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:
2250N
Explanation:
W= mg,
where W= weight
m= mass
g= acceleration due to gravity
Given that the body is 90kg, m= 90kg.
Acceleration due to gravity of planet
= 2.5(10)
= 25 m/s²
Weight of body on planet
= 90(25)
= 2250N
*Mass is the amount of matter an object has and is constant (same on earth and the planet).
Answer:
The answer would be A. - the temperature remains constant
Explanation:
An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0
Answer:
a = 2d / t²
Explanation:
d = ½ at²
Multiply both sides by 2:
2d = at²
Divide both sides by t²:
a = 2d / t²
