A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
Answer:
the answer is 104568948 degress
Explanation:
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The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:
- 5.2 meters.
- 2.6 meters.
- 1.7meters.
Given data:
Length of the fixed string = 264cms = 2.64 meters
The wavelength for standing waves is given by:
λ = 2L/n
where,
- λ is the wavelength
- L is the length of the string
For n = 1,
= 5.2 meters
For n = 2,
= 2.6 meters
For n = 3,
= 1.7 meters
To learn more about standing waves: brainly.com/question/14151246
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