What is the general term for a rock fragment present in a sedimentary rock?
1 answer:
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Explanation:
Since, it is given that one hand completes 3.19 vibrations in 8.46 sec. Therefore, in one second the number of vibrations will be as follows.
= 0.377 vibrations
Hence, frequency (f) = 0.38 Hz
Now, formula to calculate the speed is as follows.
v =
or,
=
= 1.57 cm
Thus, we can conclude that the wavelength is 1.57 cm.
Answer:
a) The takeoff speed is 10 m/s.
b) The maximum height above the ground is 1.2 m.
Explanation:
The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v =(v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial velocity.
α = jumping angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity vector at time "t"
a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:
r final = (8.7 m, 0 m)
Then at final time:
8.7 m = x0 + v0 · t · cos α
0 m = y0 + v0 · t · sin α + 1/2 · g · t²
(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:
8.7 m = v0 · t · cos α
Solving for "v0":
8.7 m /(t · cos α) = v0
Replacing v0 in the equation of the y-component, we can obtain the final time:
0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)
- 8.7 m · tan 29° / -4.9 m/s² = t²
t = 0.99 s
Now, we can calculate the initial speed:
8.7 m /t · cos α = v0
v0 = 8.7 m / (0.99 s · cos 29°)
<u>v0 = 10 m/s</u>
The takeoff speed is 10 m/s
b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:
0 = v0 · sin α + g · t
Solving for "t":
-v0 · sin α / g = t
t = - 10 m/s · sin 29° / 9.8 m/s²
t = 0.49 s
Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.
Now, let´s find the height of the kangaroo at that time:
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²
<u>y = 1.2 m</u>
The maximum height above the ground is 1.2 m.
