6a) Acceleration of the bullet: [tex]1.44\cdot 10^5 m/s^2[/tex]
6b) Final velocity of the bullet: 360 m/s
7a) Acceleration of the truck: [tex]1.50 m/s^2[/tex]
7b) Net force on the truck: 54,000 N
7c) The acceleration will triple
Explanation:
6a)
The acceleration of the bullet can be found by using Newton's second law, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:
[tex]F=ma[/tex]
where
F is the force
m is the mass
a is the acceleration
For the bullet in this problem, we have
F = 3600 N (force)
m = 0.025 kg (mass of the bullet)
Therefore, the acceleration is
[tex]a=\frac{F}{m}=\frac{3600}{0.025}=1.44\cdot 10^5 m/s^2[/tex]
6b)
The motion of the bullet is a uniformly accelerated motion, so we can use the following suvat equation:
[tex]v=u+at[/tex]
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time elapsed
In this problem, we have:
u = 0 (the bullet starts from rest)
t = 0.0025 s (time)
[tex]a=1.44\cdot 10^5 m/s^2[/tex] (acceleration)
So we can solve to find v, the final velocity of the bullet as it leaves the rifle:
[tex]v=0+(1.44\cdot 10^5)(0.0025)=360 m/s[/tex]
7a)
The acceleration of a body is defined as the rate of change of its velocity:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
a is the acceleration
For the truck in this problem, we have
u = 0 (it starts from rest)
v = 28 m/s (final velocity)
t = 18.7 s (time elapsed)
Substituting, we find the acceleration
[tex]a=\frac{28-0}{18.7}=1.50 m/s^2[/tex]
7b)
The net force on the truck can be calculated using Newton's second law:
[tex]F=ma[/tex]
where
F is the force
m is the mass
a is the acceleration
For the truck in this problem,
m = 36,000 kg is the mass of the truck
[tex]a=1.50 m/s^2[/tex] is the acceleration
So, the net force on the truck is
[tex]F=(36,000)(1.50)=54,000 N[/tex]
7c)
We said that the relationship between the net force on the truck and its acceleration is
[tex]F=ma[/tex]
The equation can be rewritten as
[tex]a=\frac{F}{m}[/tex]
We notice that the acceleration is directly proportional to the acceleration.
Here we are told that the net force on the truck is tripled, therefore
[tex]F'=3F[/tex]
As a consequence, the new acceleration of the truck would be
[tex]a'=\frac{F'}{m}=\frac{3F}{m}=3\frac{F}{m}=3a[/tex]
So, the acceleration also triples.
Learn more about force, acceleration and Newton's second law:
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