Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. place the lamp so that the source of light is
1 answer:
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Refer to the diagram shown below.
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m
Answer: The maximum height attained is 24.7 m (nearest tenth)
Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s
To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.
Answer: The ball stays in the air for 4.5 s (nearest tenth)
Complete Question
A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton
Answer:
The vector position is
Explanation:
From the question we are told that
The position of the proton is
Generally the vector location of the proton is mathematically represented as
So substituting values
Answer:
Explanation:
Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.
.
Now, let's use the ideal gas equation to the initial and the final state:
Let's recall that the term nR is a constant. That is why we can match these equations.
We can find a relation between the volumes of the initial and the final state.
Combining this equation with the first equation we have:
Now, we just need to solve this equation for T₂.
Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.
Here,
Finally, T2 will be: