ₓ = Horizontal Velocity
V₀y = Vertical Velocity
Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.
<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>
Answer:
1.6 m/s2
Explanation:
Let
be the gravitational acceleration of the moon. We know that due to the law of energy conservation, kinetic energy (and speed) of the rock when being thrown upwards from the surface and when it returns to the surface is the same. Given that
stays constant, we can conclude that the time it takes to reach its highest point, aka 0 velocity, is the same as the time it takes to fall down from that point to the surface, which is half of the total time, or 4 / 2 = 2 seconds.
So essentially it takes 2s to decelerate from 3.2 m/s to 0. We can use this information to calculate 

So the gravitational acceleration on the Moon is 1.6 m/s2