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:
The pitch will progressively lower
Explanation:
If i were bungee jumping from a bridge while blowing a hand-held air horn and someone who remains on the bridge will hear a decreased pitch or frequency as the source is moving away from the stationary listener as per the Dooplers effect. Hence, the pitch will progressively lower as the source is moving away from the observer.
<span>A. frustration-aggression theory.
Good Luck!!</span>
<span> Using conservation of energy
Potential Energy (Before) = Kinetic Energy (After)
mgh = 0.5mv^2
divide both sides by m
gh = 0.5v^2
h = (0.5V^2)/g
h = (0.5*2.2^2)/9.81
h = 0.25m
</span>
Answer:
Moment of inertia of the solid sphere:
I
s
=
2
5
M
R
2
.
.
.
.
.
.
.
.
.
.
.
(
1
)
Is=25MR2...........(1)
Here, the mass of the sphere is
M
M
