Question 2 (10 points)
1 answer:
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by
where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is
where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:
where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,
So the closest answer among the option is -5020 N.
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Answer:
The appropriate solution is "61.37 s".
Explanation:
The given values are:
Boat moves,
= 10 m/s
Water flowing,
= 1.50 m/s
Displacement,
d = 300 m
Now,
The boat is travelling,
=
=
Travelling such distance for 300 m will be:
⇒
On putting the values, we get
Throughout the opposite direction, when the boat seems to be travelling then,
=
=
Travelling such distance for 300 m will be:
⇒
On putting the values, we get
hence,
The time taken by the boat will be:
=
=
Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is,
Angle of projection is,
Time taken to land on the hill is,
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:
Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.