There is a displacement. Just because the ball is thrown up,
and not crossways, doesn't mean its location is not moving. Remember, positive
displacement is together a displacement in the direction east, right, and up.
The velocity is the distance over time. To compute that, you must look how high
the ball moved before falling back down. Acceleration is expected to be
constant at 9.80m/s^2. That is the force of gravity. But remember that you are disregarding
air friction when you are computing the acceleration.
200N
Explanation:
600N-400N = 200N
Answer:
λ = 5.65m
Explanation:
The Path Difference Condition is given as:
δ=
;
where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.
m = no of openings which is 2
∴δ= ![\frac{3*lamda}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Alamda%7D%7B2%7D)
n is the index of refraction of the medium in which the wave is traveling
To find δ we have;
δ= ![\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }](https://tex.z-dn.net/?f=%5Csqrt%7B70%5E2%2B%2833%2B%5Cfrac%7B20%7D%7B2%7D%29%5E2%20%7D-%5Csqrt%7B70%5E2%2B%2833-%5Cfrac%7B20%7D%7B2%7D%29%5E2%20%7D)
δ= ![\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}](https://tex.z-dn.net/?f=%5Csqrt%7B4900%2B%28%5Cfrac%7B66%2B20%7D%7B2%7D%29%5E2%7D-%5Csqrt%7B4900%2B%28%5Cfrac%7B66-20%7D%7B2%7D%29%5E2%7D)
δ= ![\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }](https://tex.z-dn.net/?f=%5Csqrt%7B4900%2B%28%5Cfrac%7B86%7D%7B2%7D%29%5E2%20%7D-%5Csqrt%7B4900%2B%28%5Cfrac%7B46%7D%7B2%7D%29%5E2%20%7D)
δ= ![\sqrt{4900+43^2}-\sqrt{4900+23^2}](https://tex.z-dn.net/?f=%5Csqrt%7B4900%2B43%5E2%7D-%5Csqrt%7B4900%2B23%5E2%7D)
δ= ![\sqrt{4900+1849}-\sqrt{4900+529}](https://tex.z-dn.net/?f=%5Csqrt%7B4900%2B1849%7D-%5Csqrt%7B4900%2B529%7D)
δ= ![\sqrt{6749}-\sqrt{5429}](https://tex.z-dn.net/?f=%5Csqrt%7B6749%7D-%5Csqrt%7B5429%7D)
δ= 82.15 -73.68
δ= 8.47
Again remember; to calculate the wavelength of the ocean waves; we have:
δ= ![\frac{3*lamda}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Alamda%7D%7B2%7D)
δ= 8.47
8.47 = ![\frac{3*lamda}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Alamda%7D%7B2%7D)
λ = ![\frac{8.47*2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B8.47%2A2%7D%7B3%7D)
λ = 5.65m
Answer:
41°
Explanation:
Kinetic energy at bottom = potential energy at top
½ mv² = mgh
½ v² = gh
h = v²/(2g)
h = (2.4 m/s)² / (2 × 9.8 m/s²)
h = 0.294 m
The pendulum rises to a height of above the bottom. To determine the angle, we need to use trigonometry (see attached diagram).
L − h = L cos θ
cos θ = (L − h) / L
cos θ = (1.2 − 0.294) / 1.2
θ = 41.0°
Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.