Answer:
The distance traveled by the woman is 34.1m
Explanation:
Given
The initial height of the cliff
yo = 45m final, positition y = 0m bottom of the cliff
y = yo + ut -1/2gt²
u = 20.0m/s initial speed
g = 9.80m/s²
0 = 45.0 + 20×t –1/2×9.8×t²
0 = 45 +20t –4.9t²
Solving quadratically or by using a calculator,
t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s
So this is the total time it takes for the ball to reach the ground from the height it was thrown.
The distance traveled by the woman is
s = vt
Given the speed of the woman v = 6.00m/s
Therefore
s = 6.00×5.69 = 34.14m
Approximately 34.1m to 3 significant figures.
The answer is ultraviolet rays
Answer:
Leak 1 = 3.43 m/s
Leak 2 = 2.42 m/s
Explanation:
Given that the top of the boot is 0.3 m higher than the leaks.
Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2
From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.
Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2
Find the attached files for the solution and figure
Answer:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
- PE = (7.00)(9.8)(0.40)
- PE = 27.44
Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.