Answer:
-8 Kgm/s
Explanation:
From conservation of linear momentum, the sum of momentum before collision is equal to the sum of linear momentum after collision
The sum of momentum before collision is 15+-35=-20 Kgm/s
Momentum after collision is
x+-12=-20
x=-20+12=-8 Kgm/s
Answer:
0N
Explanation:
Is the car is moving in a constant direction with constant speed, the composite force is null. We know what because any object that nonzero composite force is applied to will accelerate in the direction of the force, and this one doesn't.
Answer:
ya it is , affected!okay I think this
Answer:
a)1.51*10^-22joules b) 1.89*10^-7m
Explanation:
Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules
b) net force acting to stop the proton = 8.01*10^-16
Work done needed to stop the proton = net force acting opposite the motion * distance
Distance covered = need work done/ net force
Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m
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.