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.
21m per second. Take 210 divided by 10. Hope this helps!
Increasing the water temperature by adding hot water is called thermal heating, so the correct answer is d)
Answer:
<em>a) 37.5N</em>
<em>b) 9.375Joules</em>
Explanation:
a) According to Hooke's law
F = ke
k is the spring constant
e is the extension;
F = 150 * 0.25
F = 37.5N
b) Work done on the spring = 1/2ke^2
Work done on the spring = 1/2 * 150 * 0.25^2
Work done on the spring = 75 * 0.0625
Work done on the spring = 9.375Joules
It doesn’t because naimba barely have life