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2 years ago

An object is at rest. There are several forces acting on the object, but the net force is zero. If all the

1 answer:

3 0

It would not. Imagine four forces equal in magnitude but opposite in direction (e.g. north, east, south, and west). If these forces were to double in magnitude they would still have the same magnitude, meaning the net force is still equal to zero.

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

2 years ago

If I connect an inductor (L) to a capacitor (C), I will get an LC oscillator circuit with some natural frequency omega. If I wer

mart [117]

The new natural frequency would be ω/2.

we know that,

$f = \frac{1}{2 pi \sqrt{LC} }$ = ω. -> equation 1

now, when capacitance is quadrupled,

$f' = \frac{1}{2 pi \sqrt{L ( 4C )} }$

$f' = \frac{1}{2 pi (2)\sqrt{LC} }$ . -> equation 2

substituting value of equation 1 in equation 2 , we get,

$f' = \frac{w}{2}$

Hence, the new natural frequency of the circuit is ω/2.

what do you mean by frequency ?

The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.

Learn more about frequency here:-

brainly.com/question/12530980

#SPJ4

8 0

2 years ago

A perpendicular force is applied to a certain area and produces a pressure P. If the

Furkat [3]

Answer:

p/2

Explanation:

4 0

2 years ago