Explanation:
Formula to represent thrust is as follows.
F = 
= 
or, p = 
F = 
= 
= 201.67 N
Thus, we can conclude that the thrust is 201.67 N.
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Answer:
it will be curved as in deceleration
Explanation:
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
