The two forces acting on the object would be 1. Weight because of gravity pulling the object toward the ground and 2. Drag resisting this motion
Given:
w = z(t)= γ − β*t^2
Differentiating both sides with respect to t, we get:
α = z(t) = -2βt
Given: <span> γ = 5.35 rad/s and β = 0.810 rad/s3
</span>
so, For t = 3 sec,
angular acceleration = -2 * 0.810 * 3 = <span>-4.86</span>
Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
Answer: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
