Answer:
potential energy PE = M g h
KE at bottom = 1/2 M V^2
Regardless of the slope of the slide the change in energy is the same
1/2 V^2 = g h
V = (2 g h)^1/2 = (2 * 9.8 m/s^2 * 10 m)^1/2 = 14 m / s
Perhaps the question says that h = 55 * .1 = 5.5 m
Then V = (2 * 9.8 * 5.5) = 10.4 m/s
Answer:
53.64 m/s
Explanation:
Applying,
a = (v-u)/t............. Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
make u the subject of the equation
u = v-at............. Equation 2
From the question,
Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)
Substitute these values into equation 2
u = 0-(-5.364×10)
u = 0+53.64
u = 53.64 m/s
Answer:
The independent variable: is the amount of compost applied to each plant
The dependent variable: the height of the plants
Hypothesis: Older grass compost will produce taller plants
Three groups were created:
the control group is group C to which no compost was applied
investigated group A and B with different amounts of compost applied
Explanation:
The students are considering using the scientific method to analyze the problem.
The independent variable is the one controlled by the researcher and the dependent variable is the response of the system
The independent variable: is the amount of compost applied to each plant
The dependent variable: the height of the plants
Hypothesis: Older grass compost will produce taller plants
Three groups were created:
the control group is group C to which no compost was applied
investigated group A and B with different amounts of compost applied
The centripetal acceleration of the hummingbird is ![\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2](https://tex.z-dn.net/?f=%5Cfrac%7B%283.65%29%5E2%7D%7B0.25%7D%20%3D%2053.29%20%5C%20m%2Fs%5E2)
The given parameters;
- <em>time of motion of the hummingbird, t = 0.43 s</em>
- <em>radius of the circle, r = 0.25 m</em>
- <em>number of revolution of the hummingbird = 1 rev per 0.43 s</em>
The angular speed of the hummingbird is calculated as follows;
![\omega = \frac{1 \ rev}{0.43 \ s} \times \frac{2\pi \ rad}{1 \ rev} =14.61 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B1%20%5C%20rev%7D%7B0.43%20%5C%20s%7D%20%5Ctimes%20%5Cfrac%7B2%5Cpi%20%5C%20rad%7D%7B1%20%5C%20rev%7D%20%3D14.61%20%5C%20rad%2Fs)
The linear speed of the hummingbird is calculated as follows;
v = ωr
v = 14.61 x 0.25
v = 3.65 m/s
The centripetal acceleration of the hummingbird is calculated as follows;
![a_c = \frac{v^2}{r} \\\\a_c = \frac{(3.65)^2}{0.25} \\\\a_c = 53.29\ m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20%5C%5C%5C%5Ca_c%20%3D%20%5Cfrac%7B%283.65%29%5E2%7D%7B0.25%7D%20%5C%5C%5C%5Ca_c%20%3D%2053.29%5C%20m%2Fs%5E2)
Thus, the centripetal acceleration of the hummingbird is ![\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2](https://tex.z-dn.net/?f=%5Cfrac%7B%283.65%29%5E2%7D%7B0.25%7D%20%3D%2053.29%20%5C%20m%2Fs%5E2)
Learn more here:brainly.com/question/11700262