Respuesta :
Answer:
V
I and II
III and IV
Explanation:
The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.
Taking always east as positive direction, and labelling
u the initial velocity
v the final velocity
m = 1000 kg the mass (which is always equal)
We find:
(i)
u = 25 m/s
v = 0
[tex]|I|=m(v-u)=(1000)(0-25)=25,000 Ns[/tex]
(II)
u = 25 m/s
v = 0
[tex]|I|=m(v-u)=(1000)(0-25)=25,000 Ns[/tex]
(III)
In this case,
F = 2000 N is the force
[tex]\Delta t = 10 s[/tex] is the time
So the magnitude of the impulse is
[tex]|I| =F\Delta t = (2000N)(10)=20,000 Ns[/tex]
(IV)
F = 2000 N is the force
[tex]\Delta t = 10 s[/tex] is the time
So the magnitude of the impulse is
[tex]|I| =F\Delta t = (2000N)(10)=20,000 Ns[/tex]
(V)
u = 25 m/s
v = -25 m/s
[tex]|I|=m(v-u)=(1000)(-25-25)=50,000 Ns[/tex]
So the ranking from largest to smallest is:
V
I and II
III and IV
The rank of all situations according to the magnitude of the impulse of the net force, from the largest value to the smallest value is,
- (v) Moving east at 25 m>s; over a 30-s period, the automobile reverses direction and ends up moving west at 25 m>s.
- (i) Moving east at 25 m>s and comes to a stop in 10 s; and (ii) Moving east at 25 m>s and comes to a stop in 5 s.
- (iii) At rest, and a 2000-N net force toward the east is applied to it for 10 s and (iv) Moving east at 25 m>s, and a 2000-N net force toward the west is applied to it for 10 s.
What is impulse force?
Impulse force is the product of average force and the total time for which the force maintained is called the impulse force.
The impulse of net force can be given as,
[tex]|I|=m(v-u)[/tex]
Where, (m) is the mass, (v) is the initial velocity and (u) is the final velocity.
The impulse of net force in terms of net force can be given as,
[tex]|I|=F\Delta t[/tex]
Where, (F) is the force, and (t) is the time.
In each situation, a 1000-kg automobile is moving along a straight east–west road.
- (i) Moving east at 25 m>s and comes to a stop in 10 s;
Here, the initial velocity is 25 m/s and final velocity is zero. Thus, the impulse of net force is,
[tex]|I|=1000(0-25)\\|I|=25000\rm \; Ns[/tex]
- (ii) Moving east at 25 m>s and comes to a stop in 5 s;
Here, the initial velocity is 25 m/s and final velocity is zero. Thus, the impulse of net force is,
[tex]|I|=1000(0-25)\\|I|=25000\rm \; Ns[/tex]
- (iii) At rest, and a 2000-N net force toward the east is applied to it for 10 s;
Here, the net force is 2000 N and time taken is 10s. Thus, the impulse of net force is,
[tex]|I|=2000\times10\\|I|=20000\rm \; Ns[/tex]
- (iv) Moving east at 25 m>s, and a 2000-N net force toward the west is applied to it for 10 s;
Here, the net force is 2000 N and time taken is 10s. Thus, the impulse of net force is,
[tex]|I|=2000\times10\\|I|=20000\rm \; Ns[/tex]
- (v) Moving east at 25 m>s; over a 30-s period, the automobile reverses direction and ends up moving west at 25 m>s.
Here, the initial velocity is 25 m/s and final velocity is 25 m/s but in opposite direction. Thus, the impulse of net force is,
[tex]|I|=1000(-25-25)\\|I|=50000\rm \; Ns[/tex]
The rank of all situations according to the magnitude of the impulse of the net force, from the largest value to the smallest value is,
- (v) Moving east at 25 m>s; over a 30-s period, the automobile reverses direction and ends up moving west at 25 m>s.
- (i) Moving east at 25 m>s and comes to a stop in 10 s; and (ii) Moving east at 25 m>s and comes to a stop in 5 s.
- (iii) At rest, and a 2000-N net force toward the east is applied to it for 10 s and (iv) Moving east at 25 m>s, and a 2000-N net force toward the west is applied to it for 10 s.
Learn more about the impulse force here;
https://brainly.com/question/20586658
