Answer:
a.) 10Hz
b.) 0.1 s
c.) 187.4 m/s
d.) -412.6 m/s^2
Explanation:
Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.
(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?
from the equation given, the angular speed w = 20π
but w = 2πf
where f = frequency.
substitute w for 20π
20π = 2πf
f = 20π/2π
f = 10 Hz
(b) What is t the first time that the object is at its farthest right?
since F = 1/T
T = 1 / f
T = 1/10
T = 0.1 s
Therefore, the t of first time that the object is at its farthest right is 0.1 s
(c) At the time found in part (b), what is the object's velocity?
The velocity can be found by differentiating the equation;
x(t) = 3sin(20πt)
dx/dt = 60πcos(20πt)
where dx/dt = velocity V
V = 60πcos(20π * 0.1)
V = 187.4 m/s
(d) At the time found in part (b), what is the object's acceleration?
to get the acceleration, differentiate equation V = 60πcos(20πt)
dv/dt = -1200πSin(20πt)
dv/dt = acceleration a
a = -1200πSin(20πt)
substitute t into the equation
a = -1200πSin(20π * 0.1)
a = - 412.6 m/s^2
X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared
You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.
Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = 
Equating both, we have
So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
The labeled points which is Letter B in the given Image is the point that the axis of rotation passes through. This problem is an example of rotational dynamics, formerly an object moves in a straight line then the motion is translational but when an object at inactivity lean towards to continue at inactivity and an object in rotation be possible to continue rotating with continuous angular velocity unless bound by a net external torque to act then is rotational. In a rotational motion, the entity is not treated as a constituent part but is treated in translational motion. It points out with the study of torque that outcomes angular accelerations of the object.
