Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Answer:
Bar magnets are permanent magnets. This means that their magnetism is there all the time and cannot be turned on or off as it can with electromagnets .
Explanation:
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer: 6.47m/s
Explanation:
The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.
The time taken for 1 revolution is, 1/3.33 = 0.30s
velocity of the wheel = d/t
Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t
V = 2π * 0.309 / 0.3
V = 1.94/0.3
V = 6.47m/s
The tangential speed of the tack is 6.47m/s