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.
A machine can never be 100% efficient because some work is always lost
due to the lack of materials or equipment that would convert work by 100%. It follows
the second law of entropy. The ideal engine is known as Carnot’s engine having
a 100% efficiency. So far, no engine has ever gotten to 100%.
Answer:
Explanation:
<u>Given Data:</u>
Weight = W = 65 N
Height = h = 2 m
Time = t = 4 secs
<u>Required:</u>
Power = P = ?
Work Done in the form of Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = Wh
P = P.E. / t
<u>Solution:</u>
P.E. = (65)(2)
P.E = 130 Joules
P = P.E. / t
P = 130 / 4
P = 32.5 Watts
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Hope this helped!
<h3>~AH1807 </h3>
Resistance = (voltage) / (current)
Resistance = (6.0 v) / (2.0 A)
Resistance = 3.0 ohms
<span>Energy can be transformed from one type to another in any convection. Some of the energy is lost to the environment as
HEAT.</span>
