Answer:
15 meters
Explanation:
The inicial energy of the ball is just potencial energy, and its value is:
E = m * g * h = m * g * 20,
where m is the ball mass, and g is the value of gravity.
In the moment that the ball strickes the ground, all potencial energy transformed into kinetic energy, and 25% of this energy is lost, so the total energy at this moment will be:
E' = 0.75 * E = 0.75 * m * g * 20 = 15*m*g
This kinetic energy will make the ball goes up again, and at the maximum height, all kinetic energy is transformed back into potencial energy.
So, as the mass and the gravity are constants, we can calculate the height the ball will reach:
E' = m*g*h = 15*m*g -> h = 15 meters
Answer:
0.0619 m^3
Explanation
number of moles = n = 4.39 mol
pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa
Molar gas constant =R = 8.31 J/(mol K)
Temperature T= 385K
volume of gas = V =?
BY GENERAL GAS LAW WE HAVE
PV = nRT
or V = nRT/P
or V = (4.39×8.31×385)/(2.27×10^5)
V = 0.0618728
V = 0.0619 m^3
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
ANSWER: THE ANSWER IS SUMMER
Answer:
Zeros to the left of a decimal can be insignificant place holders, such as in 0.043 (two significant figures).
They can be significant if they are between two digits who themselves are significant, such as in 101.000 (three significant figures).
In the case of a number like 1,000 we can see there is only one significant figure. The zero digits are not between sigfigs.
