Answer:
Option D - 0.2 s
Explanation:
We are given;
Initial velocity; u = 7 m/s
Height of table; h = 1.8m
Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.
Thus;
h = ut + ½gt²
Plugging in the relevant values, we have;
1.8 = 7t + ½(9.8)t²
4.9t² + 7t - 1.8 = 0
Using quadratic formula to find the roots of the equation gives us;
t = -1.65 or 0.22
We can't have negative t value, thus we will pick the positive one.
So, t = 0.22 s
This is approximately 0.2 s
Answer:
Force exerted, F = 2.64 × 10⁷ Newton
Explanation:
It is given that,
Mass of the artillery shell, m = 1100 kg
It is accelerated at, 
We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :
F = ma

F = 26400000 Newton
or
F = 2.64 × 10⁷ Newton
So, the force exerted on the ship by the artillery shell is 2.64 × 10⁷ Newton.
It is the force applied to a machine
in order to accomplish work.
Answer:
2.06 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.
Momentum before collision
Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5
Momentum after collision
The momentum after collision will be given by (9+27)*0.9=32.4
Relating the two then 9v+13.5=32.4
9v=18.5
V=2.055555555555555555555555555555555555555 m/s
Rounded off, v is approximately 2.06 m/s
Answer:
F = 20.4 i ^
Explanation:
This exercise can be solved using the ratio of momentum and amount of movement.
I = F t = Dp
Since force and amount of movement are vector quantities, each axis must be worked separately.
X axis
Let's look for speed
cos 45 = vₓ / v
vₓ = v cos 45
vₓ = 8 cos 45
vₓ = 5,657 m / s
We write the moment
Before the crash p₀ = m vₓ
After the shock
= -m vₓ
The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ
The impulse on the x axis Fₓ t = Δp
Fₓ = 2 m vₓ / t
Fx = 2 0.450 5.657 / 0.250
Fx = 20.4 N
We perform the same calculation on the y axis
sin 45 = vy / v
vy = v sin 45
vy = 8 sin 45
vy = 5,657 m / s
We calculate the initial momentum po = m 
Final moment
= m
Variations moment Δp = m
- m
= 0
Force in the Y-axis
= 0
Therefore the total force is
F = fx i ^ + Fyj ^
F = Fx i ^
F = 20.4 i ^