Answer:
Explanation:
Since the block is at rest in an elevated position, we can assume that it only has potential energy.
U=mgh is the formula for potential energy where U=potential energy, m= mass, g=acceleration due to gravity, and h=height.
Plug in known variables....
U=4kg*9.8m/s^2*20m
U=784 joules of potential energy or letter A.
Answer:
Points downward, and its magnitude is 9.8 m/s^2
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.
- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.
The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.
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:
257.32
Explanation:
I jus worked it out on paper. Brainliest please?