The difference in the pressure between the inside and outside will be 369.36 N/m²
<h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
dP is the change in the presure=?
Using Bernoulli's Theorem;
Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²
To learn more about the pressure refer to the link;
brainly.com/question/356585
#SPJ1
-- reduce the length of a wire to 1/2 . . . cut the resistance in half
-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .
-- R2 = (R1) · (1/2) · (16) = 8 · R1
<em>-- R2 / R1 = 8</em>
The answer is D. discrimination.<span />
-- 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.
Here, K.E. = 1/2 * mv²
So, K.E. = 1/2 * (1200) * (24)²
K.E. = 1/2 * 1200 * 576
K.E. = 600 * 576
K.E. = 345,600 J
Hope this helps!
