Answer:
(1) V = 0.2 J (2) 0.05J
Explanation:
Solution
Given that:
K = 160 N/m
x = 0.05 m
Now,
(1) we solve for the initial potential energy stored
Thus,
V = 1/2 kx² = 0.5 * 160 * (0.05)²
Therefore V = 0.2 J
(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.
By using the energy conversion, we have the following
ΔV = mgh
=(0.1/9.8) * 9.8 * 1.5 = 0.15J
The internal energy = 0.2 -0.15
=0.05J
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:
It is very rare to see a solar eclipse from your home, because the Earth, Sun, and the moon need to align just right. Not everyone in the world can view a solar eclipse, only some area can. A solar eclipse is where the moon blocks out the sun. If you think about it: Let's say you live in Florida, U.S.A. You may see the moon coming in front of the sun, but if you lived in California or sumthin', the moon and the sun wouldn't be aligned to form a solar eclipse. It all depends on location... so it is rare to see one.
Answer:
PE=0.29J
Explanation:
According to the description, there is a angle and in point swung upward of 70°
So,
Appling the equation of Potential Energy we have,
<span>(a) -9.97 m/s
(b) x = 2.83
This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.
f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2
So the acceleration of the body is now expressed as
f''(x) = -2.580645161x m/s^2
Let's calculate the anti-derivative from that.
f''(x) = -2.580645161x m/s^2
f'(x) = -1.290322581x^2 + C m/s
Now let's use the known velocity value at x = 2.0 to calculate C
f'(x) = -1.290322581x^2 + C
1
1 = -1.290322581*2^2 + C
11 = -1.290322581*4 + C
11 = -5.161290323 + C
16.161290323 = C
So the velocity function is
f'(x) = -1.290322581x^2 + 16.161290323
(a) The velocity at x = 4.5
f'(x) = -1.290322581x^2 + 16.161290323
f'(4.5) = -1.290322581*4.5^2 + 16.161290323
f'(4.5) = -1.290322581*20.25 + 16.161290323
f'(4.5) = -26.12903227 + 16.161290323
f'(4.5) = -9.967741942
So the velocity is -9.97 m/s
(b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
0 = -1.290322581x^2 + 10.36129032
1.290322581x^2 = 10.36129032
x^2 = 8.029999998
x = 2.833725463</span>