I created the illustration found in the accompanying file.
There are two images included.
The upper one illustrates the impacts of:
- scaling vector A by a factor of 1.5, depicted in red with a dashed line.
- scaling vector B by -3, shown in purple with a dashed line.
The lower image displays the resultant vector: C = 1.5A - 3B.
The approach involves relocating the tail of vector -3B to the tip of vector 1.5A while maintaining the angles.
Next, an arrow is drawn from the tail of 1.5A to the position of -3B after this shift.
The arrow representing the result is vector C, marked with a black dashed line.
Respuesta:
P_(bomba) = 98,000 Pa
Explicación:
Se nos proporciona;
h2 = 30m
h1 = 20m
Densidad; ρ = 1000 kg/m³
Primero, entendemos que la suma de las presiones en el tanque y la bomba es igual a la del boquilla,
Así, se puede expresar como;
P_(tanque)+ P_(bomba) = P_(boquilla)
Ahora, la presión se daría como;
P = ρgh
Y así,
ρgh_1 + P_(bomba) = ρgh_2
<ppor lo="" tanto="">
P_(bomba) = ρg(h_2 - h_1)
<pal sustituir="" los="" valores="" pertinentes="" obtenemos="">
P_(bomba) = 1000•9.8(30 - 20)
P_(bomba) = 98,000 Pa
</pal></ppor>
This question is excellent for clearing up any confusion about the work-energy theorem.
When a force acts perpendicular to an object's motion, the work done by that force equals zero.
According to the work-energy theorem,
the net work performed by all forces equals the change in kinetic energy.
In this instance, since work done = 0,
therefore,
0 = change in kinetic energy.
This indicates that the kinetic energy remains unchanged.
I hope this clarifies things!
