Response:
The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is 
ft/ sec
Clarification:
Provided:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec. 5 = 125 ft
x(5) = 10 ft/sec. 5 = 50 ft
At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem
Now, let's calculate the derivative of distance in relation to time
By plugging in the values for h(t) and x(t) and simplifying, we arrive at,
= 
= ft / sec
Answer:
At this position, the magnetic field equals ZERO
Explanation:
The magnetic field produced by a moving charge is described as
Here, we determine the direction of the magnetic field using
Thus, we find
Leading to a magnetic field of ZERO
Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent
The question lacks clarity. The advantage of lifting with a lever is that it allows you to apply force in a more manageable direction and necessitates far less force to lift an object by balancing the torque exerted by it. For instance, if you aim to lift an object weighing 4N with a force of 2N, utilizing a class 2 lever while maintaining a distance ratio between the body and the force application point from the fulcrum of 1:2 is adequate. In any scenario, one should balance the torque to achieve the desired force.
I will assume the girl is on the right while the boy is on the left.
The net force represents the total of all forces acting on an object, factoring in negatives.
Let the force from the boy be denoted as b. We’ll apply the formula F = ma.
b + 3.5 = 0.2(2.5)
This reduces to a straightforward algebraic problem. By solving, we find that the boy is applying a force of -3N to the left.
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>
