Answer:
Force, |F| = 2100 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, 
Initial speed, v = 42 m/s
The momentum is reduced to zero, final speed, v = 0
The relation between the force and the momentum is given by :
|F| = 2100 N
So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.
The answer for the following problem is mentioned below.
The option for the question is "A" approximately.
- <u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
Explanation:
Given:
Spring constant (k) = 240 N/m
amount of the compression (x) = 0.40 m
To calculate:
Elastic potential energy (E)
We know;
<em>According to the formula;</em>
E = 
× k × x × x
<u>E = </u><u> × k ×(x)²</u>
where;
E represents the elastic potential energy
K represents the spring constant
x represents amount of the compression in the string
So therefore,
Substituting the values in the above formula;
E = × 240 × (0.40)²
E = × 240 × 0.16
E = × 38.4
E = 19.2 J or approximately 20 J
<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
A. The amount of water. 50/50
