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:As a noun relative means a person who is connected with others by blood.
Answer:
Explanation:
As the source is situated on x - axis , it must be situated in between the two listeners .
So the x coordinate of source is
(-7 + 3 )/2
= - 2 m
The equation of the wave- front will be that o a circle having centre at (-2,0)
and radius = distance between -2 and 3 , that is 5 m
equation of circle
=( x+2 )² + y² = 25
It cuts y axis when x = 0
Putting x = 0
4 + y² = 25
y² = 21
y = + √21 , or - √21
Answer:
a = √ (a_t² + a_c²)
a_t = dv / dt
, a_c = v² / r
Explanation:
In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.
a = aₓ i ^ + a_y j ^
Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains
a = √ (a_t² + a_c²)
where the tangential acceleration is
a_t = dv / dt
the centripetal acceleration is
a_c = v² / r
Hi there!
We can use the following kinematic equation:
The initial velocity is 0 m/s, so:
vf = final velocity (? m/s)
a = acceleration due to gravity (g)
d = vertical height (m)
Plug in the givens and solve:
