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Tresset [83]
3 years ago
10

How much work is needed to lift a 3kg create a vertical displacement of 22m

Physics
1 answer:
Ulleksa [173]3 years ago
6 0
I think its W= 647.239J
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What is the mechanical advantage of a lever that has an input arm of 6 meters and an output arm of 2 meters
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Answer:A

Explanation:I TOOK THE Test

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3 years ago
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
3 years ago
A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of
GREYUIT [131]

Answer:

a=190\ m/s^2

Explanation:

Mass of a hockey puck, m = 0.17 kg

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The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2

So, the acceleration of the hockey puck is 190\ m/s^2.

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Cuales son los factores para que un sobre conductor se genere y aumente la energía termíca
riadik2000 [5.3K]

daisuki

anneyong

mahal kita

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You are currently getting 26 sales opportunities per day and closing 64% of them. How many sale opportunists per day?
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Answer: 16

Explanation:

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